Ambik Baral
Grade 10
Optional Mathematics
Chapter 7: Trigonometry — Compound Angles
Exercise 7.1
• Detailed Step-by-Step Interactive Solutions
Interactive Controls:
1
Multiple Choice Questions
Tick (\(\checkmark\)) the correct option for the given questions.
A. Which one is equal to \(\sin(A + B)\)?
1 Mark
a
\(\sin A \cos B - \cos A \sin B\)
b
\(\sin A \cos B + \cos A \sin B\)
✓
c
\(\cos A \cos B - \sin A \sin B\)
d
\(\sin A \cos A + \cos B \sin B\)
Answer: (b)
By standard compound angle formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
B. Which is equal to \(\cos(A - B)\)?
1 Mark
a
\(\sin A \cos B - \cos A \sin B\)
b
\(\sin A \cos B + \cos A \sin B\)
c
\(\cos A \cos B + \sin A \sin B\)
✓
d
\(\sin A \cos A + \cos B \sin B\)
Answer: (c)
By standard compound angle formula for cosine difference:
\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]
C. If \(A = 30^\circ\) and \(B = 60^\circ\), what is the value of \(\sin(A + B)\)?
1 Mark
a
\(0\)
b
\(1\)
✓
c
\(\frac{1}{2}\)
d
\(\frac{\sqrt{3}}{2}\)
Answer: (b)
Substitute the angles:
\[ A+B = 30^\circ + 60^\circ = 90^\circ \]
\[ \sin(90^\circ) = 1 \]
D. If \(\sin(A + B) = \frac{1}{2}\) and \(B = 30^\circ\), what is the value of \(A\)?
1 Mark
a
\(0^\circ\)
✓
b
\(30^\circ\)
c
\(45^\circ\)
d
\(60^\circ\)
Answer: (a)
Substitute \(B = 30^\circ\):
\[ \sin(A+30^\circ) = \frac{1}{2} \]
For acute angles:
\[ A+30^\circ = 30^\circ \implies A = 0^\circ \]
E. Which of the following relations is correct?
1 Mark
a
\(\tan(A + B) = \tan A + \tan B\)
b
\(\tan(A + B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\)
c
\(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
✓
d
\(\tan(A + B) = \frac{\tan A + \tan B}{1 + \tan A \tan B}\)
Answer: (c)
Standard compound angle formula for tangent sum:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
2
Trigonometric Value Proofs
Prove without using a calculator or tables.
\(\text{(a)}\quad \cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}\)
Solution:
\begin{align*}
\cos 15^\circ &= \cos(45^\circ - 30^\circ) \\
&= \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \\
&= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \\
&= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \\
&= \frac{\sqrt{3}+1}{2\sqrt{2}} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}\)
Solution:
\begin{align*}
\sin 75^\circ &= \sin(45^\circ + 30^\circ) \\
&= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \\
&= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \\
&= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \\
&= \frac{\sqrt{3}+1}{2\sqrt{2}} \quad \text{Proved.}
\end{align*}
\(\text{(c)}\quad \tan 75^\circ = 2 + \sqrt{3}\)
Solution:
\begin{align*}
\tan 75^\circ &= \tan(45^\circ + 30^\circ) \\
&= \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \\
&= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \\
&= \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} \\
&= \frac{\sqrt{3}+1}{\sqrt{3}-1} \\
&= \frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)} \\
&= \frac{3 + 2\sqrt{3} + 1}{3-1} \\
&= \frac{4 + 2\sqrt{3}}{2} \\
&= 2 + \sqrt{3} \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad \cos 105^\circ = \frac{1 - \sqrt{3}}{2\sqrt{2}}\)
Solution:
\begin{align*}
\cos 105^\circ &= \cos(60^\circ + 45^\circ) \\
&= \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \\
&= \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right) \\
&= \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} \\
&= \frac{1 - \sqrt{3}}{2\sqrt{2}} \quad \text{Proved.}
\end{align*}
\(\text{(e)}\quad \tan 105^\circ = -(2 + \sqrt{3})\)
Solution:
\begin{align*}
\tan 105^\circ &= \tan(60^\circ + 45^\circ) \\
&= \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ} \\
&= \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} \\
&= \frac{\sqrt{3}+1}{1-\sqrt{3}} \\
&= \frac{(\sqrt{3}+1)^2}{(1-\sqrt{3})(1+\sqrt{3})} \\
&= \frac{3 + 2\sqrt{3} + 1}{1-3} \\
&= \frac{4+2\sqrt{3}}{-2} \\
&= -(2+\sqrt{3}) \quad \text{Proved.}
\end{align*}
\(\text{(f)}\quad \cot 15^\circ = 2 + \sqrt{3}\)
Solution:
First, let's find $\tan 15^\circ$:
\begin{align*}
\tan 15^\circ &= \tan(45^\circ - 30^\circ) \\
&= \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \\
&= \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \\
&= \frac{\sqrt{3}-1}{\sqrt{3}+1} \\
&= \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} \\
&= \frac{3 - 2\sqrt{3} + 1}{3-1} \\
&= \frac{4 - 2\sqrt{3}}{2} \\
&= 2 - \sqrt{3}
\end{align*}
Thus,
\[
\cot 15^\circ = \frac{1}{\tan 15^\circ} = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2+\sqrt{3}}{4-3} = 2 + \sqrt{3} \quad \text{Proved.}
\]
3
Compound Trigonometric Evaluations
If \(\sin A = \frac{3}{5}\) and \(\cos B = \frac{12}{13}\), find the values.
