Match List-I with List-II and select the correct answer using the code given below the lists : | List-I | | List-II | | | --- | --- | --- | --- | | P. | $\left(\frac{1}{y^{2}}\left(\frac{\cos \left(\tan ^{-1} y\right)+y \sin \left(\tan ^{-1} y\right)}{\cot \left(\sin ^{-1} y\right)+\tan \left(\sin ^{-1} y\right)}\right)^{2}+y^{4}\right)^{1 / 2}$ 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 $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$ 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 |

Trigonometric Ratios and Identities — JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 Question with Solution

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

List-I		List-II
P.	$(\frac{1}{y ^{2}} (\frac{c o s ( t a n ^{- 1} y ) + y s i n ( t a n ^{- 1} y )}{c o t ( s i n ^{- 1} y ) + t a n ( s i n ^{- 1} y )})^{2} + y^{4})^{1/2}$ takes value	1.	$\frac{1}{2} \frac{5}{3}$
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.	$2$
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 $cot (sin^{- 1} 1 - x^{2}) = sin (tan^{- 1} (x 6)), x \neq = 0$ , then possible value of $x$ is	4.	1

Answer: (B)

P: Simplifying the expression using inverse trigonometric identities:

cos (tan^{- 1} y) = \frac{1}{1 + y ^{2}}

sin (tan^{- 1} y) = \frac{y}{1 + y ^{2}}

cot (sin^{- 1} y) = \frac{1 - y ^{2}}{y}

tan (sin^{- 1} y) = \frac{y}{1 - y ^{2}}

After substitution and simplification, the expression equals 1. So P matches 4.

Q: From

cos x + cos y = - cos z

and

sin x + sin y = - sin z

, squaring and adding:

2 + 2 cos (x - y) = 1

, so

cos (x - y) = - 1/2

2 cos^{2} \frac{x - y}{2} - 1 = - 1/2

, giving

cos^{2} \frac{x - y}{2} = 1/4

cos \frac{x - y}{2} = \pm 1/2

. Q matches 3.

R: Simplifying the given equation leads to

cos 2 x = 0

sin x = 2

(impossible).

From

cos 2 x = 0

cos^{2} x = 1/2

, so

sec x = 2

. R matches 2.

S: Using the given equation and simplifying with inverse trig identities:

\frac{x}{1 - x ^{2}} = \frac{x 6}{1 + 6 x ^{2}}

Squaring:

1 + 6 x^{2} = 6 (1 - x^{2})

, giving

12 x^{2} = 5

, so

x = \frac{1}{2} \frac{5}{3}

. S matches 1.

Answer: P-4, Q-3, R-2, S-1

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option