If ^ -1 x- ^ -1 (y)/(3)=α , where -1 ≤ x ≤ 1 , -3 ≤ y ≤ 3, x ≤ (y)/(3) , then for all x, y , 9 x^2-6 x y α+y^2 is equal to

Correct answer: C. 9 ^2 α If ^ -1 (x)/(a)- ^ -1 (y)/(b)=θ , then x^2 a^2 - 2 x y ab θ+ y^2 ~b^2 = ^2 θ Given, ^ -1 x- ^ -1 (y)/(3)=α Here, a=1, ~b=3 aligned & (x^2)/(1^2)-(2 x y)/((1)(3)) α+(y^2)/(3^2)= ^2 α & x^2-(2 x y)/(3) α+(y^2)/(9)= ^2 α & 9 x^2-6 x y α+y^2=9 ^2 α aligned

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If ^ -1 x- ^ -1 (y)/(3)=α , where -1 ≤ x ≤ 1 , -3 ≤ y ≤ 3, x ≤...

\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha

, where

-1 \leq x \leq 1

-3 \leq y \leq 3, x \leq \frac{y}{3}

, then for all

x, y

9 x^2-6 x y \cos \alpha+y^2

is equal to

Options

A $\sin ^2 \alpha$
B $3\sin ^2 \alpha$
C $9\sin ^2 \alpha$
D $\frac {4}{9}\sin ^2 \alpha$

Correct answer

C. $9\sin ^2 \alpha$

Step-by-step solution

\cos ^{-1} \frac{x}{a}-\cos ^{-1} \frac{y}{b}=\theta

, then

\frac{x^2}{\mathrm{a}^2}-\frac{2 x y}{\mathrm{ab}} \cos \theta+\frac{y^2}{\mathrm{~b}^2}=\sin ^2 \theta

Given,

\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha

Here,

\mathrm{a}=1, \mathrm{~b}=3

\begin{aligned} \therefore \quad & \frac{x^2}{1^2}-\frac{2 x y}{(1)(3)} \cos \alpha+\frac{y^2}{3^2}=\sin ^2 \alpha \\ & \Rightarrow x^2-\frac{2 x y}{3} \cos \alpha+\frac{y^2}{9}=\sin ^2 \alpha \\ & \Rightarrow 9 x^2-6 x y \cos \alpha+y^2=9 \sin ^2 \alpha \end{aligned}

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