Preliminary Operations (Finding reference ratios):
Assuming \(A\) and \(B\) are acute angles (first quadrant): \[ \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \] \[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \]
\(\text{(a)}\quad \sin(A+B)\)
\[
\sin(A+B) = \sin A \cos B + \cos A \sin B = \frac{3}{5} \cdot \frac{12}{13} + \frac{4}{5} \cdot \frac{5}{13} = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}
\]
\(\text{(b)}\quad \cos(A+B)\)
\[
\cos(A+B) = \cos A \cos B - \sin A \sin B = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}
\]
\(\text{(c)}\quad \sin(A-B)\)
\[
\sin(A-B) = \sin A \cos B - \cos A \sin B = \frac{3}{5} \cdot \frac{12}{13} - \frac{4}{5} \cdot \frac{5}{13} = \frac{36}{65} - \frac{20}{65} = \frac{16}{65}
\]
\(\text{(d)}\quad \cos(A-B)\)
\[
\cos(A-B) = \cos A \cos B + \sin A \sin B = \frac{4}{5} \cdot \frac{12}{13} + \frac{3}{5} \cdot \frac{5}{13} = \frac{48}{65} + \frac{15}{65} = \frac{63}{65}
\]
\(\text{(e)}\quad \tan(A+B)\)
\[
\tan(A+B) = \frac{\sin(A+B)}{\cos(A+B)} = \frac{\frac{56}{65}}{\frac{33}{65}} = \frac{56}{33}
\]
\(\text{(f)}\quad \cot(A-B)\)
\[
\cot(A-B) = \frac{\cos(A-B)}{\sin(A-B)} = \frac{\frac{63}{65}}{\frac{16}{65}} = \frac{63}{16}
\]
4
Trigonometric Identity Proofs (Type 1)
Establish the following fundamental compound identities.
\(\text{(a)}\quad \sin(A + B) + \sin(A - B) = 2\sin A \cos B\)
Proof:
\begin{align*}
\text{LHS} &= \sin(A+B) + \sin(A-B) \\
&= (\sin A \cos B + \cos A \sin B) + (\sin A \cos B - \cos A \sin B) \\
&= 2\sin A \cos B \\
&= \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \sin(A + B) - \sin(A - B) = 2\cos A \sin B\)
Proof:
\begin{align*}
\text{LHS} &= \sin(A+B) - \sin(A-B) \\
&= (\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B) \\
&= 2\cos A \sin B \\
&= \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(c)}\quad \cos(A + B) + \cos(A - B) = 2\cos A \cos B\)
Proof:
\begin{align*}
\text{LHS} &= \cos(A+B) + \cos(A-B) \\
&= (\cos A \cos B - \sin A \sin B) + (\cos A \cos B + \sin A \sin B) \\
&= 2\cos A \cos B \\
&= \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad \cos(A + B) - \cos(A - B) = -2\sin A \sin B\)
Proof:
\begin{align*}
\text{LHS} &= \cos(A+B) - \cos(A-B) \\
&= (\cos A \cos B - \sin A \sin B) - (\cos A \cos B + \sin A \sin B) \\
&= -2\sin A \sin B \\
&= \text{RHS} \quad \text{Proved.}
\end{align*}
5
Trigonometric Identity Proofs (Type 2)
Prove without using tables or calculators.
\(\text{(a)}\quad \sin 75^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}\)
Proof:
\begin{align*}
\text{LHS} &= \sin(45^\circ+30^\circ) - \sin(45^\circ-30^\circ) \\
&= (\sin45^\circ\cos30^\circ + \cos45^\circ\sin30^\circ) - (\sin45^\circ\cos30^\circ - \cos45^\circ\sin30^\circ) \\
&= 2\cos45^\circ\sin30^\circ \\
&= 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \\
&= \frac{1}{\sqrt{2}} = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \cos 15^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}\) (Correction Applied)
Proof:
Note: The original expression was \(\cos15^\circ - \sin75^\circ\), which would equal 0. The corrected identity \(\cos 15^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}\) is evaluated here.
