Multiple Angles




\(1.\  \displaystyle \sin 2A=2\sin A\cos A \)
\( 2.\ \displaystyle \sin2A=\frac{2\tan A}{1+\tan^2A}\)
\(3.\  \displaystyle\sin2A=\frac{2\cot A}{1+\cot^2A}\)
\(4.\  \displaystyle \cos2A=\cos^2A-\sin^2A\)
\( 5.\ \displaystyle\cos2A=2\cos^2A-1\)
\(6.\ \displaystyle\cos2A=1-2\sin^2A\)
\(7.\  \displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}\)
\(8.\ \displaystyle\cos2A=\frac{\cot^2A-1}{\cot^2A+1}\)
\(9.\ \displaystyle \tan2A=\frac{2\tan A}{1-\tan^2A}\)
\(10.\ \displaystyle \cot2A=\frac{\cot^2A-1}{2\cot A}\)
\(11.\ \displaystyle \sin3A=3\sin A-4\sin^3A\)
\(12.\ \displaystyle \cos3A=4\cos^3A-3\cos A\)
\(13.\ \displaystyle \tan3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A}\)
\(14.\ \displaystyle \cot3A=\frac{3\cot A-\cot^3A}{1-3\cot^2A} \)
Question No. 1
If \(\displaystyle\sin\theta=\frac{1}{2}\left(a+\frac{1}{a}\right)\) Prove that;
\(\displaystyle \cos2\theta=-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)\)
LHS 
\(\displaystyle =\cos2\theta\)
\( \displaystyle=1-2\sin^2\theta\)
\(\displaystyle =1-2\left(\sin\theta\right)^{^2}\)
\(\displaystyle =1-2\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^2\)
\(\displaystyle =1-2\times\frac{1}{4}\left(a+\frac{1}{a}\right)^2\) 
\(\displaystyle =1-\frac{1}{2}\left(a^2+2\times a\times\frac{1}{a}+\frac{1}{a^2}\right)\)
\(\displaystyle =1-\frac{1}{2}\left(a^2+2+\frac{1}{a^2}\right)\)
\(\displaystyle =1-\frac{1}{2}\left(a^2+\frac{1}{a^2}+2\right)\)
\(\displaystyle =1-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)-\frac{1}{2}\times2\)
\(\displaystyle =1-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)-1\)
\(\displaystyle =-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)\)
\(\displaystyle =\) RHS

Question No. 2
If \(\sin\theta=\frac{1}{2}\left(a+\frac{1}{a}\right)\), show that \( \sin3\theta=-\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)\)
Solution:
LHS
\(=\sin3\theta\)
\(=3\sin\theta-4\sin^3\theta\) 
\(=3\times\frac{1}{2}\left(a+\frac{1}{a}\right)-4\times\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^3\)  
\(=\frac{3}{2}\left(a+\frac{1}{a}\right)-4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3\)  
\(=\frac{1}{2}\left\{3\left(a+\frac{1}{a}\right)-\left(a+\frac{1}{a}\right)^3\right\}\)    
\(=-\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)\right\}\)  
\(=-\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\times a\times\frac{1}{a}\left(a+\frac{1}{a}\right)\right\}\)
\(=-\frac{1}{2}\left(a^3+\frac{1}{a^3}\right) \left[\because a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\right]\)
\(=\)RHS

Question No. 3
If \( \cos\theta=\frac{1}{2}\left(a+\frac{1}{a}\right)\) prove that \(\cos2\theta=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right).\)
Solution:
LHS
\(=\cos2\theta\)
\(=2\cos^2\theta-1\)
\(=2\left(\cos\theta\right)^2-1\)
\(=2\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^2-1\)
\(=2\times\frac{1}{4}\left(a+\frac{1}{a}\right)^2-1\)
\(=\frac{1}{2}\left(a+\frac{1}{a}\right)^2-1\)
\(=\frac{1}{2}\left(a^2+2\times a\times\frac{1}{a}+\frac{1}{a^2}\right)-1\)
\(=\frac{1}{2}\left(a^2+\frac{1}{a^2}+2\right)-1\)
\(=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+\frac{1}{2}\times2-1\)
\(=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+1-1\)
\(=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)\)
\(= \) RHS

Question No. 4 
If \( \cos\theta=\frac{1}{2}\left(a+\frac{1}{a}\right)\), prove that \(\cos3\theta=\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)\)
Solution:
LHS 
\(=\cos3\theta\)
\(=4\cos^3\theta-3\cos\theta\)
\(=4\left(\cos\theta\right)^3-3\cos\theta\)
\(=4\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)^3\right\}-3\times\frac{1}{2}\left(a+\frac{1}{a}\right)^2\)
\(=4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right)\)
\(=\frac{1}{2}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right)\)
\(=\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)\right\}\)
\(=\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\times a\times\frac{1}{a}\left(a+\frac{1}{a}\right)\right\}\)
\(=\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)\)\(\left[\because x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\right]\)
\(=\)RHS

Question Number 5                                                                            {code53}
(a) Prove that:\(\cos5A=16\cos^5A-20\cos^3A+5\cos A\)
Solution:
LHS
\(=\cos5A\)
\(=\cos\left(3A+2A\right)\) 
\(=\cos3A\cos2A-\sin3A\sin2A\) 
\(=\left(4\cos^3A-3\cos A\right)\cos2A\)\(-\left(3\sin A-4\sin^3A\right)\sin2A\) 
\(=4\cos^3A\cos2A-3\cos A\cos2A\)\(-3\text{sinA}\sin2A+4\sin^3A\sin2A\) 
\(=4\cos^3A\left(2\cos^2A-1\right)-3\cos A\left(2\cos^2A-1\right)\)\(-3\sin A\left(2\sin A\cos A\right)+4\sin^3A\left(2\sin A\cos A\right)\) 
\(=8\cos^5A-4\cos^3A-6\cos^3A+3\cos A\) \(-6\sin^2A\cos A+8\sin^4A\cos A\) 
\(=8\cos^5A-10\cos^3A+3\cos A\) \(-6\left(1-\cos^2A\right)\cos A+8\left(\sin^2A\right)^2\cos A\) 
\(=8\cos^5A-10\cos^3A+3\cos A\) \(-6\cos A+6\cos^3A+\cdot8\left(1-\cos^2A\right)^2\cos A\) 
\(=8\cos^5A-4\cos^3A-3\cos A\) \(+8\left(1-2\cos^2A+\cos^4A\right)\cos A\) 
\(=8\cos^5A-4\cos^3A-3\cos A\) \(+8\cos A-16\cos^3A+8\cos^5A\) 
\(=16\cos^5A-20\cos^3A+5\cos A\) 
\(= \) RHS

