If x- 0-4 - then the derivative of tan-1 cos x- sin xcos x- sin x with respect to x is-1 mark-

Let y=tan^ 1 ( cos x+ sin xcos x sin x ) ⇒ y=tan^ 1 ( 1+tan x1 tan x ) ⇒ y=tan^ 1 ( tan ( x+π4 ) ) ⇒ y=x+π4 Differentiating w.r.t. x, we get dydx=1 ...

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Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

If x∈ 0,π4 , ...

Question

If $x \in (0, \frac{π}{4}),$ then the derivative of ${tan}^{- 1} (\frac{cos x + sin x}{cos x - sin x})$ with respect to $x$ is

[1 mark]

tan x

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1

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x + \frac{π}{4}

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1 + \frac{π}{4}

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Solution

The correct option is B $1$
Let $y = {tan}^{- 1} (\frac{cos x + sin x}{cos x - sin x})$
$\Rightarrow y = {tan}^{- 1} (\frac{1 + tan x}{1 - tan x})$
$\Rightarrow y = {tan}^{- 1} (tan (x + \frac{π}{4}))$
$\Rightarrow y = x + \frac{π}{4}$
Differentiating w.r.t. $x,$ we get
$\frac{d y}{d x} = 1$

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If x∈(0,π4), then the derivative of tan−1(cosx+sinxcosx−sinx) with respect to x is [1 mark]

The correct option is B 1Let y=tan−1(cosx+sinxcosx−sinx) ⇒y=tan−1(1+tanx1−tanx) ⇒y=tan−1(tan(x+π4)) ⇒y=x+π4 Differentiating w.r.t. x, we get dydx=1

If $x \in (0, \frac{π}{4}),$ then the derivative of ${tan}^{- 1} (\frac{cos x + sin x}{cos x - sin x})$ with respect to $x$ is

[1 mark]

The correct option is B $1$
Let $y = {tan}^{- 1} (\frac{cos x + sin x}{cos x - sin x})$
$\Rightarrow y = {tan}^{- 1} (\frac{1 + tan x}{1 - tan x})$
$\Rightarrow y = {tan}^{- 1} (tan (x + \frac{π}{4}))$
$\Rightarrow y = x + \frac{π}{4}$
Differentiating w.r.t. $x,$ we get
$\frac{d y}{d x} = 1$