\begin{align*} \text{LHS} &= \cos(45^\circ-30^\circ) - \sin(45^\circ-30^\circ) \\ &= (\cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ) - (\sin45^\circ\cos30^\circ - \cos45^\circ\sin30^\circ) \\ &= \left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right) - \left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right) \\ &= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \\ &= \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \text{RHS} \quad \text{Proved.} \end{align*}
\(\text{(c)}\quad \sin 105^\circ - \cos 75^\circ = \frac{1}{\sqrt{2}}\)
Proof:
\begin{align*}
\text{LHS} &= \sin(60^\circ+45^\circ) - \cos(45^\circ+30^\circ) \\
&= (\sin60^\circ\cos45^\circ + \cos60^\circ\sin45^\circ) - (\cos45^\circ\cos30^\circ - \sin45^\circ\sin30^\circ) \\
&= \left(\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}} + \frac{1}{2}\cdot\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right) \\
&= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \\
&= \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad \cos 105^\circ + \cos 15^\circ = \frac{1}{\sqrt{2}}\)
Proof:
\begin{align*}
\text{LHS} &= \cos(60^\circ+45^\circ) + \cos(45^\circ-30^\circ) \\
&= (\cos60^\circ\cos45^\circ - \sin60^\circ\sin45^\circ) + (\cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ) \\
&= \left(\frac{1}{2}\cdot\frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right) \\
&= \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \\
&= \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(e)}\quad \tan 15^\circ + \cot 15^\circ = 4\)
Proof:
\begin{align*}
\text{LHS} &= \tan(45^\circ-30^\circ) + \frac{1}{\tan(45^\circ-30^\circ)} \\
&= \frac{\tan45^\circ - \tan30^\circ}{1 + \tan45^\circ\tan30^\circ} + \frac{1 + \tan45^\circ\tan30^\circ}{\tan45^\circ - \tan30^\circ} \\
&= \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} + \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \\
&= \frac{\sqrt{3}-1}{\sqrt{3}+1} + \frac{\sqrt{3}+1}{\sqrt{3}-1} \\
&= \frac{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} \\
&= \frac{(3-2\sqrt{3}+1)+(3+2\sqrt{3}+1)}{3-1} \\
&= \frac{8}{2} = 4 = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(f)}\quad \cos 18^\circ - \sin 18^\circ = \sqrt{2} \sin 27^\circ\)
Proof (Method 1 - LHS to RHS):
\begin{align*}
\text{LHS} &= \cos 18^\circ - \sin 18^\circ \\
&= \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 18^\circ - \frac{1}{\sqrt{2}} \sin 18^\circ \right) \\
&= \sqrt{2} \left( \sin 45^\circ \cos 18^\circ - \cos 45^\circ \sin 18^\circ \right) \\
&= \sqrt{2} \sin(45^\circ - 18^\circ) \\
&= \sqrt{2} \sin 27^\circ = \text{RHS} \quad \text{Proved.}
\end{align*}
Proof (Method 2 - RHS to LHS):
\begin{align*}
\text{RHS} &= \sqrt{2}\sin 27^\circ \\
&= \sqrt{2} \sin(45^\circ-18^\circ) \\
&= \sqrt{2} (\sin45^\circ\cos18^\circ - \cos45^\circ\sin18^\circ) \\
&= \sqrt{2} \left(\frac{1}{\sqrt{2}}\cos18^\circ - \frac{1}{\sqrt{2}}\sin18^\circ\right) \\
&= \cos18^\circ - \sin18^\circ = \text{LHS} \quad \text{Proved.}
\end{align*}
\(\text{(g)}\quad \sin 35^\circ + \cos 35^\circ = \sqrt{2} \cos 10^\circ\)
Proof:
\begin{align*}
\text{LHS} &= \sin(45^\circ-10^\circ) + \cos(45^\circ-10^\circ) \\
&= (\sin45^\circ\cos10^\circ - \cos45^\circ\sin10^\circ) + (\cos45^\circ\cos10^\circ + \sin45^\circ\sin10^\circ) \\
&= \frac{1}{\sqrt{2}}\cos10^\circ - \frac{1}{\sqrt{2}}\sin10^\circ + \frac{1}{\sqrt{2}}\cos10^\circ + \frac{1}{\sqrt{2}}\sin10^\circ \\
&= \frac{2}{\sqrt{2}}\cos10^\circ \\
&= \sqrt{2}\cos10^\circ = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(h)}\quad \sqrt{3} \cos 20^\circ + \sin 20^\circ = 2 \sin 80^\circ\)
Proof:
\begin{align*}
\text{RHS} &= 2 \sin(60^\circ+20^\circ) \\
&= 2 (\sin60^\circ\cos20^\circ + \cos60^\circ\sin20^\circ) \\
&= 2 \left(\frac{\sqrt{3}}{2}\cos20^\circ + \frac{1}{2}\sin20^\circ\right) \\
&= \sqrt{3}\cos20^\circ + \sin20^\circ = \text{LHS} \quad \text{Proved.}
\end{align*}
6
Variable-Based Trigonometric Proofs
Verify general compound relations using arbitrary angles.
\(\text{(a)}\quad \sin(25^\circ + \alpha) \cos(65^\circ - \alpha) + \cos(25^\circ + \alpha) \sin(65^\circ - \alpha) = 1\)
Proof:
Let \(A = 25^\circ + \alpha\) and \(B = 65^\circ - \alpha\).
The equation matches the expansion of \(\sin(A+B)\):
\begin{align*}
\text{LHS} &= \sin(25^\circ + \alpha) \cos(65^\circ - \alpha) + \cos(25^\circ + \alpha) \sin(65^\circ - \alpha) \\
&= \sin\bigl((25^\circ + \alpha) + (65^\circ - \alpha)\bigr) \\
&= \sin(25^\circ + 65^\circ) \\
&= \sin(90^\circ) \\
&= 1 = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \cos(25^\circ + \alpha) \cos(65^\circ + \alpha) - \sin(25^\circ + \alpha) \sin(65^\circ + \alpha) = -\sin 2\alpha\)
Proof:
Let \(A = 25^\circ + \alpha\) and \(B = 65^\circ + \alpha\).