Question Number 6
 Prove that: \(\sin5A=16\sin^5A-20\sin^3A+5\sin A\)
Solution:
LHS
\(=\sin5A\)
\(=\sin\left(3A+2A\right)\)
\(=\sin3A\cos2A+\cos3A\sin2A\)
\(=\left(3\sin A-4\sin^3A\right)\cos2A+\left(4\cos^3A-3\cos A\right)\sin2A\)
\(=3\sin A\cos2A-4\sin^3A\cos2A+4\cos^3A\sin2A-3\cos A\sin2A\)
\(=3\sin A\left(1-2\sin^2A\right)-4\sin^3A\left(1-2\sin^2A\right)\)\(+4\cos^3A\left(2\sin A\cos A\right)-3\cos A\left(2\sin A\cos A\right)\)
\(=3\sin A-6\sin^3A-4\sin^3A+8\sin^5A\)\(+8\sin A\cos^4A-6\sin A\cos^2A\)
\(=3\sin A-10\sin^3A+8\sin^5A\)\(+8\sin A\left(\cos^2A\right)^2-6\sin A\left(1-\sin^2A\right)\)
\(=3\sin A-10\sin^3A+8\sin^5A+8\sin A\left(1-\sin^2A\right)^2-6\sin A+6\sin^3A\)
\(=8\sin^5A-4\sin^3A-3\sin A+8\sin A\left(1-2\sin^2A+\sin^4A\right)\)
\(=8\sin^5A-4\sin^3A-3\sin A+8\sin A-16\sin^3A+8\sin^5A\)
\(=16\sin^5A-20\sin^3A+5\sin A\)
\(=\) RHS
Question Number 7 
Prove that: \( \sin18^{\circ}=\frac{\sqrt{5}-1}{4}\)
Solution:
LHS
   Let, \( \theta = 18^{\circ} \)
or, \( 5\theta = 5\times 18^{\circ} \)
or, \(  2\theta + 3\theta = 90^{\circ} \)
or, \(  3\theta = 90^{\circ} - 2\theta \)
or, \(  \cos 3\theta = \cos (90^{\circ} - 2\theta )\)
or, \(  4\cos^3\theta -3\cos \theta = \sin 2\theta \)
or, \(  \cos \theta [4\cos^2 \theta - 3 ] = 2\sin \theta \cos \theta \)
or, \(  4(1-\sin^2 \theta ) -3 = 2\sin \theta \)
or, \(  4 -4\sin^2 \theta -3 -2\sin \theta =0 \)
or, \(  -4\sin^2 \theta -2\sin \theta +1 = 0 \)
or, \(  -(4\sin^2 \theta +2\sin \theta -1) =0 \)
or, \(  4\sin^2 \theta +2\sin \theta -1=0 \)
Now, comparing with  \(  ax^2+bx+c=0 \), we get
 \(  a=4, b= 2, c= -1, x = \sin \theta \)
Now,
\(  \sin \theta = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} \)
       \(  =\frac{-2\pm \sqrt{(2)^2 -4 \times 4 \times (-1)}}{2\times 4 } \)
       \(  = \frac{-2\pm \sqrt{4+16}}{8} \)
       \(  = \frac{-2\pm\sqrt{20}}{8} \)
       \(  = \frac{-2\pm 2\sqrt{5}}{8} \)
       \(  = \frac{2(-1\pm \sqrt{5})}{8}\)
       \( =\frac{-1\pm\sqrt{5}}{4}\)
Since \(  \sin 18^{\circ} \) is positive,
\(  \therefore \sin 18^{\circ} = \frac{-1+\sqrt{5}}{4}=\frac{\sqrt{5}-1}{4}\)Proved.

Question Number 8
If \( \tan \theta = \frac{b}{a} \) prove that \(a\cos 2\theta + b\sin 2\theta = a \)
Solution:
Given,
\(\tan\theta=\frac{b}{a}\)
Now,
LHS
\(=a\cos2\theta+b\sin2\theta\)
\(=a\times\frac{1-\tan^2\theta}{1+\tan^2\theta}+b\times\frac{2\tan\theta}{1+\tan^2\theta}\)
\(=a\times\frac{1-\left(\frac{b}{a}\right)^2}{1+\left(\frac{b}{a}\right)^2}+b\times\frac{2\left(\frac{b}{a}\right)}{1+\left(\frac{b}{a}\right)^2}\)
\(=a\times\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}+b\times\frac{\frac{2b}{a}}{1+\frac{b^2}{a^2}}\)
\(=a\times\frac{\frac{a^2-b^2}{a^2}}{\frac{a^2+b^2}{a^2}}+b\times\frac{\frac{2b}{a}}{\frac{a^2+b^2}{a^2}}\)
\(=a\times\frac{a^2-b^2}{a^2}\times\frac{a^2}{a^2+b^2}+b\times\frac{2b}{a}\times\frac{a^2}{a^2+b^2}\)
\(=a\times\frac{a^2-b^2}{a^2}\times\frac{a^2}{a^2+b^2}+b\times\frac{2b}{a}\times\frac{a^2}{a^2+b^2}\)
\(=\frac{a\left(a^2-b^2\right)}{a^2+b^2}+\frac{2ab^2}{a^2+b^2}\)
\(=\frac{a^3-ab^2+2ab^2}{a^2+b^2}\)
\(=\frac{a^3+ab^2}{a^2+b^2}\)
\(=\frac{a\left(a^2+b^2\right)}{\left(a^2+b^2\right)}\)
\(=a\)
\(=\) RHS


Question Number 9
If \( \displaystyle \tan \theta = \frac {1}{7} \) and \( \tan \beta = \frac{1}{3} \) prove that \( \cos 2\theta  = \sin 4 \beta \)
Given,
\(\displaystyle \tan\theta=\frac{1}{7}\) and \( \tan\beta=\frac{1}{3}\)
We know,
\(\displaystyle \tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}\)
            \(\displaystyle =\frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{3}\right)^2}\)
            \(\displaystyle =\frac{\frac{2}{3}}{1-\frac{1}{9}}\)
            \(\displaystyle =\frac{\frac{2}{3}}{\frac{9-1}{9}}\)
            \( \displaystyle=\frac{\frac{2}{3}}{\frac{8}{9}}\)
            \( \displaystyle=\frac{2}{3}\times\frac{9}{8}\)
            \( \displaystyle=\frac{3}{4}\)
\(\displaystyle \therefore \tan2\beta=\frac{3}{4}\)
 Now, 
 LHS   \( \displaystyle= \cos2\theta\)
\( \displaystyle=\frac{1-\tan^2\theta}{1+\tan^2\theta}\)
\( \displaystyle=\frac{1-\left(\frac{1}{7}\right)^2}{1+\left(\frac{1}{7}\right)^2}\)
\( \displaystyle=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}\)
\(\displaystyle =\frac{\frac{49-1}{49}}{\frac{49+1}{49}}\)
\( \displaystyle=\frac{\frac{48}{49}}{\frac{50}{49}}\)
\( \displaystyle=\frac{48}{50}\)
\( \displaystyle=\frac{24}{25}\)
RHS   \(\displaystyle =\sin4\beta\)
\( \displaystyle=\frac{2\tan2\beta}{1+\tan^22\beta}\)
\( \displaystyle=\frac{2\times\left(\frac{3}{4}\right)}{1+\left(\frac{3}{4}\right)^2}\)
\( \displaystyle=\frac{\frac{3}{2}}{1+\frac{9}{16}}\)
\(\displaystyle =\frac{\frac{3}{2}}{\frac{16+9}{16}}\)
\( \displaystyle=\frac{\frac{3}{2}}{\frac{25}{16}}\)
\(\displaystyle =\frac{3}{2}\times\frac{16}{25}\)
\( \displaystyle=\frac{24}{25}\)
\(\displaystyle \therefore\)LHS \(\displaystyle=\) RHS   Proved

