Prove that $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
Prove that $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
$L.H.S$ $=\cot 4 x(\sin 5 x+\sin 3 x)$
$=\frac{\cot 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=\left(\frac{\cos 4 x}{\sin 4 x}\right)[2 \sin 4 x \cos x]$
$=2 \cos 4 x \cos x$
$R.H.S.$ $=\cot x(\sin 5 x-\sin 3 x)$
$=\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$
$=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]$
$=2 \cos 4 x \cdot \cos x$
$L.H.S.$ $=$ $R.H.S.$
Similar Questions
$\frac{{\sec \,8\theta - 1}}{{\sec \,4\theta - 1}}$ is equal to
$\frac{{\sec \,8\theta - 1}}{{\sec \,4\theta - 1}}$ is equal to
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
- [IIT 1988]
The sines of two angles of a triangle are equal to $\frac{5}{{13}}$ & $\frac{{99}}{{101}}.$ The cosine of the third angle is :
The sines of two angles of a triangle are equal to $\frac{5}{{13}}$ & $\frac{{99}}{{101}}.$ The cosine of the third angle is :
The value of $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ $ is
The value of $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ $ is
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 $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$