Question

$si{n}^{-1}x+si{n}^{-1}\left(\frac{1}{x}\right)+co{s}^{-1}x+co{s}^{-1}\left(\frac{1}{x}\right)$=

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{2}$
• D. $2\pi$

Answer 1

Answer: (A)

$\pi$

For ${\text{sin}}^{-1}x,x\in [-1,1]$

For ${\text{sin}}^{-1}(\frac{1}{x}),x\in (-\infty ,-1]\cup [1,\infty )$

So the common solution is $x=-1$ or $1$

For $x=-1$

${\text{sin}}^{-1}x+{\text{sin}}^{-1}(\frac{1}{x})+{\text{cos}}^{-1}x+{\text{cos}}^{-1}(\frac{1}{x})={\text{sin}}^{-1}(-1)+{\text{sin}}^{-1}(-1)+{\text{cos}}^{-1}(-1)+{\text{cos}}^{-1}(-1)$

$=-\frac{\pi }{2}-\frac{\pi }{2}+\pi +\pi =\pi$

For $x=1$

${\text{sin}}^{-1}x+{\text{sin}}^{-1}(\frac{1}{x})+{\text{cos}}^{-1}x+{\text{cos}}^{-1}(\frac{1}{x})={\text{sin}}^{-1}\left(1\right)+{\text{sin}}^{-1}\left(1\right)+{\text{cos}}^{-1}\left(1\right)+{\text{cos}}^{-1}\left(1\right)$

$=\frac{\pi }{2}+\frac{\pi }{2}+0+0=\pi$

So ${\text{sin}}^{-1}x+{\text{sin}}^{-1}(\frac{1}{x})+{\text{cos}}^{-1}x+{\text{cos}}^{-1}(\frac{1}{x})=\pi$