$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $
$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $
- A
$\tan A$
- B
$\cot A$
- C
$\tan 2A$
- D
$\cot 2A$
Similar Questions
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
If $\frac{x}{{\cos \theta }} = \frac{y}{{\cos \left( {\theta - \frac{{2\pi }}{3}} \right)}} = \frac{z}{{\cos \left( {\theta + \frac{{2\pi }}{3}} \right)}},$ then $x + y + z = $
If $\frac{x}{{\cos \theta }} = \frac{y}{{\cos \left( {\theta - \frac{{2\pi }}{3}} \right)}} = \frac{z}{{\cos \left( {\theta + \frac{{2\pi }}{3}} \right)}},$ then $x + y + z = $
$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $
$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
The expression $[1 - sin (3\pi - \alpha ) + cos (3\pi + \alpha )]$ $\left[ {1\,\, - \,\,\sin \,\left( {\frac{{3\,\pi }}{2}\,\, - \,\,\alpha } \right)\,\, + \,\,\cos \,\left( {\frac{{5\,\pi }}{2}\,\, - \,\,\alpha } \right)} \right]$ when simplified reduces to :
The expression $[1 - sin (3\pi - \alpha ) + cos (3\pi + \alpha )]$ $\left[ {1\,\, - \,\,\sin \,\left( {\frac{{3\,\pi }}{2}\,\, - \,\,\alpha } \right)\,\, + \,\,\cos \,\left( {\frac{{5\,\pi }}{2}\,\, - \,\,\alpha } \right)} \right]$ when simplified reduces to :