JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q19

MathematicsSingle correct+4 / -1Question 19 of 90

$\tan\left(2\tan^{-1}\dfrac15+\sec^{-1}\dfrac{\sqrt5}2+2\tan^{-1}\dfrac18\right)$ is equal to :

\dfrac14

\dfrac54

Answer and worked solution

Official answer: B

Combine. $2\tan^{-1}\frac15+2\tan^{-1}\frac18=2\tan^{-1}\frac13=\tan^{-1}\frac34$ , and $\sec^{-1}\frac{\sqrt5}2=\tan^{-1}\frac12$ .

Total $\tan^{-1}\dfrac{3/4+1/2}{1-3/8}=\tan^{-1}2$ : value 2.

(B)

Topics:Inverse trigonometric functions

JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 19