1. Tardigrade
  2. Question
  3. Mathematics
  4. If cos -1 (x/a)- sin -1 (y/b)=θ(a, b ≠ 0), then the maximum value of b2 x2+a2 y2+2 a b x y sin θ equals

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $\cos ^{-1} \frac{x}{a}-\sin ^{-1} \frac{y}{b}=\theta(a, b \neq 0)$, then the maximum value of $b^2 x^2+a^2 y^2+2 a b x y \sin \theta$ equals

Inverse Trigonometric Functions

Solution:

We have $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\frac{\pi}{2}+\theta $
$\Rightarrow \frac{x y}{a b}-\sqrt{1-\frac{x^2}{a^2}} \sqrt{1-\frac{y^2}{b^2}}=-\sin \theta$
$\Rightarrow \frac{x y}{a b}+\sin \theta=\sqrt{1-\frac{x^2}{a^2}} \sqrt{1-\frac{y^2}{b^2}}$
On squaring both the sides, we get
$\Rightarrow \frac{x^2 y^2}{a^2 b^2}+\sin ^2 \theta+\frac{2 x y}{a b} \sin \theta=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2 y^2}{a^2 b^2} $
$\Rightarrow b^2 x^2+a^2 y^2+2 a b x \sin \theta=a^2 b^2 \cos ^2 \theta \leq a^2 b^2$