If alpha + beta - gamma = pi, then sin^2 alph | vedclass

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

Question

(A) Given $\alpha + \beta - \gamma = \pi$,which implies $\gamma = \alpha + \beta - \pi$.Using the identity $\sin^2 A - \sin^2 B = \sin(A - B) \sin(A +

Vedclass · Accepted Answer

$2 \sin \alpha \sin \beta \cos \gamma$

Vedclass · Answer

(A) Given $\alpha + \beta - \gamma = \pi$,which implies $\gamma = \alpha + \beta - \pi$.Using the identity $\sin^2 A - \sin^2 B = \sin(A - B) \sin(A + B)$:$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + (\sin^2 \beta - \sin^2 \gamma)$$= \sin^2 \alpha + \sin(\beta - \gamma) \sin(\beta + \gamma)$Since $\beta - \gamma = \pi - \alpha$,we have $\sin(\beta - \gamma) = \sin(\pi - \alpha) = \sin \alpha$.Also,$\beta + \gamma = \beta + (\alpha + \beta - \pi) = \alpha + 2\beta - \pi$.Alternatively,using $\gamma = \alpha + \beta - \pi$,then $\sin \gamma = \sin(\alpha + \beta - \pi) = -\sin(\alpha + \beta)$.So,$\sin^2 \gamma = \sin^2(\alpha + \beta)$.Expression $= \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta)$$= \sin^2 \alpha + \sin^2 \beta - (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2$$= \sin^2 \alpha + \sin^2 \beta - (\sin^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta)$$= \sin^2 \alpha(1 - \cos^2 \beta) + \sin^2 \beta(1 - \cos^2 \alpha) - 2 \sin \alpha \sin \beta \cos \alpha \cos \beta$$= \sin^2 \alpha \sin^2 \beta + \sin^2 \beta \sin^2 \alpha - 2 \sin \alpha \sin \beta \cos \alpha \cos \beta$$= 2 \sin \alpha \sin \beta (\sin \alpha \sin \beta - \cos \alpha \cos \beta)$$= 2 \sin \alpha \sin \beta (- \cos(\alpha + \beta))$Since $\alpha + \beta = \pi + \gamma$,$\cos(\alpha + \beta) = \cos(\pi + \gamma) = -\cos \gamma$.Therefore,the expression $= 2 \sin \alpha \sin \beta (-(- \cos \gamma)) = 2 \sin \alpha \sin \beta \cos \gamma$.

If $\alpha + \beta - \gamma = \pi,$ then $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = $

Similar Questions

If $\frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{a + b\tan^2 2\theta}{1 + c\tan^2 2\theta + d\tan^4 2\theta}$ (where $\theta \neq \frac{n\pi}{16}, n \in I$),then the value of $(a - b + c - d)$ is -

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as:

If the sum and difference of two angles are $\pi$ and $\frac{\pi}{12}$ respectively,then the values of the angles in degree measure are (in $^{\circ}$)

The minimum value of $3 \cos x + 4 \sin x + 5$ is

If $\sec \theta + \tan \theta = p,$ then $\tan \theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module