Question Number 10
If \( 2 \tan \alpha = 3\tan \beta \) prove that: \( \tan (\alpha - \beta ) = \frac {\sin 2\beta }{5-\cos 2\beta}\)
Given,
\(2\tan\alpha=3\tan\beta\)
or, \(\tan\alpha=\frac{3\tan\beta}{2}\)
LHS\(=\tan\left(\alpha-\beta\right)\)
   \(=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\)
   \(=\frac{\frac{3\tan\beta}{2}-\tan\beta}{1+\frac{3\tan\beta}{2}\times\tan\beta}\)
   \(=\frac{\frac{3\tan\beta-2\tan\beta}{2}}{\frac{2-3\tan^2\beta}{2}}\)
   \(=\frac{\tan\beta}{2+3\tan^2\beta}\)
   \(=\frac{\frac{\sin\beta}{\cos\beta}}{2+3\times\frac{\sin^2\beta}{\cos^2\beta}}\)
   \(=\frac{\frac{\sin\beta}{\cos\beta}}{\frac{2\cos^2\beta+3\sin^2\beta}{\cos^2\beta}}\)
   \(=\frac{\sin\beta}{\cos\beta}\times\frac{\cos^2\beta}{2\cos^2\beta+3\sin^2\beta}\)
   \(=\frac{\sin\beta\cos\beta}{2\cos^2\beta+3\sin^2\beta}\times\frac{2}{2}\)
   \(=\frac{2\sin\beta\cos\beta}{4\cos^2\beta+6\sin^2\beta}\)
   \(=\frac{\sin2\beta}{4\times\frac{1+\cos2\beta}{2}+6\times\frac{1-\cos2\beta}{2}}\)
   \(=\frac{\sin2\beta}{2+2\cos2\beta+3-3\sin2\beta}\)
   \(=\frac{\sin2\beta}{5-\cos2\beta}\)
   \(=\)RHS

Question Number 11
If \( 2 \tan \alpha = 3\tan \beta \) prove that \(\tan(\alpha+\beta)=\frac{5\sin2\beta}{5\cos2\beta-1}\)
Given, \(2\tan\alpha=3\tan\beta\)
   or, \(\tan\alpha=\frac{3\tan\beta}{2}\)
LHS\(=\tan\left(\alpha+\beta\right)\)
   \(=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)
   \(=\frac{\frac{3\tan\beta}{2}+\tan\beta}{1-\frac{3\tan\beta}{2}\times\tan\beta}\)
   \(=\frac{\frac{3\tan\beta+2\tan\beta}{2}}{\frac{2-3\tan^2\beta}{2}}\)
   \(=\frac{5\tan\beta}{2-3\tan^2\beta}\)
   \(=\frac{5\frac{\sin\beta}{\cos\beta}}{2-3\times\frac{\sin^2\beta}{\cos^2\beta}}\)
   \(=\frac{5\frac{\sin\beta}{\cos\beta}}{\frac{2\cos^2\beta-3\sin^2\beta}{\cos^2\beta}}\)
   \(=\frac{5\sin\beta}{\cos\beta}\times\frac{\cos^2\beta}{2\cos^2\beta-3\sin^2\beta}\)
   \(=\frac{5\sin\beta\cos\beta}{2\cos^2\beta-3\sin^2\beta}\times\frac{2}{2}\)
   \(=\frac{10\sin\beta\cos\beta}{4\cos^2\beta-6\sin^2\beta}\)
   \(=\frac{5\left(2\sin\beta\cos\beta\right)}{4\times\frac{1+\cos2\beta}{2}-6\times\frac{1-\cos2\beta}{2}}\)
   \(=\frac{5\sin2\beta}{2+2\cos2\beta-3+3\cos2\beta}\)
   \(=\frac{5\sin2\beta}{5\cos2\beta-1}\)
   \(=\) RHS
   
   
Question Number 12
Prove that: \( \frac{1}{\tan 3A - \tan A } - \frac{1}{\cot 3A - \cot A } = \cot 2A  \)
LHS
   \(=\frac{1}{\tan3A-\tan A}-\frac{1}{\cot3A-\cot A}\)
   \(=\frac{1}{\frac{1}{\cot3A}-\frac{1}{\cot A}}-\frac{1}{\cot3A-\cot A}\)
   \(=\frac{1}{\frac{\cot A-\cot3A}{\cot3A\cot A}}-\frac{1}{\cot3A-\cot A}\)
   \(=\frac{\cot3A\cot A}{\cot A-\cot3A}-\frac{1}{\cot3A-\cot A}\)
   \(=\frac{\cot3A\cot A}{\cot A-\cot3A}+\frac{1}{\cot A-\cot3A}\)
   \(=\frac{\cot3A\cot A+1}{\cot A-\cot3A}\)
   \(=\cot\left(3A-A\right)\)
   \(=\cot2A\)
   \(=\)RHS
   
Question Number 13
Prove that: \(\frac{1}{\tan 3A + \tan A } - \frac{1}{\cot 3A + \cot A } = \cot 4A   \)
LHS
   \(=\frac{1}{\tan3A+\tan A}-\frac{1}{\cot3A+\cot A}\)
   \(=\frac{1}{\frac{1}{\cot3A}+\frac{1}{\cot A}}-\frac{1}{\cot3A+\cot A}\)
   \(=\frac{1}{\frac{\cot A+\cot3A}{\cot3A\cot A}}-\frac{1}{\cot3A+\cot A}\)
   \(=\frac{\cot3A\cot A}{\cot A+\cot3A}-\frac{1}{\cot3A+\cot A}\)
   \(=\frac{\cot3A\cot A}{\cot A+\cot3A}-\frac{1}{\cot A+\cot3A}\)
   \(=\frac{\cot3A\cot A-1}{\cot A+\cot3A}\)
   \(=\cot\left(3A+A\right)\)
   \(=\cot4A\)
   \(=\)RHS
Question Number 14
Prove that: \(\frac{\cot A}{\cot A -\cot 3A} - \frac{\tan A}{\tan 3A -\tan A } =1 \)
LHS
   \(=\frac{\cot A}{\cot A-\cot3A}-\frac{\tan A}{\tan3A-\tan A}\)
   \(=\frac{\frac{1}{\tan A}}{\frac{1}{\tan A}-\frac{1}{\tan3A}}-\frac{\tan A}{\tan3A-\tan A}\)
   \(=\frac{\frac{1}{\tan A}}{\frac{\tan3A-\tan A}{\tan A\tan3A}}-\frac{\tan A}{\tan3A-\tan A}\)
   \(=\frac{1}{\tan A}\times\frac{\tan A\tan3A}{\tan3A-\tan A}-\frac{\tan A}{\tan3A-\tan A}\)
   \(=\frac{\tan3A}{\tan3A-\tan A}-\frac{\tan A}{\tan3A-\tan A}\)
   \(=\frac{\tan3A-\tan A}{\tan3A-\tan A}\)
  \(=1\)
   \(=\) RHS
Question Number 15
Prove that:\(\tan \left(\frac{\pi^c}{4}+A\right)+\tan \left(\frac{\pi^c}{4}-A\right)=2\sec 2A  \)
LHS
   \(=\tan\left(\frac{\pi^c}{4}+A\right)+\tan\left(\frac{\pi^c}{4}-A\right)\)
   \(=\tan\left(45^\circ+A\right)+\tan\left(45^\circ-A\right)\)
   \(=\frac{\sin\left(45^\circ+A\right)}{\cos\left(45^\circ+A\right)}+\frac{\sin\left(45^\circ-A\right)}{\cos\left(45^\circ-A\right)}\)
   \(=\frac{\sin\left(45^\circ+A\right)\cos\left(45^\circ-A\right)+\cos\left(45^\circ+A\right)\sin\left(45^\circ-A\right)}{\cos\left(45^\circ+A\right)\cos\left(45^\circ-A\right)}\)
   \(=\frac{\sin\left(45^\circ+A+45^\circ-A\right)}{\cos\left(45^\circ+A\right)\cos\left(45^\circ-A\right)}\)
   \(=\frac{\sin90^\circ}{\cos\left(45^\circ+A\right)\cos\left(45^\circ-A\right)}\)
   \(=\frac{1}{\cos\left(45^\circ+A\right)\cos\left(45^\circ-A\right)}\times\frac{2}{2}\)
   \(=\frac{2}{2\cos\left(45^\circ+A\right)\cos\left(45^\circ-A\right)}\)
   \(=\frac{2}{\cos\left(45^\circ+A+45^\circ-A\right)+\cos\left(45^\circ+A-45^\circ+A\right)}\)
   \(=\frac{2}{\cos90^\circ+\cos2A}\)
   \(=\frac{2}{0+\cos2A}\)
   \(=\frac{2}{\cos2A}\)
   \(= 2\sec2A\)
   \(= \)RHS
Question Number 16
Prove that: \(\frac{\cos^3 \alpha +\sin^3 \alpha }{\cos \alpha + \sin \alpha } = 1 -\frac{1}{2} \sin 2\alpha  \)
Solution: 
LHS
\(=\frac{\cos^3\alpha+\sin^3\alpha}{\cos\alpha+\sin\alpha}\)
\(=\frac{\left(\cos\alpha+\sin\alpha\right)\left(\cos^2\alpha-\cos\alpha\sin\alpha+\sin^2\alpha\right)}{\cos\alpha+\sin\alpha}\)
\(=\sin^2\alpha+\cos^2\alpha-\frac{1}{2}\times2\cos\alpha\sin\alpha\)
\(=1-\frac{1}{2}\sin2\alpha\)
\(=\)RHS

