Question

Prove that
$\frac{co{s}^3\theta -cos3\theta }{cos\theta }+\frac{si{n}^3\theta +sin3\theta }{sin\theta }=3$

Answer 1

To prove:- $\frac{{\cos }^3\theta -\cos 3\theta }{\cos \theta }+\frac{{\sin }^3\theta +\sin 3\theta }{\sin \theta }=0$

Proof:-

Using trigonometric identity:

$\sin 3\theta =3\sin \theta -4{\sin }^3\theta \Rightarrow {\sin }^3\theta +\sin 3\theta =3\sin \theta -3{\sin }^3\theta$

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta \Rightarrow {\cos }^3\theta -\cos 3\theta =3\cos \theta -3{\cos }^3\theta$

Taking L.H.S.-

$\frac{{\cos }^3\theta -\cos 3\theta }{\cos \theta }+\frac{{\sin }^3\theta +\sin 3\theta }{\sin \theta }$

$=\frac{3\cos \theta -3{\cos }^3\theta }{\cos \theta }+\frac{3\sin \theta -3{\sin }^3\theta }{\sin \theta }$

$=\frac{3\cos \theta (1-{\cos }^2\theta )}{\cos \theta }+\frac{3\sin \theta (1-{\sin }^2\theta )}{\sin \theta }$

$=3{\sin }^2\theta +3{\cos }^2\theta$

$=3({\sin }^2\theta +{\cos }^2\theta )$

$=3$ =R.H.S

Hence Proved.