If cos ^{-1} x - cos ^{-1} frac{y}{3} = alpha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^

Vedclass · Accepted Answer

$9 \sin ^2 \alpha$

Vedclass · Answer

(A) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^2})$,we have: $\cos ^{-1} \left( \frac{xy}{3} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{9}} \right) = \alpha$ $\frac{xy}{3} + \frac{\sqrt{1-x^2} \sqrt{9-y^2}}{3} = \cos \alpha$ $xy + \sqrt{1-x^2} \sqrt{9-y^2} = 3 \cos \alpha$ $xy - 3 \cos \alpha = -\sqrt{1-x^2} \sqrt{9-y^2}$ Squaring both sides: $(xy - 3 \cos \alpha)^2 = (1-x^2)(9-y^2)$ $x^2y^2 - 6xy \cos \alpha + 9 \cos ^2 \alpha = 9 - y^2 - 9x^2 + x^2y^2$ $9x^2 - 6xy \cos \alpha + y^2 = 9 - 9 \cos ^2 \alpha$ $9x^2 - 6xy \cos \alpha + y^2 = 9(1 - \cos ^2 \alpha) = 9 \sin ^2 \alpha$.

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$,where $-1 \leq x \leq 1$,$-3 \leq y \leq 3$,and $x \leq \frac{y}{3}$,then for all $x, y$,$9x^2 - 6xy \cos \alpha + y^2$ is equal to

Similar Questions

Prove $\tan ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

If $0 < x < \frac{1}{2}$ and $\alpha = \sin^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{\sqrt{3 - 3 x^2}}{2} \right)$,then $\tan \alpha + \cot \alpha =$

$\sin \left\{ {{\tan }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} \right) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \right\}$ is equal to

If $y = 2\sin^{-1} \sqrt{1-x} + \sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$,then $\frac{dy}{dx}$ equals-

If $\cot^{-1} \alpha + \cot^{-1} \beta = \cot^{-1} x$,then $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module