Question

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

• A. $ab$
• B. $(a+b{)}^2$
• C. $2(a+b{)}^2$
• D. $a^2{b}^2$

Answer 1

Answer: (D)

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