Match Trigonometric and Inverse Trigonometric Expressions with Evaluated Values

To match each expression in List-I with its correct value in List-II, we evaluate each part step-by-step:

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Evaluation of P:

The given expression is: $E = \left( \dfrac{1}{y^2} \left( \dfrac{\cos(\tan^{-1} y) + y \sin(\tan^{-1} y)}{\cot(\sin^{-1} y) + \tan(\sin^{-1} y)} \right)^2 + y^4 \right)^{1/2}$

1. Numerator: Let $\theta = \tan^{-1} y$, which implies: $\cos\theta = \frac{1}{\sqrt{1+y^2}} \quad \text{and} \quad \sin\theta = \frac{y}{\sqrt{1+y^2}}$ $\cos(\tan^{-1} y) + y \sin(\tan^{-1} y) = \frac{1}{\sqrt{1+y^2}} + y\left(\frac{y}{\sqrt{1+y^2}}\right) = \frac{1+y^2}{\sqrt{1+y^2}} = \sqrt{1+y^2}$

2. Denominator: Let $\phi = \sin^{-1} y$, then: $\cot(\sin^{-1} y) + \tan(\sin^{-1} y) = \cot\phi + \tan\phi = \frac{\cos\phi}{\sin\phi} + \frac{\sin\phi}{\cos\phi} = \frac{\cos^2\phi + \sin^2\phi}{\sin\phi\cos\phi} = \frac{1}{y\sqrt{1-y^2}}$

3. Substituting the numerator and denominator into the quotient: $\dfrac{\cos(\tan^{-1} y) + y \sin(\tan^{-1} y)}{\cot(\sin^{-1} y) + \tan(\sin^{-1} y)} = \sqrt{1+y^2} \cdot \left(y\sqrt{1-y^2}\right) = y\sqrt{1-y^4}$

4. Substituting back into the original expression: $E = \left( \frac{1}{y^2}\left(y\sqrt{1-y^4}\right)^2 + y^4 \right)^{1/2} = \left( \frac{1}{y^2}\cdot y^2(1-y^4) + y^4 \right)^{1/2} = \left( 1 - y^4 + y^4 \right)^{1/2} = \sqrt{1} = 1$

Thus, $\text{P} \rightarrow 4$.

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Evaluation of Q:

Given that: $\cos x + \cos y + \cos z = 0 \implies \cos x + \cos y = -\cos z$ $\sin x + \sin y + \sin z = 0 \implies \sin x + \sin y = -\sin z$

Squaring both equations and adding them: $(\cos x + \cos y)^2 + (\sin x + \sin y)^2 = (-\cos z)^2 + (-\sin z)^2$ $\cos^2 x + \cos^2 y + 2\cos x\cos y + \sin^2 x + \sin^2 y + 2\sin x\sin y = \cos^2 z + \sin^2 z$ $(1) + (1) + 2(\cos x\cos y + \sin x\sin y) = 1$ $2 + 2\cos(x-y) = 1$ $2(1 + \cos(x-y)) = 1$

Using the half-angle formula $1 + \cos(x-y) = 2\cos^2\left(\dfrac{x-y}{2}\right)$: $4\cos^2\left(\dfrac{x-y}{2}\right) = 1 \implies \cos^2\left(\dfrac{x-y}{2}\right) = \dfrac{1}{4} \implies \cos\left(\dfrac{x-y}{2}\right) = \pm\dfrac{1}{2}$

Thus, a possible value is $\dfrac{1}{2}$, so $\text{Q} \rightarrow 3$.

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Evaluation of R:

The given equation is: $\cos\left(\frac{\pi}{4}-x\right)\cos 2x + \sin x \sin 2x \sec x = \cos x \sin 2x \sec x + \cos\left(\frac{\pi}{4}+x\right)\cos 2x$

Rearranging the terms: $\left[ \cos\left(\frac{\pi}{4}-x\right) - \cos\left(\frac{\pi}{4}+x\right) \right]\cos 2x = (\cos x - \sin x)\sin 2x \sec x$

Using the identity $\cos(A-B) - \cos(A+B) = 2\sin A\sin B$: $2\sin\left(\frac{\pi}{4}\right)\sin x \cos 2x = (\cos x - \sin x)(2\sin x \cos x)\left(\frac{1}{\cos x}\right)$ $\sqrt{2}\sin x \cos 2x = 2\sin x(\cos x - \sin x)$

Assuming $\sin x \neq 0$: $\sqrt{2}\cos 2x = 2(\cos x - \sin x)$

Using $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$: $\sqrt{2}(\cos x - \sin x)(\cos x + \sin x) - 2(\cos x - \sin x) = 0$ $(\cos x - \sin x)\left[\sqrt{2}(\cos x + \sin x) - 2\right] = 0$

• If $\cos x - \sin x = 0$: $\tan x = 1 \implies \sec^2 x = 1 + \tan^2 x = 2 \implies \sec x = \pm\sqrt{2}$

• If $\sqrt{2}(\cos x + \sin x) = 2 \implies \cos x + \sin x = \sqrt{2}$: $\cos\left(x - \frac{\pi}{4}\right) = 1 \implies x = 2n\pi + \frac{\pi}{4} \implies \sec x = \sqrt{2}$

Thus, a possible value of $\sec x$ is $\sqrt{2}$, so $\text{R} \rightarrow 2$.

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Evaluation of S:

The given equation is: $\cot\left(\sin^{-1}\sqrt{1-x^2}\right) = \sin\left(\tan^{-1}(x\sqrt{6})\right), \quad x \neq 0$

1. For the LHS, let $\alpha = \sin^{-1}\sqrt{1-x^2}$ where $\sin\alpha = \sqrt{1-x^2}$. Then: $\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - (1-x^2)} = |x| = x \quad (\text{for } x > 0)$ $\cot\alpha = \frac{\cos\alpha}{\sin\alpha} = \frac{x}{\sqrt{1-x^2}}$

2. For the RHS, let $\beta = \tan^{-1}(x\sqrt{6})$ where $\tan\beta = x\sqrt{6}$. Then: $\sin\beta = \frac{\tan\beta}{\sqrt{1+\tan^2\beta}} = \frac{x\sqrt{6}}{\sqrt{1 + 6x^2}}$

Equating LHS and RHS: $\frac{x}{\sqrt{1-x^2}} = \frac{x\sqrt{6}}{\sqrt{1+6x^2}}$

Since $x \neq 0$, divide both sides by $x$: $\frac{1}{\sqrt{1-x^2}} = \frac{\sqrt{6}}{\sqrt{1+6x^2}}$

Squaring both sides: $\frac{1}{1-x^2} = \frac{6}{1+6x^2}$ $1 + 6x^2 = 6(1-x^2)$ $1 + 6x^2 = 6 - 6x^2$ $12x^2 = 5 \implies x^2 = \frac{5}{12}$ $x = \sqrt{\frac{5}{12}} = \frac{1}{2}\sqrt{\frac{5}{3}}$

Thus, $\text{S} \rightarrow 1$.

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Conclusion:

The correct matching is: $\text{P} \rightarrow 4, \quad \text{Q} \rightarrow 3, \quad \text{R} \rightarrow 2, \quad \text{S} \rightarrow 1$

Therefore, the correct option is B.