3 Marks Question

Question 13 Marks

Prove that:
$\frac{\sin9\text{A}-\sin\text7{A}}{\cos7\text{A}-\cos9\text{A}}=\cot8\text{A}$

Answer

We have,
$\text{LHS}=\frac{\sin9\text{A}-\sin7\text{A}}{\cos7\text{A}-\cos9\text{A}}$
$=\ \frac{2\sin\Big(\frac{9\text{A}-7\text{A}}{2}\Big)\cos\Big(\frac{9\text{A}+7\text{A}}{2}\Big)}{-2\sin\Big(\frac{7\text{A}+9\text{A}}{2}\Big)\sin\Big(\frac{7\text{A}-9\text{A}}{2}\Big)}$
$=\ \frac{-\sin\text{A}\cos8\text{A}}{\sin8\text{A}\sin(-\text{A})}$
$=\ \frac{-\sin\text{A}\cos8\text{A}}{-\sin\text{A}\times\sin8\text{A}}$ $[\because\ \sin(-\theta)=-\sin\theta]$
$=\ \frac{\cos8\text{A}}{\sin8\text{A}}$
$=\ \cot8\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\sin9\text{A}-\sin7\text{A}}{\cos7\text{A}-\cos9\text{A}}=\cot8\text{A}.$ Hence proved.

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Question 23 Marks

$\text{If}\ \text{cosec}\text{A}+\sec\text{A}=\text{cosecB}+\sec\text{B},$ prove that:
$\tan\text{A}\tan\text{B}=\cot\frac{\text{A+B}}{2}$

Answer

We have,
$\text{cosec}\text{A}+\sec\text{A}=\text{cosec}\text{B}+\sec\text{B}$
$\Rightarrow\ \sec\text{A}-\sec\text{B}=\text{cosec}\text{B}-\text{cosec}\text{A}$
$\Rightarrow\ \frac{1}{\cos\text{A}}-\frac{1}{\cos\text{B}}=\frac{1}{\sin\text{B}}-\frac{1}{\sin\text{A}}$
$\Rightarrow\ \frac{\cos\text{B}-\cos\text{A}}{\cos\text{A}\cos\text{B}}=\frac{\sin\text{A}-\sin\text{B}}{\sin\text{A}\sin\text{B}}$
$\Rightarrow\ \frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}=\frac{\sin\text{A}-\sin\text{B}}{\cos\text{B}-\cos\text{A}}$
$\Rightarrow\ \tan\text{A}\tan\text{B}=\frac{2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)}{-2\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)\sin\Big(\frac{\text{B}+\text{A}}{2}\Big)}$
$\Rightarrow\ \tan\text{A}\tan\text{B}=\frac{-\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)}{-\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)}$ $[\because\ \sin(-\theta)=-\sin\theta]$
$\Rightarrow\ \tan\text{A}\tan\text{B}=\cos\Big(\frac{\text{A+B}}{2}\Big)$ Hence proved.

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Question 33 Marks

Prove that:
$\frac{\sin\text{A}-\sin\text{B}}{\cos\text{A}+\cos\text{B}}=\tan\frac{\text{A}-\text{B}}{2}$

Answer

We have,
$\text{LHS}=\frac{\sin\text{A}-\sin\text{B}}{\cos\text{A}+\cos\text{B}}$
$=\ \frac{2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}{2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}$
$=\ \frac{\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}{\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}$
$= \tan\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\ \text{RHS}$
$\therefore\ \frac{\sin\text{A}-\sin\text{B}}{\cos\text{A}+\cos\text{B}}=\tan\Big(\frac{\text{A}-\text{B}}{2}\Big).$ Hence proved.

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