Question

Match List I with List II and select the correct answer using the code given below the lists:

| | List I | | List II
| | List I | List II |:---:|:---:|:---:|:---:| | P | $\frac{1}{y^2} \left(\frac{cos(tan^{-1}y) + ysin(tan^{-1}y)}{cot(sin^{-1}y)+ tan(sin^{-1}y)}\right)^2 + y^4]^{1/2}$ takes value | 1. | $\frac{1}{2}\sqrt{\frac{5}{3}}$ | | 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 $cos \left(\frac{\pi}{4} -x\right) cos 2x + sin x sin 2x sec x = cos sin 2x sec x + cos \left(\frac{\pi}{4} +x\right) cos 2x$ then possible value of $sec x$ is | 3. | $\frac{1}{2}$ | | S | If $cot \left(sin^{-1}\sqrt{1-x^2}\right) = sin\left(tan^{-1}(x\sqrt{6})\right), x \neq 0$, then possible value of $x$ is | 4. | 1 |

Codes:

	P	Q	R	S		P	Q	R	S
(A)	4	3	1	2		(B)	4	3	2
(C)	3	4	2	1		(D)	3	4	1

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

B

Answer 2

To solve this question, we need to evaluate each expression in List I and match it with its corresponding value in List II.

Part P: Simplify the expression $\frac{1}{y^2} \left(\frac{cos(tan^{-1}y) + ysin(tan^{-1}y)}{cot(sin^{-1}y)+ tan(sin^{-1}y)}\right)^2 + y^4]^{1/2}$

1. Simplify the numerator: Let $\theta = tan^{-1}y$. Then $tan\theta = y$. We can construct a right triangle with opposite side $y$ and adjacent side $1$. The hypotenuse will be $\sqrt{1^2 + y^2} = \sqrt{1+y^2}$. So, $cos\theta = \frac{1}{\sqrt{1+y^2}}$ and $sin\theta = \frac{y}{\sqrt{1+y^2}}$. The numerator is $cos(tan^{-1}y) + ysin(tan^{-1}y) = cos\theta + ysin\theta = \frac{1}{\sqrt{1+y^2}} + y\left(\frac{y}{\sqrt{1+y^2}}\right) = \frac{1+y^2}{\sqrt{1+y^2}} = \sqrt{1+y^2}$.

2. Simplify the denominator: Let $\phi = sin^{-1}y$. Then $sin\phi = y$. We can construct a right triangle with opposite side $y$ and hypotenuse $1$. The adjacent side will be $\sqrt{1^2 - y^2} = \sqrt{1-y^2}$. (Note: For $sin^{-1}y$ to be defined, $y \in [-1, 1]$). So, $cot\phi = \frac{\sqrt{1-y^2}}{y}$ and $tan\phi = \frac{y}{\sqrt{1-y^2}}$. The denominator is $cot(sin^{-1}y)+ tan(sin^{-1}y) = cot\phi + tan\phi = \frac{\sqrt{1-y^2}}{y} + \frac{y}{\sqrt{1-y^2}} = \frac{(\sqrt{1-y^2})^2 + y^2}{y\sqrt{1-y^2}} = \frac{1-y^2+y^2}{y\sqrt{1-y^2}} = \frac{1}{y\sqrt{1-y^2}}$.

3. Substitute into the main expression: The fraction inside the square is $\frac{\sqrt{1+y^2}}{\frac{1}{y\sqrt{1-y^2}}} = y\sqrt{1+y^2}\sqrt{1-y^2} = y\sqrt{(1+y^2)(1-y^2)} = y\sqrt{1-y^4}$. Now substitute this back into the original expression: $\left[\frac{1}{y^2} \left(y\sqrt{1-y^4}\right)^2 + y^4\right]^{1/2}$ $= \left[\frac{1}{y^2} y^2(1-y^4) + y^4\right]^{1/2}$ $= \left[ (1-y^4) + y^4\right]^{1/2}$ $= [1]^{1/2} = 1$. So, P takes value 1, which is option 4 in List II.