This expression matches the expansion of \(\cos(A+B)\):
\begin{align*}
\text{LHS} &= \cos(25^\circ + \alpha) \cos(65^\circ + \alpha) - \sin(25^\circ + \alpha) \sin(65^\circ + \alpha) \\
&= \cos\bigl((25^\circ + \alpha) + (65^\circ + \alpha)\bigr) \\
&= \cos(90^\circ + 2\alpha) \\
&= -\sin 2\alpha = \text{RHS} \quad \text{Proved.} \quad (\because \cos(90^\circ + \theta) = -\sin \theta)
\end{align*}
\(\text{(c)}\quad \cos(60^\circ + \beta) - \cos \beta + \cos(60^\circ - \beta) = 0\)
Proof:
\begin{align*}
\text{LHS} &= \cos(60^\circ + \beta) - \cos \beta + \cos(60^\circ - \beta) \\
&= \left(\cos 60^\circ \cos \beta - \sin 60^\circ \sin \beta\right) - \cos \beta + \left(\cos 60^\circ \cos \beta + \sin 60^\circ \sin \beta\right) \\
&= 2\cos 60^\circ \cos \beta - \cos \beta \\
&= 2 \cdot \frac{1}{2}\cos \beta - \cos \beta \\
&= \cos \beta - \cos \beta \\
&= 0 = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad \frac{\tan(45^\circ + \phi) - \tan \phi}{1 + \tan(45^\circ + \phi) \tan \phi} = 1\)
Proof:
Let \(A = 45^\circ + \phi\) and \(B = \phi\).
Using the tangent difference identity \(\frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan(A-B)\):
\begin{align*}
\text{LHS} &= \tan\bigl((45^\circ + \phi) - \phi\bigr) \\
&= \tan 45^\circ \\
&= 1 = \text{RHS} \quad \text{Proved.}
\end{align*}
7
Proofs Given \(A + B = \left(\frac{\pi}{4}\right)^c\)
If the sum of two angles is \(\frac{\pi}{4}\) radian (\(45^\circ\)), prove the given equations.
\(\text{(a)}\quad \tan A + \tan B + \tan A \tan B = 1\)
Proof:
Given: \(A+B = \frac{\pi}{4}\)
Taking \(\tan\) on both sides:
\begin{align*}
&\tan(A+B) = \tan \frac{\pi}{4} \\
\implies &\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1 \\
\implies &\tan A + \tan B = 1 - \tan A \tan B \\
\implies &\tan A + \tan B + \tan A \tan B = 1 \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad (1+\tan A)(1+\tan B)=2\)
Proof (Method 1 - Adding 1 to both sides):
From the proven identity in part (a):
\begin{align*}
&\tan A + \tan B + \tan A \tan B = 1 \\
\text{Add 1 on both sides:} & \\
&1 + \tan A + \tan B + \tan A \tan B = 1 + 1 \\
\implies &(1 + \tan A) + \tan B(1 + \tan A) = 2 \\
\implies &(1 + \tan A)(1 + \tan B) = 2 \quad \text{Proved.}
\end{align*}
Proof (Method 2 - Direct Expansion):
\begin{align*}
\text{LHS} &= (1+\tan A)(1+\tan B) \\
&= 1 + \tan A + \tan B + \tan A \tan B \\
&= 1 + (\tan A + \tan B + \tan A \tan B) \\
&= 1 + 1 \quad [\text{using the identity from part (a)}] \\
&= 2 = \text{RHS} \quad \text{Proved.}
\end{align*}
8
Proofs Involving Tangent Conditions
Calculate composite angles using specified trigonometric values.
\(\text{(a)}\quad \text{If } \tan X = \frac{2}{3} \text{ and } \tan Y = \frac{1}{5}, \text{ prove } X + Y = 45^\circ\)
Proof:
\begin{align*}
\tan(X+Y) &= \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \\
&= \frac{\frac{2}{3} + \frac{1}{5}}{1 - \frac{2}{3}\cdot\frac{1}{5}} \\
&= \frac{\frac{10+3}{15}}{1 - \frac{2}{15}} \\
&= \frac{\frac{13}{15}}{\frac{13}{15}} = 1 \\
\implies \tan(X+Y) &= \tan 45^\circ \\
\implies X+Y &= 45^\circ \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \text{If } \tan A = m \text{ and } \tan B = \frac{1}{m}, \text{ prove } A + B = \left(\frac{\pi}{2}\right)^c\)
Proof:
\begin{align*}
\tan(A+B) &= \frac{\tan A + \tan B}{1 - \tan A \tan B} \\
&= \frac{m + \frac{1}{m}}{1 - m\cdot\frac{1}{m}} \\
&= \frac{m + \frac{1}{m}}{1 - 1} \\
&= \frac{m + \frac{1}{m}}{0} = \infty \quad (\text{Undefined}) \\
\implies \tan(A+B) &= \tan 90^\circ \\
\implies A+B &= 90^\circ = \frac{\pi^c}{2} \quad \text{Proved.}
\end{align*}
9
Rational Trigonometric Simplifications
Transform sum/difference trigonometric fractions into standard tangents.
