Match List I with List II: List I P....

MathematicsSolving Inverse Trigonometric EquationsJEE Advanced 2013Moderate

Match List I with List II:

List-I

(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

(Q)

cos x + cos y + cos z = 0 = sin x + sin y + sin z

then possible value of

cos \frac{x - y}{2}

(R)

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

(S)

cot (sin^{- 1} 1 - x^{2}) = sin (tan^{- 1} (x 6))

x \neq = 0

, then possible value of

x

List-II

(1)

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

(2)

2

(3)

\frac{1}{2}

(4)

Answer:P → 4, Q → 3, R → 2, S → 1

Visualized Solution (English)

Introduction to the Match List Problem

We need to match four trigonometric expressions in List I with their numerical values in List II.
The problem covers Inverse Trigonometric Functions, Trigonometric Identities, and System of Equations.
Let's denote the parts as (P), (Q), (R), and (S) and solve them sequentially.

Part (P): Simplifying the Numerator

Let $tan^{- 1} y = θ ⟹ tan θ = y$ .
Using a right triangle: $cos θ = \frac{1}{1 + y ^{2}}$ and $sin θ = \frac{y}{1 + y ^{2}}$ .
Numerator: $cos θ + y sin θ = \frac{1}{1 + y ^{2}} + y (\frac{y}{1 + y ^{2}}) = \frac{1 + y ^{2}}{1 + y ^{2}} = 1 + y^{2}$ .

Part (P): Simplifying the Denominator

Let $sin^{- 1} y = ϕ ⟹ sin ϕ = y$ .
Then $cot ϕ = \frac{1 - y ^{2}}{y}$ and $tan ϕ = \frac{y}{1 - y ^{2}}$ .
Denominator: $\frac{1 - y ^{2}}{y} + \frac{y}{1 - y ^{2}} = \frac{( 1 - y ^{2} ) + y ^{2}}{y 1 - y ^{2}} = \frac{1}{y 1 - y ^{2}}$ .

Part (P): Final Evaluation

Expression: $[\frac{1}{y ^{2}} (\frac{1 + y ^{2}}{1/ ( y 1 - y ^{2} )})^{2} + y^{4}]^{1/2}$
$= [\frac{1}{y ^{2}} (y 1 - y^{2} 1 + y^{2})^{2} + y^{4}]^{1/2}$
$= [1 - y^{4} + y^{4}]^{1/2} = 1$ .
Therefore, (P) $\to$ 4.

Part (Q): Squaring and Adding

Given: $cos x + cos y = - cos z$ and $sin x + sin y = - sin z$ .
Squaring and adding: $(cos x + cos y)^{2} + (sin x + sin y)^{2} = (- cos z)^{2} + (- sin z)^{2}$ .
$(cos^{2} x + sin^{2} x) + (cos^{2} y + sin^{2} y) + 2 (cos x cos y + sin x sin y) = cos^{2} z + sin^{2} z$ .
$1 + 1 + 2 cos (x - y) = 1 ⟹ 2 cos (x - y) = - 1$ .

Part (Q): Finding the Half-Angle Value

$cos (x - y) = - \frac{1}{2}$ .
Using $cos 2 θ = 2 cos^{2} θ - 1$ : $2 cos^{2} (\frac{x - y}{2}) - 1 = - \frac{1}{2}$ .
$2 cos^{2} (\frac{x - y}{2}) = \frac{1}{2} ⟹ cos^{2} (\frac{x - y}{2}) = \frac{1}{4}$ .
$cos (\frac{x - y}{2}) = \pm \frac{1}{2}$ . Match with (Q) $\to$ 3.

Part (R): Rearranging the Equation

Rearrange: $cos 2 x [cos (\frac{π}{4} - x) - cos (\frac{π}{4} + x)] = sin 2 x sec x (cos x - sin x)$ .
Using $cos (A - B) - cos (A + B) = 2 sin A sin B$ :
$cos 2 x [2 sin \frac{π}{4} sin x] = (2 sin x cos x) \frac{1}{c o s x} (cos x - sin x)$ .
$2 sin x cos 2 x = 2 sin x (cos x - sin x)$ .

Part (R): Solving for sec x

Assume $sin x \neq = 0$ and $cos x - sin x \neq = 0$ .
$2 (cos x - sin x) (cos x + sin x) = 2 (cos x - sin x)$ .
$2 (cos x + sin x) = 2 ⟹ cos x + sin x = 2$ .
This implies $x = \frac{π}{4}$ , so $sec x = 2$ . Match with (R) $\to$ 2.

Part (S): Algebraic Conversion

Left Side: $cot (sin^{- 1} 1 - x^{2}) = cot (cos^{- 1} x) = \frac{x}{1 - x ^{2}}$ .
Right Side: Let $tan^{- 1} (x 6) = β ⟹ tan β = x 6$ .
Then $sin β = \frac{x 6}{1 + ( x 6 ) ^{2}} = \frac{x 6}{1 + 6 x ^{2}}$ .
Equation: $\frac{x}{1 - x ^{2}} = \frac{x 6}{1 + 6 x ^{2}}$ .

Part (S): Solving for x

$\frac{1}{1 - x ^{2}} = \frac{6}{1 + 6 x ^{2}} ⟹ 1 + 6 x^{2} = 6 - 6 x^{2}$ .
$12 x^{2} = 5 ⟹ x^{2} = \frac{5}{12}$ .
$x = \frac{5}{12} = \frac{1}{2} \frac{5}{3}$ .
Therefore, (S) $\to$ 1.

Final Matching and Summary

Final Mapping:
(P) $\to$ 4 (Value: $1$ )
(Q) $\to$ 3 (Value: $\frac{1}{2}$ )
(R) $\to$ 2 (Value: $2$ )
(S) $\to$ 1 (Value: $\frac{1}{2} \frac{5}{3}$ )
The correct option is (b).

00:00 / 00:00

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Match List I with List II:

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List-II

Visualized Solution (English)

Introduction to the Match List Problem

Part (P): Simplifying the Numerator

Part (P): Simplifying the Denominator

Part (P): Final Evaluation

Part (Q): Squaring and Adding

Part (Q): Finding the Half-Angle Value

Part (R): Rearranging the Equation

Part (R): Solving for sec x

Part (S): Algebraic Conversion

Part (S): Solving for x

Final Matching and Summary

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