Question

Evaluate:

(i) $\cot \left({\sin }^{-1}\frac{3}{4}+{\sec }^{-1}\frac{4}{3}\right)$
(ii) $\sin \left({\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}\right)\mathrm{for}x<0$
(iii) $\sin \left({\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}\right)\mathrm{for}x>0$
(iv) $\cot \left({\tan }^{-1}a+{\cot }^{-1}a\right)$
(v) $\cos \left({\sec }^{-1}x+{\mathrm{cosec}}^{-1}x\right),\left|x\right|\ge 1$

Answer 1

(i)
$$\begin{gathered} \cot \left({\sin }^{-1}\frac{3}{4}+{\sec }^{-1}\frac{4}{3}\right) \\ =\cot \left({\sin }^{-1}\frac{3}{4}+{\cos }^{-1}\frac{3}{4}\right)\left[\because {\sec }^{-1}x={\cos }^{-1}\frac{1}{x}\right] \\ =\cot \left(\frac{\mathrm{\pi }}{2}\right)\left[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =0 \end{gathered}$$

(ii)
$$\begin{gathered} \sin \left({\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}\right)=\sin \left[{\tan }^{-1}\left(-x\right)+{\tan }^{-1}\left(-\frac{1}{x}\right)\right]\left[\because x<0\right] \\ =\sin \left[-{\tan }^{-1}\left(x\right)-{\tan }^{-1}\left(\frac{1}{x}\right)\right] \\ =\sin \left\{-\left[{\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(\frac{1}{x}\right)\right]\right\} \\ =\sin \left[-\left({\tan }^{-1}x+{\cot }^{-1}x\right)\right]\left[\because {\tan }^{-1}\frac{1}{x}={\cot }^{-1}x\right] \\ =-\sin \left({\tan }^{-1}x+{\cot }^{-1}x\right) \\ =-\sin \left(\frac{\mathrm{\pi }}{2}\right)\left[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =-1 \end{gathered}$$

(iii)
$$\begin{gathered} \sin \left({\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}\right) \\ =\sin \left({\tan }^{-1}x+{\cot }^{-1}x\right)\left[\because {\tan }^{-1}x={\cot }^{-1}\frac{1}{x}\right] \\ =\sin \left(\frac{\mathrm{\pi }}{2}\right)\left[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =1 \end{gathered}$$

(iv)
$$\begin{gathered} \cot \left({\tan }^{-1}a+{\cot }^{-1}a\right) \\ =\cot \left(\frac{\mathrm{\pi }}{2}\right)\left[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =0 \end{gathered}$$

(v)
$$\begin{gathered} \cos \left({\sec }^{-1}x+{\mathrm{cosec}}^{-1}x\right) \\ =\cos \left(\frac{\mathrm{\pi }}{2}\right)\left[\because {\sec }^{-1}x+{\mathrm{cosec}}^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =0 \end{gathered}$$