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Q. Let the function $f(x)=\frac{\operatorname{arccot} \frac{x}{2}+\operatorname{arccot} \frac{x}{3}}{\arctan \frac{x}{2}+\arctan \frac{x}{3}}$, then $f(1)$ is equal to

Inverse Trigonometric Functions

Solution:

We have, $f(1)=\frac{\cot ^{-1} \frac{1}{2}+\cot ^{-1} \frac{1}{3}}{\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}}=\frac{\tan ^{-1} 2+\tan ^{-1} 3}{\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}}=\frac{\frac{3 \pi}{4}}{\frac{\pi}{4}}=3$