D-dx - tan-1 -1-cos x-1 - cos x- -

Explanation for correct option:Step 1: Simplify the given data.Given, d/dx(tan^ 1(√(1 cos(x)/1+cos(x))))=d/dx(tan^ 1(√(2sin^2(x/2)/2cos^2(x/2))))0ex0ex=d/dx(tan ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

d/dx [ tan^-1...

Question

$\frac{d}{d x} (\tan^{- 1} (\sqrt{\frac{1 - \cos (x)}{1 + \cos (x)}})) =$

$\frac{1}{2}$

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$0$

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$1$

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$- \frac{1}{2}$

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Solution

The correct option is A
$\frac{1}{2}$

Explanation for correct option:
Step-1: Simplify the given data.
Given, $\frac{d}{d x} (\tan^{- 1} (\sqrt{\frac{1 - \cos (x)}{1 + \cos (x)}}))$
$= \frac{d}{d x} (\tan^{- 1} (\sqrt{\frac{2 \sin^{2} (\frac{x}{2})}{2 \cos^{2} (\frac{x}{2})}})) = \frac{d}{d x} (\tan^{- 1} (\tan (\frac{x}{2}))) = \frac{d}{d x} (\frac{x}{2})$
Step-2: Differentiate w.r.t $x$
$= \frac{1}{2} \times 1$
$= \frac{1}{2}$
Hence, correct answer is option $(A)$

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Similar questions

Q. If

f (x) = {sin}^{- 1} {\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}}}, - \frac{1}{2} \leq x \leq 1,

then

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Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

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^{- 1}

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$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

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ddxtan-11-cosx1+cosx=

The correct option is A 12Explanation for correct option:Step-1: Simplify the given data.Given, ddxtan-11-cosx1+cosx=ddxtan-12sin2x22cos2x2=ddxtan-1tanx2=ddxx2Step-2: Differentiate w.r.t x=12×1=12Hence, correct answer is option A

$\frac{d}{d x} (\tan^{- 1} (\sqrt{\frac{1 - \cos (x)}{1 + \cos (x)}})) =$