If derivative of tan -1((2 x/1-x2)) w.r.t. cos -1((1-x2/1+x2)) at x=50 is equal to a and derivative of tan -1((2 x/1-x2)) w.r.t. cos -1((2 x/1+x2)) at x=-50 is equal to b. Then (a+b) equals

Question

If derivative of tan -1((2 x/1-x2)) w.r.t. cos -1((1-x2/1+x2)) at x=50 is equal to a and derivative of tan -1((2 x/1-x2)) w.r.t. cos -1((2 x/1+x2)) at x=-50 is equal to b. Then (a+b) equals (A) 0 (B) 1 (C) - 1 (D) 2

Tardigrade Admin · Accepted Answer

a = 1 and b = 1 ∴ (a + b) = 2.

Q. If derivative of $tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ w.r.t. $cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})$ at $x = 50$ is equal to a and derivative of $tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ w.r.t. $cos^{- 1} (\frac{2 x}{1 + x ^{2}})$ at $x = - 50$ is equal to $b$ . Then $(a + b)$ equals

0

1

- 1

2

$a = 1$ and $b = 1$
$∴ (a + b) = 2.$

Q. If derivative of tan−1(1−x22x​) w.r.t. cos−1(1+x21−x2​) at x=50 is equal to a and derivative of tan−1(1−x22x​) w.r.t. cos−1(1+x22x​) at x=−50 is equal to b. Then (a+b) equals

0

1

- 1

2

a=1 and b=1 ∴(a+b)=2.

Q. If derivative of $tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ w.r.t. $cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})$ at $x = 50$ is equal to a and derivative of $tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ w.r.t. $cos^{- 1} (\frac{2 x}{1 + x ^{2}})$ at $x = - 50$ is equal to $b$ . Then $(a + b)$ equals

$a = 1$ and $b = 1$
$∴ (a + b) = 2.$