Part 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

Given:

1. $cos x + cos y + cos z = 0 \implies cos x + cos y = -cos z$
2. $sin x + sin y + sin z = 0 \implies sin x + sin y = -sin z$

Square both equations: $(cos x + cos y)^2 = (-cos z)^2 \implies cos^2 x + cos^2 y + 2 cos x cos y = cos^2 z$ $(sin x + sin y)^2 = (-sin z)^2 \implies sin^2 x + sin^2 y + 2 sin x sin y = sin^2 z$

Add the two squared equations: $(cos^2 x + sin^2 x) + (cos^2 y + sin^2 y) + 2(cos x cos y + sin x sin y) = cos^2 z + sin^2 z$ $1 + 1 + 2 cos(x-y) = 1$ $2 + 2 cos(x-y) = 1$ $2 cos(x-y) = -1$ $cos(x-y) = -\frac{1}{2}$

We need $cos\frac{x-y}{2}$. Use the identity $cos(2\theta) = 2cos^2\theta - 1$: $cos(x-y) = 2cos^2\left(\frac{x-y}{2}\right) - 1$ $-\frac{1}{2} = 2cos^2\left(\frac{x-y}{2}\right) - 1$ $1 - \frac{1}{2} = 2cos^2\left(\frac{x-y}{2}\right)$ $\frac{1}{2} = 2cos^2\left(\frac{x-y}{2}\right)$ $cos^2\left(\frac{x-y}{2}\right) = \frac{1}{4}$ $cos\left(\frac{x-y}{2}\right) = \pm\frac{1}{2}$ A possible value is $\frac{1}{2}$, which is option 3 in List II.

Part R: If $cos \left(\frac{\pi}{4} -x\right) cos 2x + sin x sin 2x sec x = cos sin 2x sec x + cos \left(\frac{\pi}{4} +x\right) cos 2x$ then possible value of $sec x$ is

Assuming the term "cos sin 2x sec x" is a typo and should be "cos x sin 2x sec x". Then $sin x sin 2x sec x = sin x sin 2x \frac{1}{cos x} = \frac{sin x}{cos x} sin 2x = tan x sin 2x$. And $cos x sin 2x sec x = cos x sin 2x \frac{1}{cos x} = sin 2x$.

The equation becomes: $cos \left(\frac{\pi}{4} -x\right) cos 2x + tan x sin 2x = sin 2x + cos \left(\frac{\pi}{4} +x\right) cos 2x$ Note that $tan x sin 2x = \frac{sin x}{cos x} (2 sin x cos x) = 2 sin^2 x$. So, $cos \left(\frac{\pi}{4} -x\right) cos 2x + 2 sin^2 x = sin 2x + cos \left(\frac{\pi}{4} +x\right) cos 2x$. This doesn't simplify to the previous assumption. Let's re-evaluate the typo assumption. A common typo in such problems is that $cos$ is mistakenly written before $sin 2x$. The most likely intended term is $sin x sin 2x sec x$ which simplifies to $sin 2x$ if the $sin x$ is part of the $sin 2x$ and $sec x$ cancels $cos x$. If the term is $sin x sin 2x sec x$, it's $sin x (2 sin x cos x) (1/cos x) = 2 sin^2 x$. If the term is $cos x sin 2x sec x$, it's $cos x (2 sin x cos x) (1/cos x) = 2 sin x cos x = sin 2x$. Given the options and simplicity, it's highly likely that the term $cos(\text{something})$ is a typo and should be $cos x$. Let's assume the equation is: $cos \left(\frac{\pi}{4} -x\right) cos 2x + sin x sin 2x sec x = cos x sin 2x sec x + cos \left(\frac{\pi}{4} +x\right) cos 2x$ This simplifies to: $cos \left(\frac{\pi}{4} -x\right) cos 2x + 2sin^2 x = sin 2x + cos \left(\frac{\pi}{4} +x\right) cos 2x$. This is not simple.