Question Number 17
Prove that: \( \sin8\theta=8\sin\theta\cos\theta\cos2\theta\cos4\theta \)
Solution:
LHS 
   \(=\sin8\theta\)
   \(=2\sin4\theta\cos4\theta\)   \(\left[\because\sin2A=2\sin A\cos A\right]\)      
   \(=2\times2\sin2\theta\cos2\theta\cos4\theta\)  
   \(=2\times2\times2\sin\theta\cos\theta\cos2\theta\cos4\theta\)
   \(=8\sin\theta\cos\theta\cos2\theta\cos4\theta\)
   \(=\) RHS
Question Number 18
Prove that: \( \cos4\theta=8\cos^4\theta-8\cos^2\theta+1 \)
LHS 
   \(=\cos4\theta\)
   \(=2\cos^22\theta-1\) \(\left[\because\cos2A=2\cos^2A-1\right]\)
   \(=2\left(\cos2\theta\right)^3-1\)
   \(=2\left(2\cos^2\theta-1\right)^2-1\)
   \(=2\left[\left(2\cos\theta\right)^2-2\times2\cos^2\theta\times1+1^2\right]-1\)
   \(=2\left(4\cos^2\theta-4cos^2\theta+1\right)-1\)
   \(=8\cos^2\theta-8\cos\theta+2-1\) 
   \(=8\cos^4\theta-8\cos^2\theta+1\)
   \(=\) RHS
Question Number 19
Prove that:\( \cos 4\theta = 1 - 8\sin^2 \theta + 8\sin^4 \theta\)
LHS
   \(=\cos4\theta\)
   \(=2cos^22\theta-1\)
   \(=2\left(\cos2\theta\right)^2-1\)
   \(=2\left(1-2\sin^2\theta\right)^2-1\)
   \(=2\left\{1^2-2\times1\times2\sin^2\theta+\left(2\sin^2\theta\right)^2\right\}-1\)
   \(=2\left(1-4\sin^2\theta+4\sin^4\theta\right)-1\)
   \(=2-8\sin^2\theta+8\sin^4\theta-1\)
   \(=1-8\sin^2\theta+8\sin^4\theta\)
   \(=\) RHS
Question Number 20
Prove that: \( \frac{\sin8\theta}{8\sin\theta}=\cos\theta\cos2\theta\cos4\theta\)
Solution:
LHS
   \(=\frac{2\sin4\theta\cos4\theta}{8\sin\theta}\)
   \(=2\times\frac{2\sin2\theta\cos2\theta\cos4\theta}{8\sin\theta}\)
   \(=2\times2\times\frac{2\sin\theta\cos\theta\cos2\theta\cos4\theta}{8\sin\theta}\)
   \(=\cos\theta\cos2\theta\cos4\theta\)
   \(=\) RHS
Question Number 21
Prove that: \( \frac{1}{4}\sin4\theta.\sec\theta=\sin\theta\cos2\theta\)
LHS
   \(=\frac{1}{4}\sin4\theta.\sec\theta\)
   \(=\frac{1}{4}\times2\sin2\theta\cos2\theta.\sec\theta\)
   \(=\frac{1}{2}\times2\sin\theta\cos\theta.\cos2\theta\times\frac{1}{\cos\theta}\)
   \(=\sin\theta\cos2\theta\)
   \(=\) RHS
Question Number 22
Prove that: \( 2\sin^2\left(\frac{\pi^c}{4}-A\right)=\frac{\left(1-\tan A\right)^2}{1+\tan^2A}\)
Solution:
LHS
\(=2\sin^2\left(\frac{\pi^c}{4}-A\right)\)
\(=2\sin^2\left(45^\circ-A\right)\)
\(=1-\cos2\left(45^\circ-A\right)\)
\(=1-\cos\left(90^\circ-2A\right)\)
\(=1-\sin2A\)
\(=1-\frac{2\tan A}{1+\tan^2A}\)
\(=\frac{1+\tan^2A-2\tan A}{1+\tan^2A}\)
\(=\frac{\left(1-\tan A\right)^2}{1+\tan^2A}\)
\(=\) RHS

Question Number 23
Prove that: \(\frac{1}{2}\sqrt{\frac{1-\cos4\theta}{2}}=\sin\theta\cos\theta\)
LHS
   \(=\frac{1}{2}\sqrt{\frac{1-\cos4\theta}{2}}\)
   \(=\frac{1}{2}\sqrt{\frac{1-\left(1-2\sin^22\theta\right)}{2}}\)
   \(=\frac{1}{2}\sqrt{\frac{1-1+2\sin^22\theta}{2}}\)
   \(=\frac{1}{2}\sqrt{\frac{2\sin^22\theta}{2}}\)
   \(=\frac{1}{2}\sqrt{\sin^22\theta}\)
   \(=\frac{1}{2}\sin2\theta\)
   \(=\frac{1}{2}\times2\sin\theta\cos\theta\)
   \(=\sin\theta\cos\theta\)
   \(=\) RHS
Question Number 24
Prove that: \(\tan2\alpha+\sin2\alpha=\frac{4\tan\alpha}{1-\tan^4\alpha}\)
LHS
   \(=\tan2\alpha+\sin2\alpha\)
   \(=\frac{2\tan\alpha}{1-\tan^2\alpha}+\frac{2\tan\alpha}{1+\tan^2\alpha}\)
   \(=\frac{2\tan\alpha\left(1+\tan^2\alpha\right)+2\tan\alpha\left(1-\tan^2\alpha\right)}{\left(1-\tan^2\alpha\right)\left(1+\tan^2\alpha\right)}\)
   \(=\frac{2\tan\alpha+\tan^3\alpha+2\tan\alpha-\tan^3\alpha}{1^2-\left(\tan^2\alpha\right)^2}\)
   \(=\frac{4\tan\alpha}{1-\tan^4\alpha}\)
   \(=\)RHS

