Solveeit Logo
Login

Question

Mathematics Question on Inverse Trigonometric Functions

tan⁡(cos⁡−1(152)−sin⁡−1(417))\tan\left(\cos^{-1}\left( \frac{1}{5\sqrt{2}}\right)-\sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right) is

A

293\frac{\sqrt{29}}{3}

B

293\frac{29}{3}

C

329\frac{\sqrt{3}}{29}

D

329\frac{3}{29}

Answer

329\frac{3}{29}

Explanation

Solution

tan⁡(cos⁡−1(152)−sin⁡−1(417)) \tan\left(\cos^{-1}\left( \frac{1}{5\sqrt{2}}\right)-\sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right)
=tan⁡(π2−sin⁡−1(152)−sin⁡−1(417))= \tan\left(\frac{\pi}{2} - \sin^{-1}\left( \frac{1}{5\sqrt{2}}\right)-\sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right)
=tan⁡(π2−(sin⁡−1417+sin⁡−1152))= \tan\left(\frac{\pi}{2} - \left(\sin^{-1} \frac{4}{\sqrt{17}}+\sin^{-1}\frac{1}{5\sqrt{2}}\right)\right)
=tan⁡(π2−sin⁡−1(4171−150+1521−1617))= \tan\left(\frac{\pi}{2} - \sin^{-1} \left(\frac{4}{\sqrt{17}} \sqrt{1-\frac{1}{50} }+\frac{1}{5\sqrt{2}} \sqrt{1-\frac{16}{17}}\right)\right)
=tan⁡(π2−sin⁡−1(417.752+152.117))= \tan\left(\frac{\pi}{2} - \sin^{-1} \left(\frac{4}{\sqrt{17}} . \frac{7}{5\sqrt{2}} +\frac{1}{5\sqrt{2}}. \frac{1}{\sqrt{17}}\right)\right)
=cot⁡(sin⁡−12952.17)=cot⁡(cot⁡−1329)=\cot\left(\sin^{-1} \frac{29}{5\sqrt{2}.\sqrt{17}}\right)=\cot\left(\cot^{-1} \frac{3}{29}\right)
=329= \frac{3}{29}.