2 votes 2 votes The value of $\tan \left(\sin^{-1}\left(\frac{3}{5}\right)+\cot^{-1}\left(\frac{3}{2}\right)\right)$ is $\frac{1}{18}$ $\frac{11}{6}$ $\frac{13}{6}$ $\frac{17}{6}$ Geometry isi2018-dcg trigonometry inverse non-gate + – gatecse asked Sep 18, 2019 • edited Nov 19, 2019 by Lakshman Bhaiya gatecse 279 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes $\tan \bigg(\sin^{-1}\big(\frac{3}{5}\big)+\cot^{-1}\big(\frac{3}{2}\big)\bigg)$ 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/ Lakshman Bhaiya answered Nov 28, 2019 • edited Nov 28, 2019 by Lakshman Bhaiya Lakshman Bhaiya comment Share Follow See all 0 reply Please log in or register to add a comment.