Question Number 25
Prove that: \(4\left(\cos^310^\circ+\sin^320^\circ\right)=3\left(\cos10^\circ+\sin20^\circ\right)\)
LHS
   \(=4\left(\cos^310^\circ+\sin^320^\circ\right)\)
   \(=4\cos^310^\circ+4\sin^320^\circ\)
   \(=3\cos10^\circ+\cos3\left(10^\circ\right)+3\sin20^\circ-\sin3\left(20^\circ\right)\)
   \(=3\cos10^\circ+\cos30^\circ+3\sin20^\circ-\sin60^\circ\)
   \(=3\cos10^\circ+\frac{\sqrt{3}}{2}+3\sin20^\circ-\frac{\sqrt{3}}{2}\)
   \(=3\cos10^\circ+3\cos20^\circ\)
   \(=3\left(\cos10^\circ+\cos20^\circ\right)\)
   \(=\) RHS

Question Number 26   
Prove that:\(4\left(\cos^320^{\circ}+\sin^350^{\circ}\right)=3\left(\cos20^{\circ}+\sin50^{\circ}\right)\)
Solution:
LHS
   \(=4\left(\cos^320^{\circ}+\sin^350^{\circ}\right)\)
   \(=4\cos^320^\circ+4\sin^350^\circ\)
   \(=3\cos20^\circ+\cos3\left(20^\circ\right)+3\sin50^\circ-\sin3\left(50^\circ\right)\)
   \(=3\cos20^{\circ}+\cos60^{\circ}+3\sin50^{\circ}-\sin150^{\circ}\)
   \(=3\cos20^{\circ}+\frac{1}{2}+3\sin50^{\circ}-\frac{1}{2}\)
   \(=3\left(\cos20^\circ+\sin50^\circ\right)\)
   \(=\)RHS

Question Number 27   
Prove that: \(2\cos2\theta+1=\left(2\cos\theta+1\right)\left(2\cos\theta-1\right)\)
LHS
   \(=2\cos2\theta+1\)
   \(=2\left(2\cos^2\theta-1\right)+1\)
   \(=4\cos^2\theta-2+1\)
   \(=4\cos^2\theta-1\)
   \(=\left(2\cos\theta\right)^2-1^2\)
   \(=\left(2\cos\theta-1\right)\left(2\cos\theta+1\right)\)
   \(=\) RHS

Question Number 28
Prove that: \((2\cos\theta+1)(2\cos\theta-1)\left(2\cos2\theta-1\right)=2\cos4\theta+1\)
Solution:
LHS
   \(=(2\cos\theta+1)(2\cos\theta-1)\left(2\cos2\theta-1\right)\)
   \(=\left\{\left(2\cos\theta\right)^2-1^2\right\}\left(2\cos2\theta-1\right)\)
   \(=\left(4\cos^2\theta-1\right)\left(2\cos2\theta-1\right)\)
   \(=\left(4\times\frac{1+\cos2\theta}{2}-1\right)\left(2\cos2\theta-1\right)\)
   \(=\left(2+2\cos2\theta-1\right)\left(2\cos2\theta-1\right)\)
   \(=\left(2\cos2\theta+1\right)\left(2\cos2\theta-1\right)\)
   \(=\left(2\cos2\theta\right)^2-1^2\)
   \(=4\cos^22\theta-1\)
   \(=4\times\frac{1+\cos4\theta}{2}-1\)
   \(=2\left(1+\cos4\theta\right)-1\)
   \(=2+2\cos4\theta-1\)
   \(=2\cos4\theta+1\)
   \(=\) RHS

Question Number 29   
Prove that: \(\frac{2\cos8\theta+1}{2\cos\theta+1}=\left(2\cos\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos4\theta-1\right)\)
Solution:
LHS
\(=\frac{2\cos8\theta+1}{2\cos\theta+1}\)
\(=\frac{2\left(2\cos^24\theta-1\right)+1}{2\cos\theta+1}\)
\(=\frac{4\cos^24\theta-2+1}{2\cos\theta+1}\)
\(=\frac{4\cos^24\theta-1}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta\right)^2-1^2}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(2\cos4\theta+1\right)}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left\{2\left(2\cos^22\theta-1\right)+1\right\}}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(4\cos^22\theta-2+1\right)}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(4\cos^22\theta-1\right)}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left\{\left(2\cos2\theta\right)^2-1^2\right\}}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos2\theta+1\right)}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left\{2\left(2\cos^2\theta-1\right)+1\right\}}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(4\cos^2\theta-2+1\right)}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(4\cos^2\theta-1\right)}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left\{\left(2\cos\theta\right)^2-1^2\right\}}{2\cos\theta+1}\)
\(=\frac{\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos\theta-1)(2\cos\theta+1)\right)}{2\cos\theta+1}\)
\(=\left(2\cos4\theta-1\right)\left(2\cos2\theta-1\right)\left(2\cos\theta-1)\right) \)
\(=\left(2\cos\theta-1)\right)\left(2\cos2\theta-1\right)\left(2\cos4\theta-1\right) \)
\( = \) RHS

Question Number 30
Prove that:\( \frac{\sec4\theta-1}{\sec2\theta-1}=\tan4\theta\cot\theta\)
Solution:
LHS
   \(=\frac{\sec4\theta-1}{\sec2\theta-1}\)
   \(= \dfrac{\dfrac{1}{\cos4\theta}-1}{\dfrac{1}{\cos2\theta}-1}\)
   \(= \dfrac{\dfrac{1-\cos4\theta}{\cos4\theta}}{\dfrac{1-\cos2\theta}{\cos2\theta}}\)
   \(=\frac{1-\cos4\theta}{\cos4\theta}\times\frac{\cos2\theta}{1-\cos2\theta}\)
   \(=\frac{1-\left(1-2\sin^22\theta\right)}{\cos4\theta}\times\frac{\cos2\theta}{1-\left(1-2\sin^2\theta\right)}\)
   \(=\frac{2\sin^22\theta}{\cos4\theta}\times\frac{\cos2\theta}{2\sin^2\theta}\)
   \(=\frac{2\sin2\theta\cos2\theta}{\cos4\theta}\times\frac{\sin2\theta}{2\sin^2\theta}\)
   \(=\frac{\sin4\theta}{\cos4\theta}\times\frac{2\sin\theta\cos\theta}{2\sin^2\theta}\)
   \(=\tan4\theta\times\frac{\cos\theta}{\sin\theta}\)
   \(=\tan4\theta.\cot\theta\)
   \(=\)RHS

Question Number 31
Prove that: \( \frac{\sec8A-1}{\sec4A-1}=\frac{\tan8A}{\tan2A}\) 
Solution:
LHS
   \(=\frac{\sec8A-1}{\sec4A-1}\)
   \(=\frac{\frac{1}{\cos8A}-1}{\frac{1}{\cos4A}-1}\)
   \(=\frac{\frac{1-\cos8A}{\cos8A}}{\frac{1-\cos4A}{\cos4A}}\)
   \(=\frac{1-\cos8A}{\cos8A}\times\frac{\cos4A}{1-\cos4A}\)
   \(=\frac{1-\left(1-2\sin^24A\right)}{\cos8A}\times\frac{\cos4A}{1-\left(1-2\sin^22A\right)}\)
   \(=\frac{2\sin^24A}{\cos8A}\times\frac{\cos4A}{2\sin^22A}\)
   \(=\frac{2\sin4A\cos4A}{\cos8A}\times\frac{\sin4A}{2\sin^22A}\)
   \(=\frac{\sin8A}{\cos8A}\times\frac{2\sin2A\cos2A}{2\sin^22A}\)
   \(=\tan8A\times\frac{\cos2A}{\sin2A}\)
   \(=\tan8A.\cot2A\)
   \(=\frac{\tan8A}{\tan2A}\)
   \(=\)RHS

