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INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS3 Marks
Question
Show that $\cos\Big(2\tan^{-1}\frac{1}{7}\Big)=\sin\Big(4\tan^{-1}\frac{1}{3}\Big).$
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Answer

We have, $\cos\Big(2\tan^{-1}\frac{1}{7}\Big)=\sin\Big(4\tan^{-1}\frac{1}{3}\Big)$
$\Rightarrow\ \cos\begin{bmatrix}\cos^{-1}\begin{pmatrix}\frac{1-\big(\frac{1}{7}\big)^2}{1+\big(\frac{1}{7}\big)^2} \end{pmatrix}\end{bmatrix}=\sin\Big[2.2\tan^{-1}\frac{1}{3}\Big]$
$\Big[\because\ 2\tan^{-1}\text{x}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]$
$\Rightarrow\ \cos\begin{bmatrix}\cos^{-1}\begin{pmatrix}\frac{\frac{48}{49}}{\frac{50}{49}} \end{pmatrix}\end{bmatrix}=\sin\begin{bmatrix}2.\begin{pmatrix}\tan^{-1}\frac{\frac{2}{3}}{1-\big(\frac{1}{3}\big)^2} \end{pmatrix}\end{bmatrix}$
$\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\Rightarrow\ \cos\Big[\cos^{-1}\Big(\frac{24}{25}\Big)\Big]=\sin\Big(2\tan^{-1}\frac{3}{4}\Big)$
$\Rightarrow\ \cos\Big[\cos^{-1}\Big(\frac{24}{25}\Big)\Big]=\sin\Bigg(\sin^{-1}\frac{2\times\frac{3}{4}}{1+\frac{9}{16}}\Bigg)$
$\Big[\because\ 2\tan^{-1}\text{x}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big]$
$\Rightarrow\ \frac{24}{25}=\sin\Bigg(\sin^{-1}\frac{\frac{3}{2}}{\frac{25}{16}}\Bigg)$
$\Rightarrow\ \frac{24}{25}=\frac{48}{50}$
$\Rightarrow\ \frac{24}{25}=\frac{24}{25}$
$\therefore\ \text{LHS}=\text{RHS}$

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