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INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS3 Marks
Question
Prove that $\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=7$
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Answer

We have, $\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=7$
$\Rightarrow\ \Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=\cot^{-1}7$
$\Rightarrow\ (2\cot^{-1}3)=\frac{\pi}{4}-\cot^{-1}7$
$\Rightarrow\ 2\tan^{-1}\frac{1}{3}=\frac{\pi}{4}-\tan^{-1}\frac{1}{7}$
$\Big[\because\ \cot^{-1}\text{x}=\tan^{-1}\frac{1}{\text{x}}\Big]$
$\Rightarrow\ 2\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}=\frac{\pi}{4}$
$\Rightarrow\ \tan^{-1}\frac{\frac{2}{3}}{1-\big(\frac{1}{3}\big)^2}+\tan^{-1}\frac{1}{7}=\frac{\pi}{4}$
$\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}\Big]$
$\Rightarrow\ \tan^{-1}\frac{3}{4}+\tan^{-1}\frac{1}{7}=\frac{\pi}{4}$
$\Rightarrow\ \tan^{-1}\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}.{\frac{1}{7}}}=\frac{\pi}{4}$
$\Big[\because\ \tan^{-1}\frac{\text{A}+\text{B}}{1-\text{AB}}\Big]$
$\Rightarrow\ \tan^{-1}\frac{25}{25}=\frac{\pi}{4}$
$\Rightarrow\ 1=\tan\frac{\pi}{4}$
$\Rightarrow\ 1=1$
$\Rightarrow\ \text{LHS}=\text{RHS}$

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