If y = 2 a 2 - b 2 tan - 1 [ a - b a + b tan x 2 ] , a , b 0 , d y d x x = 2 =

Question

If $y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2}], a, b > 0,$ $\frac{d y}{d x} |_{x = \frac{π}{2}}$ $=$

Smart Achievers · Accepted Answer

(2)

$y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} (\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2})$

$\frac{d y}{d x} = \frac{2}{\sqrt{a^{2} - b^{2}}} \frac{1}{(1 + \frac{a - b}{a + b} \tan^{2} \frac{x}{2})} \sec^{2} \frac{x}{2} \cdot \frac{1}{2} \sqrt{\frac{a - b}{a + b}}$

$\frac{d y}{d x} |_{x = \frac{π}{2}}$ $= \frac{1}{a}$

Solution

Q.
If $y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2}], a, b > 0,$ $\frac{d y}{d x} |_{x = \frac{π}{2}}$ $=$

1 $\frac{1}{b}$

2 $\frac{1}{a}$

3 $\frac{2}{a}$

4 $\frac{2}{b}$

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Solution

Q. If y=2a2-b2tan-1[a-ba+btanx2], a,b>0, dydx|x=π2=

1 1b

2 1a

3 2a

4 2b