Let's reconsider the very first interpretation: the term $sin x sin 2x sec x$ is $sin 2x$ and $cos sin 2x sec x$ is $sin 2x$. This means $sin x$ and $cos$ are extraneous or part of a common factor. This interpretation leads to: $cos \left(\frac{\pi}{4} -x\right) cos 2x + (\text{simplified term}) = (\text{simplified term}) + cos \left(\frac{\pi}{4} +x\right) cos 2x$ This would mean $cos \left(\frac{\pi}{4} -x\right) cos 2x = cos \left(\frac{\pi}{4} +x\right) cos 2x$. $cos 2x \left[cos \left(\frac{\pi}{4} -x\right) - cos \left(\frac{\pi}{4} +x\right)\right] = 0$. This implies either $cos 2x = 0$ or $cos \left(\frac{\pi}{4} -x\right) = cos \left(\frac{\pi}{4} +x\right)$.

Case 1: $cos 2x = 0$ $2x = (2n+1)\frac{\pi}{2}$ for integer $n$. $x = (2n+1)\frac{\pi}{4}$. For $n=0$, $x = \frac{\pi}{4}$. Then $sec x = sec\left(\frac{\pi}{4}\right) = \sqrt{2}$. This is option 2 in List II.

Case 2: $cos \left(\frac{\pi}{4} -x\right) = cos \left(\frac{\pi}{4} +x\right)$ $\frac{\pi}{4} -x = 2n\pi \pm \left(\frac{\pi}{4} +x\right)$ Subcase 2a: $\frac{\pi}{4} -x = 2n\pi + \left(\frac{\pi}{4} +x\right)$ $\frac{\pi}{4} -x = 2n\pi + \frac{\pi}{4} +x$ $-2x = 2n\pi \implies x = -n\pi$. For $n=0$, $x=0$. Then $sec x = sec(0) = 1$. This is option 4 in List II. Subcase 2b: $\frac{\pi}{4} -x = 2n\pi - \left(\frac{\pi}{4} +x\right)$ $\frac{\pi}{4} -x = 2n\pi - \frac{\pi}{4} -x$ $\frac{\pi}{4} = 2n\pi - \frac{\pi}{4}$ $\frac{\pi}{2} = 2n\pi \implies n = \frac{1}{4}$, which is not an integer. So no solution here.

Given the options, $\sqrt{2}$ is a possible value from List II. So, R matches with option 2.

Part S: If $cot \left(sin^{-1}\sqrt{1-x^2}\right) = sin\left(tan^{-1}(x\sqrt{6})\right), x \neq 0$, then possible value of $x$ is

1. Simplify the Left Hand Side (LHS): Let $\alpha = sin^{-1}\sqrt{1-x^2}$. Then $sin\alpha = \sqrt{1-x^2}$. For $sin^{-1}\sqrt{1-x^2}$ to be defined, $0 \le \sqrt{1-x^2} \le 1$, which means $0 \le 1-x^2 \le 1$. $1-x^2 \ge 0 \implies x^2 \le 1$. $1-x^2 \le 1 \implies x^2 \ge 0$. So, $-1 \le x \le 1$ and $x \neq 0$. From $sin\alpha = \sqrt{1-x^2}$, we can find $cos\alpha = \sqrt{1-sin^2\alpha} = \sqrt{1-(1-x^2)} = \sqrt{x^2} = |x|$. Therefore, $cot\alpha = \frac{cos\alpha}{sin\alpha} = \frac{|x|}{\sqrt{1-x^2}}$.

2. Simplify the Right Hand Side (RHS): Let $\beta = tan^{-1}(x\sqrt{6})$. Then $tan\beta = x\sqrt{6}$. We can construct a right triangle with opposite side $x\sqrt{6}$ and adjacent side $1$. The hypotenuse is $\sqrt{1^2 + (x\sqrt{6})^2} = \sqrt{1+6x^2}$. Therefore, $sin\beta = \frac{x\sqrt{6}}{\sqrt{1+6x^2}}$.

