Match the following. Column-I Column-IIA(1x2(cos(tan−1x)+xsin(tan−1x)cot(sin−1x)+tan(sin−1x))2+x4)12 takes the value(s)P1253BIf cot(sin−11−x2)=sin(tan−1(x6)),x≠0 then possible values of x is/areQ12CIf cosα+cosβ+cosγ=0=sinα+sinβ+sinγ, then the possible value of cos(α−β2) isR2DIf cos(π4−θ)cos2θ+sinθsin2θsecθ=cosθsin2θsecθ+cos(π4+θ)cos2θ,(θ≠nπ2,n∈I) , then the value of secθ isS1

A)1x2cos⁡tan−1⁡x+ysin⁡tan−1⁡xcot⁡sin−1⁡x+tan⁡sin−1⁡x2+x41/2tan−1⁡x=θ⇒x=tan⁡θ1x2x1−x211+x22+x41/2⇒1x2x21−x4+x41/2=1B) x1−x2=x66x2+1x=±512C) cos⁡α+cos⁡β=−cos⁡γ, sin⁡α+sin⁡β=−sin⁡γsquaring and adding2+2cos⁡(α−β)=1⇒4cos2⁡α−β2=1,⇒cos⁡α−β2=1/2D)cos⁡2θ 2sin⁡π4sin⁡θ=sec⁡θsin⁡2θ(cos⁡θ−sin⁡θ)cos⁡θ+sin⁡θ=2θ=π4 sec⁡θ=2

Match the following. Column-I Column-IIA(1x2(cos(tan−1x)+xsin(tan−1x)cot(sin−1x)+tan(sin−1x))2+x4)12 takes the value(s)P1253BIf cot(sin−11−x2)=sin(tan−1(x6)),x≠0 then possible values of x is/areQ12CIf cosα+cosβ+cosγ=0=sinα+sinβ+sinγ, then the possible value of cos(α−β2) isR2DIf cos(π4−θ)cos2θ+sinθsin2θsecθ=cosθsin2θsecθ+cos(π4+θ)cos2θ,(θ≠nπ2,n∈I) , then the value of secθ isS1

A)1x2cos⁡tan−1⁡x+ysin⁡tan−1⁡xcot⁡sin−1⁡x+tan⁡sin−1⁡x2+x41/2tan−1⁡x=θ⇒x=tan⁡θ1x2x1−x211+x22+x41/2⇒1x2x21−x4+x41/2=1B) x1−x2=x66x2+1x=±512C) cos⁡α+cos⁡β=−cos⁡γ, sin⁡α+sin⁡β=−sin⁡γsquaring and adding2+2cos⁡(α−β)=1⇒4cos2⁡α−β2=1,⇒cos⁡α−β2=1/2D)cos⁡2θ 2sin⁡π4sin⁡θ=sec⁡θsin⁡2θ(cos⁡θ−sin⁡θ)cos⁡θ+sin⁡θ=2θ=π4 sec⁡θ=2

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Match the following.
Column-I Column-II
A ${(\frac{1}{x^{2}} {(\frac{\cos (t a n^{- 1} x) + x \sin (t a n^{- 1} x)}{\cot (s i n^{- 1} x) + t a n (s i n^{- 1} x)})}^{2} + x^{4})}^{\frac{1}{2}}$ takes the value(s) P $\frac{1}{2} \sqrt{\frac{5}{3}}$
B
If $c o t (s i n^{- 1} \sqrt{1 - x^{2}}) = \sin (t a n^{- 1} (x \sqrt{6})), x \neq 0$ then possible values of
x is/are
Q $\frac{1}{2}$
C If $c o s α + c o s β + \cos γ = 0 = \sin α + \sin β + \sin γ$ , then the possible value of $\cos (\frac{α - β}{2})$ is R $\sqrt{2}$
D If $\cos (\frac{π}{4} - θ) \cos 2 θ + \sin θ \sin 2 θ \sec θ = \cos θ \sin 2 θ \sec θ + \cos (\frac{π}{4} + θ) \cos 2 θ$ $, (θ \neq \frac{n π}{2}, n \in I)$ , then the value of $\sec θ$ is S 1

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$A - S; B - P; C - R; D - Q$

$A - P; B - S; C - Q; D - R$

$A - S; B - P; C - Q; D - R$

$A - P; B - S; C - R; D - Q$

answer is C.

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Detailed Solution

A) $\begin{aligned} {[\frac{1}{x^{2}} {(\frac{\cos (\tan^{- 1} x) + y \sin (\tan^{- 1} x)}{\cot (\sin^{- 1} x) + \tan (\sin^{- 1} x)})}^{2} + x^{4}]}^{1 / 2} \\ \tan^{- 1} x = θ \Rightarrow x = \tan θ \end{aligned}$

${[\frac{1}{x^{2}} {(\frac{x \sqrt{1 - x^{2}}}{1} \sqrt{1 + x^{2}})}^{2} + x^{4}]}^{1 / 2}$

$\Rightarrow {[\frac{1}{x^{2}} x^{2} (1 - x^{4}) + x^{4}]}^{1 / 2} = 1$

B) $\frac{x}{\sqrt{1 - x^{2}}} = \frac{x \sqrt{6}}{\sqrt{6 x^{2} + 1}}$

$x = \pm \sqrt{\frac{5}{12}}$

C) $\cos α + \cos β = - \cos γ, \sin α + \sin β = - \sin γ$

squaring and adding

$2 + 2 \cos (α - β) = 1$

$\Rightarrow 4 \cos^{2} (\frac{α - β}{2}) = 1, \Rightarrow \cos (\frac{α - β}{2}) = 1 / 2$

D) $\cos 2 θ 2 \sin \frac{π}{4} \sin θ = \sec θ \sin 2 θ (\cos θ - \sin θ)$

$\cos θ + \sin θ = \sqrt{2}$

$θ = \frac{π}{4} \sec θ = \sqrt{2}$

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