\(\text{(a)}\quad \frac{\cos 35^\circ - \sin 35^\circ}{\cos 35^\circ + \sin 35^\circ} = \tan 10^\circ\)
Proof:
Divide both numerator and denominator by $\cos 35^\circ$:
\begin{align*}
\text{LHS} &= \frac{\frac{\cos 35^\circ}{\cos 35^\circ} - \frac{\sin 35^\circ}{\cos 35^\circ}}{\frac{\cos 35^\circ}{\cos 35^\circ} + \frac{\sin 35^\circ}{\cos 35^\circ}} \\
&= \frac{1 - \tan 35^\circ}{1 + \tan 35^\circ} \\
&= \frac{\tan 45^\circ - \tan 35^\circ}{1 + \tan 45^\circ \tan 35^\circ} \quad (\because \tan 45^\circ = 1) \\
&= \tan(45^\circ - 35^\circ) \\
&= \tan 10^\circ = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ} = \tan 62^\circ\)
Proof:
Divide both numerator and denominator by $\cos 17^\circ$:
\begin{align*}
\text{LHS} &= \frac{1 + \tan 17^\circ}{1 - \tan 17^\circ} \\
&= \frac{\tan 45^\circ + \tan 17^\circ}{1 - \tan 45^\circ \tan 17^\circ} \quad (\because \tan 45^\circ = 1) \\
&= \tan(45^\circ + 17^\circ) \\
&= \tan 62^\circ = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(c)}\quad \frac{\cos 8^\circ + \sin 8^\circ}{\cos 8^\circ - \sin 8^\circ} = \tan 53^\circ\)
Proof:
Divide both numerator and denominator by $\cos 8^\circ$:
\begin{align*}
\text{LHS} &= \frac{1 + \tan 8^\circ}{1 - \tan 8^\circ} \\
&= \frac{\tan 45^\circ + \tan 8^\circ}{1 - \tan 45^\circ \tan 8^\circ} \quad (\because \tan 45^\circ = 1) \\
&= \tan(45^\circ + 8^\circ) \\
&= \tan 53^\circ = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad \frac{\cos 10^\circ - \sin 10^\circ}{\cos 10^\circ + \sin 10^\circ} = \tan 35^\circ\)
Proof:
Divide both numerator and denominator by $\cos 10^\circ$:
\begin{align*}
\text{LHS} &= \frac{1 - \tan 10^\circ}{1 + \tan 10^\circ} \\
&= \frac{\tan 45^\circ - \tan 10^\circ}{1 + \tan 45^\circ \tan 10^\circ} \\
&= \tan(45^\circ - 10^\circ) \\
&= \tan 35^\circ = \text{RHS} \quad \text{Proved.}
\end{align*}
10
Proofs of Compound Tan/Cot Identities
Solve algebraic proof patterns containing multiple complex angles.
\(\text{(a)}\quad \frac{\tan 5A - \tan 4A}{1 + \tan 5A \tan 4A} = \tan A\)
Proof:
Using standard tangent difference formula:
\[ \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi} = \tan(\theta - \phi) \]
Substituting \(\theta = 5A\) and \(\phi = 4A\):
\begin{align*}
\text{LHS} &= \tan(5A - 4A) \\
&= \tan A = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \frac{\tan 7A - \tan 4A}{1 + \tan 7A \tan 4A} = \tan 3A\)
Proof:
Substituting \(\theta = 7A\) and \(\phi = 4A\) in tangent difference identity:
\begin{align*}
\text{LHS} &= \tan(7A - 4A) \\
&= \tan 3A = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(c)}\quad \frac{\tan (A + B) - \tan B}{1 + \tan (A + B) \tan B} = \tan A\)
Proof:
Let \(\theta = A + B\) and \(\phi = B\):
\begin{align*}
\text{LHS} &= \tan\bigl((A+B) - B\bigr) \\
&= \tan A = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad \tan (A + B) \tan (A - B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\)
Proof:
\begin{align*}
\text{LHS} &= \tan(A+B) \cdot \tan(A-B) \\
&= \left(\frac{\tan A + \tan B}{1 - \tan A \tan B}\right) \left(\frac{\tan A - \tan B}{1 + \tan A \tan B}\right) \\
&= \frac{(\tan A + \tan B)(\tan A - \tan B)}{(1 - \tan A \tan B)(1 + \tan A \tan B)} \\
&= \frac{\tan^2 A - \tan^2 B}{1 - (\tan A \tan B)^2} \\
&= \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B} = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(e)}\quad \frac{\sin (A + B)}{\sin A \sin B} = \cot B + \cot A\)
Proof:
\begin{align*}
\text{LHS} &= \frac{\sin(A+B)}{\sin A \sin B} \\
&= \frac{\sin A \cos B + \cos A \sin B}{\sin A \sin B} \\
&= \frac{\sin A \cos B}{\sin A \sin B} + \frac{\cos A \sin B}{\sin A \sin B} \\
&= \frac{\cos B}{\sin B} + \frac{\cos A}{\sin A} \\
&= \cot B + \cot A = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(f)}\quad \frac{\cos (A + B)}{\sin A \sin B} = \cot A \cot B - 1\)
Proof:
\begin{align*}
\text{LHS} &= \frac{\cos(A+B)}{\sin A \sin B} \\
&= \frac{\cos A \cos B - \sin A \sin B}{\sin A \sin B} \\
&= \frac{\cos A \cos B}{\sin A \sin B} - \frac{\sin A \sin B}{\sin A \sin B} \\
&= \cot A \cot B - 1 = \text{RHS} \quad \text{Proved.}
\end{align*}
11
Proofs of Numerical Angle Relations
Trigonometric proofs based on numeric sums totaling $45^\circ$.
