In any triangle $ABC ,$ ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to
In any triangle $ABC ,$ ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to
- A
$1 - 2\,\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
- B
$1 - 2\,\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
- C
$1 - 2\,\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
- D
$1 - 2\,\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$
Similar Questions
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
If $\alpha $ and $\beta $ are solutions of $sin^2\,x + a\, sin\, x + b = 0$ as well that of $cos^2\,x + c\, cos\, x + d = 0$ , then $sin\,(\alpha + \beta )$ is equal to
If $\alpha $ and $\beta $ are solutions of $sin^2\,x + a\, sin\, x + b = 0$ as well that of $cos^2\,x + c\, cos\, x + d = 0$ , then $sin\,(\alpha + \beta )$ is equal to
If $\tan A = \frac{{1 - \cos B}}{{\sin B}},$ find $\tan 2A$ in terms of $\tan B$ and show that
If $\tan A = \frac{{1 - \cos B}}{{\sin B}},$ find $\tan 2A$ in terms of $\tan B$ and show that
- [IIT 1983]
Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$
Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$
$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $
$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $