If operatornamecosec-1 x+ sec -1 x2+(π/2)=0 is satisfied by x=α then sin -1 α- cos -1 α is equal to

Question

If operatornamecosec-1 x+ sec -1 x2+(π/2)=0 is satisfied by x=α then sin -1 α- cos -1 α is equal to  (A) -(3 π/2) (B) -(π/2) (C) 0 (D) (π/2)

Tardigrade Admin · Accepted Answer

Θ operatornamecosec-1 x + sec -1 x 2+(π/2)=0 ⇒ operatornamecosec-1 x =-(π/2) text and sec -1 x 2=0 ⇒ x =-1 text and x 2=1 ⇒ x = ± 1 ∴ α=-1 ∴ sin -1 α- cos -1 α=-(π/2)-π=-(3 π/2)

Q. If $cosec^{- 1} x + sec^{- 1} x^{2} + \frac{π}{2} = 0$ is satisfied by $x = α$ then $sin^{- 1} α - cos^{- 1} α$ is equal to $

-\frac{3 \pi}{2}$

$- \frac{π}{2}$

0

$\frac{π}{2}$

$Θ cosec^{- 1} x + sec^{- 1} x^{2} + \frac{π}{2} = 0$
$\Rightarrow cosec^{- 1} x = - \frac{π}{2} and sec^{- 1} x^{2} = 0$
$\Rightarrow x = - 1 and x^{2} = 1 \Rightarrow x = \pm 1 < b r / > ∴ α = - 1$
$∴ sin^{- 1} α - cos^{- 1} α = - \frac{π}{2} - π = - \frac{3 π}{2} < b r / >$

Q. If cosec−1x+sec−1x2+2π​=0 is satisfied by x=α then sin−1α−cos−1α is equal to $

-\frac{3 \pi}{2}$

−2π​

0

2π​

Θcosec−1x+sec−1x2+2π​=0 ⇒cosec−1x=−2π​ and sec−1x2=0 ⇒x=−1 and x2=1⇒x=±1<br/>∴α=−1 ∴sin−1α−cos−1α=−2π​−π=−23π​<br/>

Q. If $cosec^{- 1} x + sec^{- 1} x^{2} + \frac{π}{2} = 0$ is satisfied by $x = α$ then $sin^{- 1} α - cos^{- 1} α$ is equal to $

$- \frac{π}{2}$

$\frac{π}{2}$

$Θ cosec^{- 1} x + sec^{- 1} x^{2} + \frac{π}{2} = 0$
$\Rightarrow cosec^{- 1} x = - \frac{π}{2} and sec^{- 1} x^{2} = 0$
$\Rightarrow x = - 1 and x^{2} = 1 \Rightarrow x = \pm 1 < b r / > ∴ α = - 1$
$∴ sin^{- 1} α - cos^{- 1} α = - \frac{π}{2} - π = - \frac{3 π}{2} < b r / >$