Question Number 32
Prove that:\(\left(\sec2\theta+1\right)\left(\sec4\theta+1\right)\left(\sec8\theta+1\right)=\tan8\theta.\cot\theta\)
LHS
\(=\left(\sec2\theta+1\right)\left(\sec4\theta+1\right)\left(\sec8\theta+1\right)\)
\(=\left(\frac{1}{\cos2\theta+1}\right)\left(\frac{1}{\cos4\theta}+1\right)\left(\frac{1}{\cos8\theta+1}\right)\)
\(=\frac{1+\cos2\theta}{\cos2\theta}\times\frac{1+\cos4\theta}{\cos4\theta}\times\frac{1+\cos8\theta}{\cos8\theta}\)
\(=\frac{2\cos^2\theta}{\cos2\theta}\times\frac{2\cos^22\theta}{\cos4\theta}\times\frac{2\cos^24\theta}{\cos8\theta}\)
\(=\frac{8}{\cos8\theta}\left(\cos^2\theta\cos2\theta\cos4\theta\right)\)
\(=\frac{\cos\theta}{\sin\theta}\times\frac{4}{\cos8\theta}\times\left(2\sin\theta.\cos\theta\right)\cos2\theta.\cos4\theta\)
\(=\cot\theta\times\frac{4}{\cos8\theta}\times\sin2\theta\cos2\theta.\cos4\theta\)
\(=\cot\theta\times\frac{2}{\cos8\theta}\times\left(2\sin2\theta\cos2\theta\right)\cos4\theta\)
\(=\cot\theta\times\frac{2}{\cos8\theta}\times\sin4\theta\cos4\theta\)
\(=\cot\theta\times\frac{\sin8\theta}{\cos8\theta}\)
\(=\cot\theta.\tan8\theta\)
\(=\tan8\theta\cot\theta\)
\(=\)RHS

Question Number 33
Prove that: \(\sin^2\theta+\sin^2(\theta+120^\circ)+\sin^2(\theta-120^\circ)=\frac{3}{2}\)
Solution:
LHS
\(=\sin^2\theta+\sin^2(\theta+120^{\circ})+\sin^2(\theta-120^{\circ})\)
\(=\frac{1-\cos2\theta}{2}+\frac{1-\cos2\left(\theta+120^\circ\right)}{2}+\frac{1-\cos2\left(\theta-120^\circ\right)}{2}\)
\(=\frac{1-\cos2\theta+1-\cos\left(2\theta+240^\circ\right)+1-\cos\left(2\theta-240^\circ\right)}{2}\)
\(=\frac{3-\cos2\theta-\left\{\cos\left(2\theta+240^\circ\right)+\cos\left(2\theta-240^\circ\right)\right\}}{2}\)
\(=\frac{3-\cos2\theta-2\cos2\theta\cos240^\circ}{2}\)
\(=\frac{3-\cos2\theta-2\cos2\theta\times\left(-\frac{1}{2}\right)}{2}\)\(\left[\because\cos240^\circ=\cos\left(180^\circ+60^\circ\right)=-\cos60^\circ=-\frac{1}{2}\right]\)
\(=\frac{3-\cos2\theta+\cos2\theta}{2}\)
\(=\frac{3}{2}\)
\(=\) RHS

Question Number 34
Prove that:\( \cos^2\theta+\cos^2(\theta+120^{\circ})+\cos^2(\theta-120^{\circ})=\frac{3}{2}\)
Solution:
LHS
\(=\cos^2\theta+\cos^2(\theta+120^{\circ})+\cos^2(\theta-120^{\circ})\)
\(=\frac{1+\cos2\theta}{2}+\frac{1+\cos2\left(\theta+120^\circ\right)}{2}+\frac{1+\cos2\left(\theta-120^\circ\right)}{2}\)
\(=\frac{1+\cos2\theta+1+\cos\left(2\theta+240^\circ\right)+1+\cos\left(2\theta-240^\circ\right)}{2}\)
\(=\frac{3+\cos2\theta+\left\{\cos\left(2\theta+240^{\circ}\right)+\cos\left(2\theta-240^{\circ}\right)\right\}}{2}\)
\(=\frac{3+\cos2\theta+2\cos2\theta\cos240^\circ}{2}\)
\(=\frac{3+\cos2\theta+2\cos2\theta\left(-\frac{1}{2}\right)}{2}\)\(\left[\because\cos240^{\circ}=\cos\left(180^{\circ}+60^{\circ}\right)=-\cos60^{\circ}=-\frac{1}{2}\right]\)
\(=\frac{3+\cos2\theta-\cos2\theta}{2}\)
\(=\frac{3}{2}\)
\(=\)

Question Number 35
Prove that: \(-\sin^3A+\sin^3(60^{\circ}+A)+\sin^3(240^{\circ}-A)=\frac{3}{4}\sin3A\)
Solution:
LHS
\(=-\sin^3A+\sin^3(60^{\circ}+A)+\sin^3(240^{\circ}-A)\)
\(=-\sin^3A+\sin^3(60^{\circ}+A)+\sin^3(180^\circ+60^\circ-A)\)
\(=-\sin^3A+\sin^3(60^{\circ}+A)-\sin^3(60^{\circ}-A)\)
\(=\frac{-\left(3\sin A-\sin3A\right)}{4}+\frac{3\sin\left(60^{\circ}+A\right)+\sin3\left(60^{\circ}+A\right)}{4}-\frac{3\sin\left(60^{\circ}-A\right)-\sin3\left(60^{\circ}-A\right)}{4}\)
\(=\frac{-3\sin A+\sin3A+3\sin\left(60^{\circ}+A\right)-\sin\left(180^{\circ}+3A\right)-3\sin\left(60^{\circ}-A\right)+\sin\left(180^{\circ}-3A\right)}{4}\)
\(=\frac{-3\sin A+\sin3A+3\sin\left(60^{\circ}+A\right)+\sin3A-3\sin\left(60^{\circ}-A\right)+\sin3A}{4}\)
\(=\frac{-3\sin A+3\sin3A+3\left[\sin\left(60^{\circ}+A\right)-\sin\left(60^{\circ}-A\right)\right]}{4}\)
\(=\frac{-3\sin A+3\sin3A+3\left(2\cos60^\circ\sin A\right)}{4}\)
\(=\frac{-3\sin A+3\sin3A+3\left(2\times\frac{1}{2}\sin A\right)}{4}\)
\(=\frac{-3\sin A+3\sin3A+3\sin A}{4}\)
\(=\frac{3}{4}\sin3A\)
\(=\)RHS

