Prove that: $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)$
$L.H.S.$ $=\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$
$=\frac{\left[2 \sin \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \sin \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}{\left[2 \cos \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \cos \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}$
$=\frac{[2 \sin 6 x \cdot \cos x]+[2 \sin 6 x \cdot \cos 3 x]}{[2 \cos 6 x \cdot \cos x]+[2 \cos 6 x \cdot \cos 6 x]}$
$=\frac{2 \sin 6 x[\cos x+\cos 3 x]}{2 \cos 6 x[\cos x+\cos 3 x]}$
$=\tan 6 x$
$= R . H.S.$
If $\cos x + {\cos ^2}x = 1,$ then the value of ${\sin ^2}x + {\sin ^4}x$ is
If $\sin {\theta _1} + \sin {\theta _2} + \sin {\theta _3} = 3,$ then $\cos {\theta _1} + \cos {\theta _2} + \cos {\theta _3} = $
If $x = \sec \,\phi - \tan \phi ,y = {\rm{cosec}}\phi + \cot \phi ,$ then
Let the function $:(0, \pi) \rightarrow R$ be defined by
$f (\theta)=(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .
If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then
$(A)$ $\tan ^2 x=\frac{2}{3}$ $(B)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C)$ $\tan ^2 x=\frac{1}{3}$ $(D)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$