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

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

Question

(C) Given the equation $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} (uv + \sqrt{1-u^

Vedclass · Accepted Answer

$9\sin ^2 \alpha$

Vedclass · Answer

(C) Given the equation $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} (uv + \sqrt{1-u^2}\sqrt{1-v^2})$,we have $\cos ^{-1} (\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}}) = \alpha$. Taking cosine on both sides: $\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha$. Rearranging: $\sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha - \frac{xy}{3}$. Squaring both sides: $(1-x^2)(1-\frac{y^2}{9}) = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$. $1 - \frac{y^2}{9} - x^2 + \frac{x^2 y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$. $1 - x^2 - \frac{y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha$. $1 - \cos^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$. $\sin^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$. Multiplying by $9$: $9 \sin^2 \alpha = 9x^2 - 6xy \cos \alpha + y^2$.

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 $x, y$,$9 x^2-6 x y \cos \alpha+y^2$ is equal to

Similar Questions

The value of $x$ which satisfies $\sin \left(\cot ^{-1} x\right)=\cos \left(\tan ^{-1}(1+x)\right)$ is

$\cos \left(\cos ^{-1}\left(-\frac{1}{4}\right)+\sin ^{-1}\left(-\frac{1}{4}\right)\right) = $ . . . . . . .

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then the value of $x^{2025}+x^{2026}+x^{2027}$ is

The value of $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ is

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then the value of $x^2+y^2+z^2-2xyz$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module