Question Number 35
Prove that: \( \text{cosec }2A+\cot4A=\cot A-\text{cosec }4A \) 
Solution:
LHS
\(=\text{cosec }2A+\cot4A\)
\(=\frac{1}{\sin2A}+\frac{\cos4A}{\sin4A}\)
\(=\frac{1}{\sin2A}+\frac{2\cos^22A-1}{2\sin2A\cos2A}\)
\(=\frac{2\cos2A+2\cos^22A-1}{2\sin2A\cos2A}\)
\(=\frac{2\cos2A\left(1+\cos2A\right)-1}{\sin4A}\)
\(=\frac{2\cos2A\left(1+\cos2A\right)}{2\sin2A\cos2A}-\frac{1}{2\sin2A\cos2A}\)
\(=\frac{1+\cos2A}{\sin2A}-\frac{1}{\sin4A}\)
\(=\frac{1+2\cos^2A-1}{2\sin A\cos A}-\text{cosec }4A\)
\(=\frac{2\cos^2A}{2\sin A\cos A}-\text{cosec }4A\)
\(=\frac{\cos A}{\sin A}-\text{cosec }4A\)
\(=\cot A-\text{cosec }4A\)
\(=\)RHS

Question Number 37
Prove that:\( \text{cosec }4\theta+\cot8\theta=\cot2\theta-\text{cosec }8\theta\)

Question Number 38
Prove that: \( \text{cosec }2\theta+\text{cosec }4\theta=\cot\theta-\cot4\theta\)
Solution:
LHS
\(=\text{cosec }2\theta+\text{cosec }4\theta\)
\(=\cot\theta-\cot\theta+\text{cosec }2\theta+\text{cosec }4\theta\)
\(=\cot\theta-\frac{\cos\theta}{\sin\theta}+\frac{1}{\sin2\theta}+\text{cosec }4\theta\)
\(=\cot\theta-\left(\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin2\theta}\right)+\text{cosec }4\theta\)
\(=\cot\theta-\left(\frac{\cos\theta}{\sin\theta}-\frac{1}{2\sin\theta\cos\theta}\right)+\text{cosec }4\theta\)
\(=\cot\theta-\frac{\cos^2\theta-1}{2\sin\theta\cos\theta}+\text{cosec }4\theta\)
\(=\cot\theta-\frac{\cos2\theta}{\sin2\theta}+\frac{1}{\sin4\theta}\)
\(=\cot\theta-\left(\frac{\cos2\theta}{\sin2\theta}-\frac{1}{2\sin2\theta\cos2\theta}\right)\)
\(=\cot\theta-\frac{2\cos^22\theta-1}{2\sin2\theta\cos2\theta}\)
\(=\cot\theta-\frac{\cos4\theta}{\sin4\theta}\)
\(=\cot\theta-\cot4\theta\)
\(=\)RHS


Question Number 39
Prove that: \( \frac{\cos^3\theta-\cos3\theta}{\cos\theta}+\frac{\sin^3\theta+\sin3\theta}{\sin\theta}=3\)
Solution:
LHS
\(=\frac{\cos^3\theta-\cos3\theta}{\cos\theta}+\frac{\sin^3\theta+\sin3\theta}{\sin\theta}\)
\(=\frac{\cos^3\theta-\left(4\cos^3\theta-3\cos\theta\right)}{\cos\theta}+\frac{\sin^3\theta+\left(3\sin\theta-4\sin^3\theta\right)}{\sin\theta}\)
\(=\frac{\cos^3\theta-4\cos^3\theta+3\cos\theta}{\cos\theta}+\frac{\sin^3\theta+3\sin\theta-4\sin^3\theta}{\cos\theta}\)
\(=\frac{3\cos\theta-3\cos^3\theta}{\cos\theta}+\frac{3\sin\theta-3\sin^3\theta}{\sin\theta}\)
\(=\frac{3\cos\theta}{\cos\theta}-\frac{3\cos^3\theta}{\cos\theta}+\frac{3\sin\theta}{\sin\theta}-\frac{3\sin^3\theta}{\sin\theta}\)
\(=3-3\cos^2\theta+3-3\sin^2\theta\)
\(=6-3\left(\sin^2\theta+\cos^2\theta\right)\)
\(=6-3\left(1\right)\)
\(=6-3\)
\(=3\)
\(=\)RHS


Question Number 40
Prove that:\(\cos^3A\cos3A+\sin^3A\sin3A=\cos^32A\)
Solution:
LHS
\(=\cos^3\theta.\cos3\theta+\sin^3\theta.\sin3\theta\)
\(=\frac{3\cos\theta+\cos3\theta}{4}\times\cos3\theta+\frac{3\sin\theta-\sin3\theta}{4}\times\sin3\theta\)
\(=\frac{3\cos3\theta\cos\theta+\cos^23\theta}{4}+\frac{3\sin3\theta\sin\theta-\sin^23\theta}{4}\)
\(=\frac{3\cos3\theta\cos\theta+\cos^23\theta+3\sin3\theta\sin\theta-\sin^23\theta}{4}\)
\(=\frac{3\left(\cos3\theta\cos\theta+\sin3\theta\sin\theta\right)+\left(\cos^23\theta-\sin^23\theta\right)}{4}\)
\(=\frac{3\cos\left(3\theta-\theta\right)+\cos2\left(3\theta\right)}{4}\)
\(=\frac{3\cos\left(3\theta-\theta\right)+\cos6\theta}{4}\)
\(=\frac{3\cos2\theta+\cos3\left(2\theta\right)}{4}\)
\(=\frac{3\cos2\theta+4\cos^3\theta-3\cos\theta}{4}\)
\(=\frac{4\cos^3\theta}{4}\)
\(=\cos^3\theta\)
\(=\)RHS

Question Number 41
Prove that: \(\sin^3\theta\cos3\theta+\cos^3\theta\sin3\theta=\frac{3}{4}\sin4\theta\)
Solution:
LHS
\(=\sin^3\theta\cos3\theta+\cos^3\theta\sin3\theta\)
\(=\frac{3\sin\theta-\sin3\theta}{4}\times\cos3\theta+\frac{3\cos\theta+\cos3\theta}{4}\times\sin3\theta\)
\(=\frac{3\cos3\theta\sin\theta-\sin3\theta\cos3\theta}{4}+\frac{3\sin3\theta\cos\theta+\sin3\theta\cos3\theta}{4}\)
\(=\frac{3\cos3\theta\sin\theta-\sin3\theta\cos3\theta+3\sin3\theta\cos\theta+\sin3\theta\cos3\theta}{4}\)
\(=\frac{3\left(\cos3\theta\sin\theta+\sin3\theta\cos\theta\right)}{4}\)
\(=\frac{3\sin\left(3\theta+\theta\right)}{4}\)
\(=\frac{3}{4}\sin4\theta\)
\(=\)RHS

Question Number 42
Prove that: \( \cos^2\alpha+\sin^2\alpha.\cos2\beta=\cos^2\beta+\sin^2\beta.\cos2\alpha\)
Solution:
LHS
   \(=\cos^2\alpha+\sin^2\alpha.\cos2\beta\)
   \(=\cos^2\alpha+\sin^2\alpha\left(1-2\sin^2\beta\right)\)
   \(=\cos^2\alpha+\sin^2\alpha-2\sin^2\alpha\sin^2\beta\)
   \(=1-2\sin^2\alpha\sin^2\beta\)
   \(=\cos^2\beta+\sin^2\beta-2\sin^2\alpha\sin^2\beta\)
   \(=\cos^2\beta+\sin^2\beta\left(1-2\sin^2\alpha\right)\)
   \(=\cos^2\beta+\sin^2\beta\cos2\alpha\)
   \(=\) RHS