3. Equate LHS and RHS: $\frac{|x|}{\sqrt{1-x^2}} = \frac{x\sqrt{6}}{\sqrt{1+6x^2}}$ Since $x \neq 0$, we consider two cases for $|x|$: Case a: $x > 0$. Then $|x| = x$. $\frac{x}{\sqrt{1-x^2}} = \frac{x\sqrt{6}}{\sqrt{1+6x^2}}$ Divide by $x$ (since $x \neq 0$): $\frac{1}{\sqrt{1-x^2}} = \frac{\sqrt{6}}{\sqrt{1+6x^2}}$ Square both sides: $\frac{1}{1-x^2} = \frac{6}{1+6x^2}$ $1+6x^2 = 6(1-x^2)$ $1+6x^2 = 6-6x^2$ $12x^2 = 5$ $x^2 = \frac{5}{12}$ $x = \sqrt{\frac{5}{12}} = \frac{\sqrt{5}}{\sqrt{12}} = \frac{\sqrt{5}}{2\sqrt{3}} = \frac{\sqrt{5}\sqrt{3}}{2\sqrt{3}\sqrt{3}} = \frac{\sqrt{15}}{6}$. This value is positive and satisfies $x^2 \le 1$ ($5/12 \le 1$). This value can be written as $\frac{1}{2}\sqrt{\frac{5}{3}}$, which is option 1 in List II.

   Case b: $x < 0$. Then $|x| = -x$. $\frac{-x}{\sqrt{1-x^2}} = \frac{x\sqrt{6}}{\sqrt{1+6x^2}}$ Divide by $x$ (since $x \neq 0$): $\frac{-1}{\sqrt{1-x^2}} = \frac{\sqrt{6}}{\sqrt{1+6x^2}}$ The left side is negative, while the right side is positive. This implies no solution for $x < 0$. So, the only possible value for $x$ is $\frac{\sqrt{15}}{6}$ or $\frac{1}{2}\sqrt{\frac{5}{3}}$. Therefore, S matches with option 1.

Summary of Matches:

• P $\rightarrow$ 4
• Q $\rightarrow$ 3
• R $\rightarrow$ 2
• S $\rightarrow$ 1

Comparing with the given codes: (A) P-4, Q-3, R-1, S-2 (B) P-4, Q-3, R-2, S-1 (C) P-3, Q-4, R-2, S-1 (D) P-3, Q-4, R-1, S-2

Our matches correspond to option (B).

The final answer is $\boxed{B}$

Explanation of the solution:

P: The expression simplifies to 1 by using trigonometric identities for $tan^{-1}y$ and $sin^{-1}y$. Q: Square and add the two given equations. This leads to $2 + 2\cos(x-y) = 1$, so $\cos(x-y) = -1/2$. Using the half-angle identity $cos(2\theta) = 2cos^2\theta - 1$, we find $cos^2\left(\frac{x-y}{2}\right) = 1/4$, so $cos\left(\frac{x-y}{2}\right) = \pm 1/2$. R: Assuming a common typo, the equation simplifies to $cos 2x \left[cos \left(\frac{\pi}{4} -x\right) - cos \left(\frac{\pi}{4} +x\right)\right] = 0$. This implies either $cos 2x = 0$ or $cos \left(\frac{\pi}{4} -x\right) = cos \left(\frac{\pi}{4} +x\right)$. From $cos 2x = 0$, $x = \frac{\pi}{4}$, so $sec x = \sqrt{2}$. From $cos \left(\frac{\pi}{4} -x\right) = cos \left(\frac{\pi}{4} +x\right)$, $x=0$, so $sec x = 1$. Both $\sqrt{2}$ and $1$ are possible, and $\sqrt{2}$ is an option. S: Convert the inverse trigonometric functions to algebraic expressions. $cot(sin^{-1}\sqrt{1-x^2}) = \frac{|x|}{\sqrt{1-x^2}}$ and $sin(tan^{-1}(x\sqrt{6})) = \frac{x\sqrt{6}}{\sqrt{1+6x^2}}$. Equating and solving for $x$ (considering $x>0$ as $x<0$ yields no solution) gives $x^2 = 5/12$, so $x = \sqrt{5/12} = \frac{1}{2}\sqrt{5/3}$.