JEE Main · Mathematics · Inverse Trigonometric Functions
If ^ -1 (1)/(5)+(1)/(2) ^ -1 x+ ^ -1 (1)/(8)=(π)/(8) , then x^2=
If tan−151+21sec−1x+tan−181=8π, then x2=
Options
A712
B4950
C1213
D21
Correct answer
B. 4950
Step-by-step solution
We have, tan−151+21sec−1x+tan−181=8π⇒(tan−151+tan−181)+21sec−1x=8π\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \times \frac{1}{8}}\right]+\frac{1}{2} \sec ^{-1}x=\frac{\pi}{8}⇒ and tan−13913+21sec−1x=8π⇒ and 2tan−13913+sec−1x=4π⇒ and 2tan−131+sec−1x=4π⇒ and tan−11−9132+sec−1x=4π⇒ and tan−143+sec−1x=4π⇒⇒tan−143+tan−1x2−1=4πtan−1[1−43x2−143+x2−1]=4π⇒ and 4−3x2−13+4x2−1=1⇒ and 3+4x2−1=4−3x2−1⇒7x2−1=1⇒ and x2−1=71\Rightarrowx2−1=491\Rightarrowx2=4950