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INVERSE TRIGNOMETRIC FUNCTIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINVERSE TRIGNOMETRIC FUNCTIONS3 Marks
Question
Find the values:
$\tan\bigg(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\bigg)$
✓

Answer

Putting $\sin^{-1}\frac{3}{5}=\text{x}\ \text{and}\ \cot^{-1}\frac{3}{2}=\text{x}$ so that $\sin \text{x}=\frac{3}{5}\ \text{and}\ \cot \text{y}=\frac{3}{2}$ Now, $\cos \text{x}=\sqrt{1-\sin^2\text{x}}=\sqrt{1-\frac{9}{25}}$$=\sqrt{\frac{16}{25}=\frac{4}{5}} $
And $\tan \text{x}=\frac{\sin \text{x}}{\cos \text{x}}=\frac{3}{4} \ \text{and}\tan \text{y}=\frac{1}{\cot \text{y}}=\frac{2}{3}$ $\therefore \tan\bigg(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\bigg)=\tan\left(\text{x}+\text{y}\right)$ $=\frac{\tan \text{x}\tan \text{y}}{1-\tan \text{x}\tan \text{y}}=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times\frac{2}{3}}=\frac{\frac{17}{12}}{\frac{1}{2}}=\frac{17}{6}$

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