\(\text{(a)}\quad \tan 25^\circ + \tan 20^\circ + \tan 25^\circ \tan 20^\circ = 1\)
Proof:
\begin{align*}
&\text{We know, } 25^\circ + 20^\circ = 45^\circ \\
\text{Taking } \tan \text{ on both sides:} & \\
&\tan(25^\circ + 20^\circ) = \tan 45^\circ \\
\implies &\frac{\tan 25^\circ + \tan 20^\circ}{1 - \tan 25^\circ \tan 20^\circ} = 1 \\
\implies &\tan 25^\circ + \tan 20^\circ = 1 - \tan 25^\circ \tan 20^\circ \\
\implies &\tan 25^\circ + \tan 20^\circ + \tan 25^\circ \tan 20^\circ = 1 \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad 1 - \tan 27^\circ - \tan 18^\circ - \tan 27^\circ \tan 18^\circ = 0\)
Proof:
\begin{align*}
&\text{We know, } 27^\circ + 18^\circ = 45^\circ \\
\text{Taking } \tan \text{ on both sides:} & \\
&\tan(27^\circ + 18^\circ) = \tan 45^\circ \\
\implies &\frac{\tan 27^\circ + \tan 18^\circ}{1 - \tan 27^\circ \tan 18^\circ} = 1 \\
\implies &\tan 27^\circ + \tan 18^\circ = 1 - \tan 27^\circ \tan 18^\circ \\
\implies &1 - \tan 27^\circ - \tan 18^\circ - \tan 27^\circ \tan 18^\circ = 0 \quad \text{Proved.}
\end{align*}
\(\text{(c)}\quad \tan 65^\circ - \tan 20^\circ = 1 + \tan 65^\circ \tan 20^\circ\)
Proof:
\begin{align*}
&\text{We know, } 65^\circ - 20^\circ = 45^\circ \\
\text{Taking } \tan \text{ on both sides:} & \\
&\tan(65^\circ - 20^\circ) = \tan 45^\circ \\
\implies &\frac{\tan 65^\circ - \tan 20^\circ}{1 + \tan 65^\circ \tan 20^\circ} = 1 \\
\implies &\tan 65^\circ - \tan 20^\circ = 1 + \tan 65^\circ \tan 20^\circ \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad 1 + \cot 18^\circ + \cot 27^\circ = \cot 18^\circ \cot 27^\circ\)
Proof:
\begin{align*}
&\text{We know, } 18^\circ + 27^\circ = 45^\circ \\
\text{Taking } \cot \text{ on both sides:} & \\
&\cot(18^\circ + 27^\circ) = \cot 45^\circ \\
\implies &\frac{\cot 18^\circ \cot 27^\circ - 1}{\cot 18^\circ + \cot 27^\circ} = 1 \\
\implies &\cot 18^\circ \cot 27^\circ - 1 = \cot 18^\circ + \cot 27^\circ \\
\implies &1 + \cot 18^\circ + \cot 27^\circ = \cot 18^\circ \cot 27^\circ \quad \text{Proved.}
\end{align*}
12
Advanced Multi-Angle Proofs
Develop advanced proof equations containing 3 separate angles or variable parameters.
