Question

sin theta - 2sin cube theta/2cos cube theta-cos theta = tan theta​

Answer 1

Answer:

solution refers to the attachment

hope it helps u dear

Answer 2

$\dfrac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta-\cos \theta}=\tan \theta$, proved.

Step-by-step explanation:

To prove that, $\dfrac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta-\cos \theta}=\tan \theta$.

L.H.S. = $\dfrac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta-\cos \theta}$

$=\dfrac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta-1)}$

$=\dfrac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2(1 - \sin^2 \theta)-1)}$

Using the trigonometric identity,

$\sin^2 A+\cos^2 A=1$

⇒ $\cos^2 A=1-\sin^2 A$

$=\dfrac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2- 2\sin^2 \theta-1)}$

$=\dfrac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(1- 2\sin^2 \theta)}$

$=\dfrac{\sin \theta}{\cos \theta}$

= $\tan \theta$

= R.H.S., proved.

Thus, $\dfrac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta-\cos \theta}=\tan \theta$, proved.