Question

If $f\left(x\right)=sin\left\{co{t}^{-1}\right(x+1\left)\right\}-cos\left(ta{n}^{-1}x\right)andv=cos\left(ta{n}^{-1}\right(sin\left(co{t}^{-1}x\right)\left)\right)$ then the value of $v^2$ for f(x) = 0 is equal to

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{5}{9}$
• D. $\frac{9}{5}$

Answer 1

The correct option is C $\frac{5}{9}$

$f\left(x\right)=0\frac{1}{\sqrt{1+(x+1{)}^2}}=\frac{1}{\sqrt{1+x^2}}\Rightarrow x=-\frac{1}{2}Now{v}^2=\frac{x^2+1}{x^2+2}=\frac{5}{9}$