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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions3 Marks
Question
Evaluate:
$\cos\Big(\tan^{-1}\frac{3}{4}\Big)$
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Answer

We have
$\cos\Big(\tan^{-1}\frac{3}{4}\Big)$
$=\cos\Bigg[\frac{1}{2}\cos^{-1}\Bigg(\frac{1-\big(\frac{3}{4}\big)^2}{1+\big(\frac{3}{4}\big)^2}\Bigg)\Bigg]$ $\Big[\because\ 2\tan^{-1}\text{x}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]$
$=\cos\Big[\frac{1}{2}\cos^{-1}\Big(\frac{7}{25}\Big)\Big]$
Let
$\text{y}=\cos^{-1}\Big(\frac{7}{25}\Big)$
$\Rightarrow\cos\text{y}=\frac{7}{25}$
Now,
$=\cos\Big[\frac{1}{2}\cos^{-1}\Big(\frac{7}{25}\Big)\Big]=\cos\Big[\frac{1}{2}\text{y}\Big]$
$=\sqrt{ \frac{\cos\text{y}+1}{2}}$ $\big[\because\ \cos2\text{x}=2\cos^2\text{x}-1\big]$
$=\sqrt{\frac{\frac{7}{25}+1}{2}}$
$=\sqrt{\frac{32}{50}}$
$=\frac{4}{5}$
$\therefore\ \cos\Big[\tan^{-1}\Big(\frac{3}{4}\Big)\Big]=\frac{4}{5}$

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