IIT-JEE 1989Question 1
The greater of the two angles $A=2\tan^{-1}\left(2\sqrt2-1\right)$ and $B=3\sin^{-1}(1/3)+\sin^{-1}(3/5)$ is ______.
Inverse Trigonometric Functions
Answer$A$
Solution
$\tan\frac A2=2\sqrt2-1\approx1.83$ gives $A\approx2.137$. $B=3\sin^{-1}\frac13+\sin^{-1}\frac35\approx3(0.3398)+0.6435\approx1.663$. So $A>B$.