Question

The value of $\cos {\tan }^{-1}\frac{3}{4}+\cos {\tan }^{-1}x$ is equal to

• A. $\frac{4}{5}+\frac{x}{\sqrt{1+x^2}}$
• B. $\frac{3}{5}+\frac{1}{\sqrt{1+x^2}}$
• C. $\frac{4}{5}+\frac{1}{\sqrt{1+x^2}}$
• D. $\frac{3}{5}+\frac{x}{\sqrt{1x^2}}$

Answer 1

Answer: (C)

$\cos {\tan }^{-1}\frac{3}{4}+\cos {\tan }^{-1}x$
$=\cos {\cos }^{-1}\frac{4}{5}+\cos {\cos }^{-1}\frac{1}{\sqrt{1+x^2}}$
$=\frac{4}{5}+\frac{1}{\sqrt{1+x^2}}$