Question Number 43
Prove that: \(\sin^2\theta-\cos^2\theta\cos2\beta=\sin^2\beta-\cos^2\beta\cos2\theta\)
Solution:
LHS
\(=\sin^2\theta-\cos^2\theta\cos2\beta\)
\(=\sin^2\theta-\cos^2\theta\left(2\cos^2\beta-1\right)\)
\(=\sin^2\theta-2\cos^2\theta\cos^2\beta+\cos^2\theta\)
\(=\sin^2\theta+\cos^2\theta-2\cos^2\theta\cos^2\beta\)
\(=1-2\cos^2\theta\cos^2\beta\)
\(=\sin^2\beta+\cos^2\beta-2\cos^2\theta\cos^2\beta\)
\(=\sin^2\beta-2\cos^2\theta\cos^2\beta+\cos^2\beta\)
\(=\sin^2\beta-2\cos^2\beta\left(2\cos^2\theta-1\right)\)
\(=\sin^2\beta-2\cos^2\beta\cos2\theta\)
\(=\)RHS

Question Number 44
Prove that:\(\tan\theta+2\tan2\theta+4\cot4\theta=\cot\theta\)
Solution:
LHS
\(=\tan\theta+2\tan2\theta+4\cot4\theta\)
\(=\tan\theta+2\tan2\theta+\frac{4}{\tan4\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{4}{\frac{2\tan2\theta}{1-\tan^22\theta}}\)
\(=\tan\theta+2\tan2\theta+\frac{4\left(1-\tan^22\theta\right)}{2\tan2\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{2\left(1-\tan^22\theta\right)}{\tan2\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{2-2\tan^22\theta}{\tan2\theta}\)
\(=\tan\theta+\frac{2\tan^22\theta+2-2\tan^22\theta}{\tan2\theta}\)
\(=\tan\theta+\frac{2}{\tan2\theta}\)
\(=\tan\theta+\frac{2}{\frac{2\tan\theta}{1-\tan^2\theta}}\)
\(=\tan\theta+\frac{2\left(1-\tan^2\theta\right)}{2\tan\theta}\)
\(=\tan\theta+\frac{1-\tan^2\theta}{\tan\theta}\)
\(=\frac{\tan^2\theta+1-\tan^2\theta}{\tan\theta}\)
\(=\frac{1}{\tan\theta}\)
\(=\cot\theta\)
\(=\) RHS

Question Number 45
Prove that:\(\tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta=\cot\theta\)
Solution:
LHS
\(=\tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta\)
\(=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8}{\tan8\theta}\)
\(=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8}{\frac{2\tan4\theta}{1-\tan^24\theta}}\)
\(=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8\left(1-\tan^24\theta\right)}{2\tan4\theta}\)
\(=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{4\left(1-\tan^24\theta\right)}{\tan4\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{4\tan^24\theta+4-4\tan^24\theta}{\tan4\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{4}{\tan4\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{4}{\frac{2\tan2\theta}{1-\tan^22\theta}}\)
\(=\tan\theta+2\tan2\theta+\frac{4\left(1-\tan^22\theta\right)}{2\tan2\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{2\left(1-\tan^22\theta\right)}{\tan2\theta}\)
\(=\tan\theta+2\tan2\theta+\frac{2-2\tan^22\theta}{\tan2\theta}\)
\(=\tan\theta+\frac{2\tan^22\theta+2-2\tan^22\theta}{\tan2\theta}\)
\(=\tan\theta+\frac{2}{\tan2\theta}\)
\(=\tan\theta+\frac{2}{\frac{2\tan\theta}{1-\tan^2\theta}}\)
\(=\tan\theta+\frac{2\left(1-\tan^2\theta\right)}{2\tan\theta}\)
\(=\tan\theta+\frac{1-\tan^2\theta}{\tan\theta}\)
\(=\frac{\tan^2\theta+1-\tan^2\theta}{\tan\theta}\)
\(=\frac{1}{\tan\theta}\)
\(=\cot\theta\)
\(=\)RHS

Question Number 46
Prove that: \(\frac{2\sin\theta}{\cos3\theta}+\frac{2\sin3\theta}{\cos9\theta}+\frac{2\sin9\theta}{\cos27\theta}=\tan27\theta-\tan\theta\)
Solution:
LHS
\(=\frac{2\sin\theta}{\cos3\theta}+\frac{2\sin3\theta}{\cos9\theta}+\frac{2\sin9\theta}{\cos27\theta}\)
\(=\frac{2\sin\theta\cos\theta}{\cos3\theta\cos\theta}+\frac{2\sin3\theta\cos3\theta}{\cos9\theta\cos3\theta}+\frac{2\sin9\theta\cos9\theta}{\cos27\theta\cos9\theta}\)
\(=\frac{\sin2\theta}{\cos3\theta\cos\theta}+\frac{\sin6\theta}{\cos9\theta\cos3\theta}+\frac{\sin18\theta}{\cos27\theta\cos9\theta}\)
\(=\frac{\sin\left(3\theta-\theta\right)}{\cos3\theta\cos\theta}+\frac{\sin\left(9\theta-3\theta\right)}{\cos9\theta\cos3\theta}+\frac{\sin\left(27\theta-9\theta\right)}{\cos27\theta\cos9\theta}\)
\(=\frac{\sin3\theta\cos\theta-\cos3\theta\sin\theta}{\cos3\theta\cos\theta}\)\(+\frac{\sin9\theta\cos3\theta-\cos9\theta\sin3\theta}{\cos9\theta\cos3\theta}\)\(+\frac{\sin27\theta\cos9\theta-\cos27\theta\sin9\theta}{\cos27\theta\cos9\theta}\)
\(=\frac{\sin3\theta\cos\theta}{\cos3\theta\cos\theta}-\frac{\cos3\theta\sin\theta}{\cos3\theta\cos\theta}+\frac{\sin9\theta\cos3\theta}{\cos9\theta\cos3\theta}-\frac{\cos9\theta\sin3\theta}{\cos9\theta\cos3\theta}+\frac{\sin27\theta\cos9\theta}{\cos27\theta\cos9\theta}-\frac{\cos27\theta\sin9\theta}{\cos27\theta\cos9\theta}\)
\(=\frac{\sin3\theta}{\cos3\theta}-\frac{\sin\theta}{\cos\theta}+\frac{\sin9\theta}{\cos9\theta}-\frac{\sin3\theta}{\cos3\theta}+\frac{\sin27\theta}{\cos27\theta}-\frac{\sin9\theta}{\cos9\theta}\)
\(=\frac{\sin27\theta}{\cos27\theta}-\frac{\sin\theta}{\cos\theta}\)
\(=\tan27\theta-\tan\theta\)
\(=\)RHS

Question No 47
\( \text{Prove that }:\cos^2 \theta+\sin^2 \theta.\cos2\beta=cos^2\beta+\sin^2\beta\cos2\theta \)
\( \text{Solution:} \)
\( \begin{align*} \text{LHS }&=\cos^2 \theta+\sin^2 \theta.\cos2\beta\\ &=\cos^2 \theta+\sin^2 \theta.(1-2\sin^2\beta)\\ &=\cos^2 \theta+\sin^2 \theta-2\sin^2\theta\sin^2\beta\\ &=1-2\sin^2\theta\sin^2\beta\\ &=\cos^2 \beta+\sin^2 \beta-2\sin^2\theta\sin^2\beta\\ &=\cos^2 \beta+\sin^2 \beta (1-2\sin^2\theta)\\ &=cos^2\beta+\sin^2\beta\cos2\theta\\ &=RHS \end{align*} \) 

 

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