\(\text{(a)}\quad \tan 50^\circ + \tan 60^\circ + \tan 70^\circ = \tan 50^\circ \tan 60^\circ \tan 70^\circ\)
Proof:
\begin{align*}
&\text{We know, } 50^\circ + 60^\circ + 70^\circ = 180^\circ \\
\implies &50^\circ + 60^\circ = 180^\circ - 70^\circ \\
\text{Taking } \tan \text{ on both sides:} & \\
&\tan(50^\circ + 60^\circ) = \tan(180^\circ - 70^\circ) \\
\implies &\frac{\tan 50^\circ + \tan 60^\circ}{1 - \tan 50^\circ \tan 60^\circ} = -\tan 70^\circ \quad (\because \tan(180^\circ - \theta) = -\tan \theta) \\
\implies &\tan 50^\circ + \tan 60^\circ = -\tan 70^\circ(1 - \tan 50^\circ \tan 60^\circ) \\
\implies &\tan 50^\circ + \tan 60^\circ = -\tan 70^\circ + \tan 50^\circ \tan 60^\circ \tan 70^\circ \\
\implies &\tan 50^\circ + \tan 60^\circ + \tan 70^\circ = \tan 50^\circ \tan 60^\circ \tan 70^\circ \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \tan 10A - \tan 6A - \tan 4A = \tan 10A \tan 6A \tan 4A\)
Proof:
\begin{align*}
&\text{We know, } 10A - 6A = 4A \\
\text{Taking } \tan \text{ on both sides:} & \\
&\tan(10A - 6A) = \tan 4A \\
\implies &\frac{\tan 10A - \tan 6A}{1 + \tan 10A \tan 6A} = \tan 4A \\
\implies &\tan 10A - \tan 6A = \tan 4A(1 + \tan 10A \tan 6A) \\
\implies &\tan 10A - \tan 6A = \tan 4A + \tan 10A \tan 6A \tan 4A \\
\implies &\tan 10A - \tan 6A - \tan 4A = \tan 10A \tan 6A \tan 4A \quad \text{Proved.}
\end{align*}
\(\text{(c)}\quad \tan 20^\circ \tan 30^\circ + \tan 30^\circ \tan 40^\circ + \tan 40^\circ \tan 20^\circ = 1\)
Proof:
\begin{align*}
&\text{We know, } 20^\circ + 30^\circ + 40^\circ = 90^\circ \\
\implies &20^\circ + 30^\circ = 90^\circ - 40^\circ \\
\text{Taking } \tan \text{ on both sides:} & \\
&\tan(20^\circ + 30^\circ) = \tan(90^\circ - 40^\circ) \\
\implies &\frac{\tan 20^\circ + \tan 30^\circ}{1 - \tan 20^\circ \tan 30^\circ} = \cot 40^\circ \\
\implies &\frac{\tan 20^\circ + \tan 30^\circ}{1 - \tan 20^\circ \tan 30^\circ} = \frac{1}{\tan 40^\circ} \\
\implies &\tan 40^\circ(\tan 20^\circ + \tan 30^\circ) = 1 - \tan 20^\circ \tan 30^\circ \\
\implies &\tan 20^\circ \tan 40^\circ + \tan 30^\circ \tan 40^\circ = 1 - \tan 20^\circ \tan 30^\circ \\
\implies &\tan 20^\circ \tan 30^\circ + \tan 30^\circ \tan 40^\circ + \tan 40^\circ \tan 20^\circ = 1 \quad \text{Proved.}
\end{align*}
\(\text{(d)}\quad \cot 5A \cot 4A - \cot 9A \cot 5A - \cot 9A \cot 4A = 1\)
Proof:
\begin{align*}
&\text{We know, } 5A + 4A = 9A \\
\text{Taking } \cot \text{ on both sides:} & \\
&\cot(5A + 4A) = \cot 9A \\
\implies &\frac{\cot 5A \cot 4A - 1}{\cot 5A + \cot 4A} = \cot 9A \\
\implies &\cot 5A \cot 4A - 1 = \cot 9A(\cot 5A + \cot 4A) \\
\implies &\cot 5A \cot 4A - 1 = \cot 9A \cot 5A + \cot 9A \cot 4A \\
\implies &\cot 5A \cot 4A - \cot 9A \cot 5A - \cot 9A \cot 4A = 1 \quad \text{Proved.}
\end{align*}
13
Proofs of Specific Tangent Equations
Proof analysis for compound and complementary numeric angle expansions.
\(\text{(a)}\quad \tan 50^\circ - \tan 40^\circ = 2 \tan 10^\circ\)
Proof:
\begin{align*}
&\text{We know, } 50^\circ - 40^\circ = 10^\circ \\
\text{Taking } \tan \text{ on both sides:} & \\
&\tan(50^\circ - 40^\circ) = \tan 10^\circ \\
\implies &\frac{\tan 50^\circ - \tan 40^\circ}{1 + \tan 50^\circ \tan 40^\circ} = \tan 10^\circ \\
\implies &\frac{\tan 50^\circ - \tan 40^\circ}{1 + \tan 50^\circ \tan(90^\circ - 50^\circ)} = \tan 10^\circ \\
\implies &\frac{\tan 50^\circ - \tan 40^\circ}{1 + \tan 50^\circ \cot 50^\circ} = \tan 10^\circ \\
\implies &\frac{\tan 50^\circ - \tan 40^\circ}{1 + 1} = \tan 10^\circ \\
\implies &\frac{\tan 50^\circ - \tan 40^\circ}{2} = \tan 10^\circ \\
\implies &\tan 50^\circ - \tan 40^\circ = 2 \tan 10^\circ \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \text{Prove whether: } \tan 50^\circ = 2 \tan 30^\circ + \tan 20^\circ\) (Incorrect Identity Analyzed)
Evaluation & Correct Identity:
Correct Standard Identity associated with this layout: \[ \tan 50^\circ = \tan 40^\circ + 2\tan 10^\circ \] Which is derived straight from part (a): \[ \tan 50^\circ - \tan 40^\circ = 2\tan 10^\circ \implies \tan 50^\circ = \tan 40^\circ + 2\tan 10^\circ \quad \text{Proved.} \]
Analysis: The identity as originally printed (\(\tan 50^\circ = 2\tan 30^\circ + \tan 20^\circ\)) is numerically incorrect.
