For alpha, beta, gamma neq 0. If sin ^{-1} al | vedclass

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

Question

(A) Let $\sin ^{-1} \alpha = A, \sin ^{-1} \beta = B, \sin ^{-1} \gamma = C$. Then $A+B+C = \pi$,which implies $\sin A = \alpha, \sin B = \beta, \sin

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(A) Let $\sin ^{-1} \alpha = A, \sin ^{-1} \beta = B, \sin ^{-1} \gamma = C$. Then $A+B+C = \pi$,which implies $\sin A = \alpha, \sin B = \beta, \sin C = \gamma$. The given equation is $(\alpha+\beta+\gamma)(\alpha+\beta-\gamma) = 3\alpha\beta$. This can be written as $(\alpha+\beta)^2 - \gamma^2 = 3\alpha\beta$. Expanding this,we get $\alpha^2 + \beta^2 + 2\alpha\beta - \gamma^2 = 3\alpha\beta$,which simplifies to $\alpha^2 + \beta^2 - \gamma^2 = \alpha\beta$. Dividing by $2\alpha\beta$,we get $\frac{\alpha^2 + \beta^2 - \gamma^2}{2\alpha\beta} = \frac{1}{2}$. Using the law of cosines in a triangle with sides $a, b, c$ where $\sin A = \alpha, \sin B = \beta, \sin C = \gamma$,we have $\cos C = \frac{\alpha^2 + \beta^2 - \gamma^2}{2\alpha\beta} = \frac{1}{2}$. Since $\cos C = \frac{1}{2}$,we have $C = \frac{\pi}{3}$. Therefore,$\gamma = \sin C = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.

For $\alpha, \beta, \gamma \neq 0$. If $\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$ and $(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$,then $\gamma$ is equal to

Similar Questions

The number of solutions of the equation $2\tan^{-1}(\cos^2 x) = \tan^{-1}(2\csc^2 x)$ in the interval $[0, 5\pi]$ is $m$. Then which of the following is true?

$\lim _{n \rightarrow \infty} \sum_{r=1}^n \cot ^{-1}\left(r^2+\frac{3}{4}\right)=$

Let $\cos ^{-1}(x) + \cos ^{-1} (2x) + \cos ^{-1}(3x) = \pi.$ If $x$ satisfies the cubic equation $ax^3 + bx^2 + cx - 1 = 0,$ then the value of $(a + b + c)$ is -

If $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=k$ then $\sin ^{-1}\left(\sqrt{\frac{k}{2}}\right)+\cos ^{-1}\left(\frac{k}{3}\right)=$

If $\sin^{-1}(\tan \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$, then $x$ is a root of the equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module