ISI2018-DCG-20

Question

The value of $\tan \left(\sin^{-1}\left(\frac{3}{5}\right)+\cot^{-1}\left(\frac{3}{2}\right)\right)$ is

• A. $\frac{1}{18}$
• B. $\frac{11}{6}$
• C. $\frac{13}{6}$
• D. $\frac{17}{6}$

Answer 1

$\tan \bigg(\sin^{-1}\big(\frac{3}{5}\big)+\cot^{-1}\big(\frac{3}{2}\big)\bigg)$

Let $\alpha = \sin^{-1}\big(\frac{3}{5}\big)\implies \sin\alpha = \frac{3}{5}$

From Pythagorean theorem, $\cos \alpha = \frac{4}{5}$

$\therefore \tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

and $\beta = \cot^{-1}\big(\frac{3}{2}\big)\implies \cot\beta = \frac{3}{2}\implies \tan\beta = \frac{2}{3}$

Now$,\tan(\alpha + \beta) = \dfrac{\tan \alpha + \tan\beta}{1-\tan\alpha\cdot\tan\beta}$

$\implies \tan(\alpha + \beta) = \dfrac{\frac{3}{4} + \frac{2}{3}}{1-\bigg(\frac{3}{4}\cdot\frac{2}{3}\bigg)}  = \dfrac{17}{6}$

So, the correct answer is $(D).$

References:

• https://www.liverpool.ac.uk/~maryrees/homepagemath191/trigid.pdf
• https://brilliant.org/wiki/basic-trigonometric-functions/
• https://brilliant.org/wiki/trigonometry/