Let's test numerically: \[ \tan 50^\circ \approx 1.1918 \] \[ 2\tan 30^\circ + \tan 20^\circ \approx 2(0.5774) + 0.3640 \approx 1.1548 + 0.3640 \approx 1.5188 \] Since \(1.1918 \ne 1.5188\), the identity is invalid.Correct Standard Identity associated with this layout: \[ \tan 50^\circ = \tan 40^\circ + 2\tan 10^\circ \] Which is derived straight from part (a): \[ \tan 50^\circ - \tan 40^\circ = 2\tan 10^\circ \implies \tan 50^\circ = \tan 40^\circ + 2\tan 10^\circ \quad \text{Proved.} \]
14
Identity Proofs using Compound Formulas
Convert complex fraction models to single trigonometric parameters.
\(\text{(a)}\quad \frac{\sin(\beta+\gamma)}{\sin(\beta-\gamma)}=\frac{\tan\beta+\tan\gamma}{\tan\beta-\tan\gamma}\)
Proof:
\begin{align*}
\text{LHS} &= \frac{\sin(\beta+\gamma)}{\sin(\beta-\gamma)} \\
&= \frac{\sin\beta\cos\gamma+\cos\beta\sin\gamma}{\sin\beta\cos\gamma-\cos\beta\sin\gamma} \\
\text{Divide both } &\text{numerator and denominator by } \cos\beta\cos\gamma: \\
&= \frac{\frac{\sin\beta\cos\gamma}{\cos\beta\cos\gamma} + \frac{\cos\beta\sin\gamma}{\cos\beta\cos\gamma}}{\frac{\sin\beta\cos\gamma}{\cos\beta\cos\gamma} - \frac{\cos\beta\sin\gamma}{\cos\beta\cos\gamma}} \\
&= \frac{\tan\beta+\tan\gamma}{\tan\beta-\tan\gamma} = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \frac{\sin(A+B)\sin(A-B)}{\cos^{2}A\cos^{2}B}=\tan^{2}A-\tan^{2}B\)
Proof:
\begin{align*}
\text{LHS} &= \frac{\sin(A+B)\sin(A-B)}{\cos^{2}A\cos^{2}B} \\
&= \frac{(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)}{\cos^{2}A\cos^{2}B} \\
&= \frac{\sin^{2}A\cos^{2}B-\cos^{2}A\sin^{2}B}{\cos^{2}A\cos^{2}B} \\
&= \frac{\sin^{2}A\cos^{2}B}{\cos^{2}A\cos^{2}B} - \frac{\cos^{2}A\sin^{2}B}{\cos^{2}A\cos^{2}B} \\
&= \frac{\sin^{2}A}{\cos^{2}A} - \frac{\sin^{2}B}{\cos^{2}B} \\
&= \tan^{2}A-\tan^{2}B = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(c)}\quad \frac{\sin(A+B)+\sin(A-B)}{\cos(A+B)+\cos(A-B)}=\tan A\)
Proof:
Using expansions of \(\sin(A \pm B)\) and \(\cos(A \pm B)\):
\begin{align*}
\text{LHS} &= \frac{(\sin A\cos B+\cos A\sin B) + (\sin A\cos B-\cos A\sin B)}{(\cos A\cos B-\sin A\sin B) + (\cos A\cos B+\sin A\sin B)} \\
&= \frac{2\sin A\cos B}{2\cos A\cos B} \\
&= \frac{\sin A}{\cos A} \\
&= \tan A = \text{RHS} \quad \text{Proved.}
\end{align*}
15
Fractional Sum Identities
Advanced multi-term algebraic proofs utilizing cyclic angle values.
\(\text{(a)}\quad \frac{\sin(A-B)}{\cos A\cos B}+\frac{\sin(B-C)}{\cos B\cos C}+\frac{\sin(C-A)}{\cos C\cos A}=0\)
Proof:
Let us simplify each term individually:
\begin{align*}
\text{First Term: } \frac{\sin(A-B)}{\cos A\cos B} &= \frac{\sin A\cos B - \cos A\sin B}{\cos A\cos B} \\
&= \frac{\sin A\cos B}{\cos A\cos B} - \frac{\cos A\sin B}{\cos A\cos B} \\
&= \tan A - \tan B
\end{align*}
Similarly:
\[
\text{Second Term: } \frac{\sin(B-C)}{\cos B\cos C} = \tan B - \tan C
\]
\[
\text{Third Term: } \frac{\sin(C-A)}{\cos C\cos A} = \tan C - \tan A
\]
Adding all three terms:
\begin{align*}
\text{LHS} &= (\tan A-\tan B) + (\tan B-\tan C) + (\tan C-\tan A) \\
&= \tan A - \tan A - \tan B + \tan B - \tan C + \tan C \\
&= 0 = \text{RHS} \quad \text{Proved.}
\end{align*}
\(\text{(b)}\quad \sin(A-B)\cos B + \cos(A-B)\sin B = \sin A\) (Truncated LaTeX completed gracefully)
Proof:
Let \(\theta = A-B\) and \(\phi = B\).
The expression matches the standard expansion of \(\sin(\theta + \phi)\):
\begin{align*}
\text{LHS} &= \sin(A-B)\cos B + \cos(A-B)\sin B \\
&= \sin\bigl((A-B) + B\bigr) \\
&= \sin A = \text{RHS} \quad \text{Proved.}
\end{align*}
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