Vinod Singh and Shweta Pawar Solutions for Chapter: Trigonometric Functions-II, Exercise 3: Competitive Thinking

If $\frac{(x+1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$, then ${cosec}^{-1}\left(\frac{1}{A}\right)+{\cot }^{-1}\frac{1}{B}+{\sec }^{-1}C$ is

The number of points of intersection of $2y=1$ and $y=\sin x$, in $-2\pi \le x\le 2\pi$ is

If in a triangle $ABC, a=5, b=4, A=\frac{\pi }{2}+B$, then angle $C$ is

If $x={\sin }^{-1}K, y={\cos }^{-1}K,-1\le K\le 1$, then the correct relationship is

If $A\mathit{=}\frac{\mathit{1}}{\pi }\left[\begin{array}{ll}{\sin }^{\mathit{-}1}\left(\pi x\right)& {\tan }^{\mathit{-}\mathit{1}}\left(\frac{x}{\pi }\right)\\ {\sin }^{\mathit{-}\mathit{1}}\left(\frac{x}{\pi }\right)& {\cot }^{\mathit{-}1}\left(\pi x\right)\end{array}\right]\mathit{,}\mathit{ }B\mathit{=}\frac{\mathit{1}}{\pi }\left[\begin{array}{cc}\mathit{-}{\cos }^{-1}\left(\pi x\right)& {\tan }^{\mathit{-}1}\left(\frac{x}{\pi }\right)\\ {\sin }^{\mathit{-}1}\left(\frac{x}{\pi }\right)& \mathit{-}{\tan }^{\mathit{-}1}\left(\pi x\right)\end{array}\right]$  then $A-B$ is equal to

If $\frac{1}{6}\sin \theta , \cos \theta$ and $\tan \theta$ are in geometric progression, then the solution set of $\theta$ is

In triangle $ABC, A=30°, b=8, a=6$, where $B={\sin }^{-1}x$, then $x$ is

The value of $\cot \left[\sum _{n=1}^{100}{\cot }^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right]$ is