List-IList-II(P)(1y(cos(tan−1y)+ysin(tan−1y)cot(sin−1y)+tan(sin−1y))+y4)1/2 takes value(1)1252(Q)If cosx+cosy+cosz=0=sinx+siny+sinz then possible value of cosx−y2 is(2)2(R)If cos(π4−x)cos2x+sinxsin2xsecx=cosxsin2xsecx+cos(π4+x)cos2x then possible value of secx is(3)12(S)If cot(sin−11−x2)=sin(tan−1(x6)),x≠0, then possible value of x is(4)1 (5)23

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

List-IList-II(P)(1y(cos(tan−1y)+ysin(tan−1y)cot(sin−1y)+tan(sin−1y))+y4)1/2 takes value(1)1252(Q)If cosx+cosy+cosz=0=sinx+siny+sinz then possible value of cosx−y2 is(2)2(R)If cos(π4−x)cos2x+sinxsin2xsecx=cosxsin2xsecx+cos(π4+x)cos2x then possible value of secx is(3)12(S)If cot(sin−11−x2)=sin(tan−1(x6)),x≠0, then possible value of x is(4)1 (5)23

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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List-I
List-II
(P) ${(\frac{1}{y} (\frac{\cos (\tan^{- 1} y) + y \sin (\tan^{- 1} y)}{\cot (\sin^{- 1} y) + \tan (\sin^{- 1} y)}) + y^{4})}^{1 / 2}$ takes value (1) $\frac{1}{2} \sqrt{\frac{5}{2}}$
(Q) If $\cos x + \cos y + \cos z = 0 = \sin x + \sin y + \sin z$ then possible value of $\cos \frac{x - y}{2}$ is (2) $\sqrt{2}$
(R) If $\cos (\frac{π}{4} - x) \cos 2 x + \sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x + \cos (\frac{π}{4} + x) \cos 2 x$ then possible value of $\sec x$ is (3) $\frac{1}{2}$
(S) If $\cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \sin (\tan^{- 1} (x \sqrt{6})), x \neq 0,$ then possible value of x is (4) 1
(5) $\frac{2}{\sqrt{3}}$

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P $\to$ 4; Q $\to$ 3; R $\to$ 5; S $\to$ 2

P $\to$ 4; Q $\to$ 3; R $\to$ 2; S $\to$ 1

P $\to$ 3; Q $\to$ 4; R $\to$ 2; S $\to$ 1

P $\to$ 3; Q $\to$ 4; R $\to$ 5; S $\to$ 2

answer is B.

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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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