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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Evaluate the following:
$\tan\Big\{2\tan^{-1}\frac{1}{5}-\frac{\pi}{4}\Big\}$
✓

Answer

$\tan\Big(2\tan^{-1}\frac{1}{5}-\frac{\pi}{4}\Big)=\tan\Big(2\tan^{-1}\frac{1}{5}-\tan^{-1}1\Big)$
$=\tan\begin{bmatrix}\tan^{-1}\begin{Bmatrix}\frac{2\times\frac{1}{5}}{1-\Big(\frac{1}{5}\Big)^2}\end{Bmatrix}-\tan^{-1}1\end{bmatrix}$
$ \Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big\{\frac{2\text{x}}{1-\text{x}^2}\Big\}\Big]$
$=\tan\Bigg[\tan^{-1}\Bigg\{\frac{\frac{2}{5}}{\frac{24}{25}}\Bigg\}-\tan^{-1}1\Bigg]$
$ =\tan\Big[\tan^{-1}\frac{5}{12}+\tan^{-1}1\Big]$
$=\tan^{-1}\Bigg[\tan^{-1}\Bigg(\frac{\frac{15}{12}-1}{1+\frac{5}{12}}\Bigg)\Bigg]$
$\Big[\because\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
$ =\tan\Bigg[\tan^{-1}\Bigg(\frac{\frac{-7}{12}}{\frac{17}{12}}\Bigg)\Bigg]$
$=\tan\Big[\tan^{-1}\frac{-7}{17}\Big]$
$=\frac{-7}{17}$

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