If y= tan -1 x+ cot -1 x+ sec -1 x+ operatornamecosec-1 x, then (d y/d x) is equal to

Question

If y= tan -1 x+ cot -1 x+ sec -1 x+ operatornamecosec-1 x, then (d y/d x) is equal to (A) (x2-1/x2+1) (B) π (C) 0 (D) 1

Tardigrade Admin · Accepted Answer

y= tan -1 x+ cot -1 x+ sec -1 x+ operatornamecosec-1 x =(π/2)+(π/2) (∵ tan -1 x+ cot -1 x=(π/2)) ( sec -1 x+ operatornamecosec-1 x=(π/2)) ∴ y=π ⇒ (d y/d x)=0

Q. If $y = tan^{- 1} x + cot^{- 1} x + sec^{- 1} x + cosec^{- 1} x$ , then $\frac{d y}{d x}$ is equal to

$\frac{x ^{2} - 1}{x ^{2} + 1}$

$π$

0

1

$y = tan^{- 1} x + cot^{- 1} x + sec^{- 1} x + cosec^{- 1} x$
$= \frac{π}{2} + \frac{π}{2} (∵ tan^{- 1} x + cot^{- 1} x = \frac{π}{2})$
$(sec^{- 1} x + cosec^{- 1} x = \frac{π}{2})$
$∴ y = π \Rightarrow \frac{d y}{d x} = 0$

Q. If y=tan−1x+cot−1x+sec−1x+cosec−1x, then dxdy​ is equal to

x2+1x2−1​

π

0

1

y=tan−1x+cot−1x+sec−1x+cosec−1x =2π​+2π​(∵tan−1x+cot−1x=2π​) (sec−1x+cosec−1x=2π​) ∴y=π⇒dxdy​=0

Q. If $y = tan^{- 1} x + cot^{- 1} x + sec^{- 1} x + cosec^{- 1} x$ , then $\frac{d y}{d x}$ is equal to

$\frac{x ^{2} - 1}{x ^{2} + 1}$

$π$

$y = tan^{- 1} x + cot^{- 1} x + sec^{- 1} x + cosec^{- 1} x$
$= \frac{π}{2} + \frac{π}{2} (∵ tan^{- 1} x + cot^{- 1} x = \frac{π}{2})$
$(sec^{- 1} x + cosec^{- 1} x = \frac{π}{2})$
$∴ y = π \Rightarrow \frac{d y}{d x} = 0$