Question

Prove the following trigonometric identities.

(i) $\cot \theta -\tan \theta =\frac{2{\cos }^2\theta -1}{\sin \theta \cos \theta }$

(ii) $\tan \theta -\cot \theta =\frac{2{\sin }^2\theta -1}{\sin \theta \cos \theta }$

(iii) $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$

Answer 1

(i) We have to prove

We know that,

So,

(ii) We have to prove

We know that,

So,

(iii)
Given
$\frac{\mathrm{sinA}-2{\sin }^3\mathrm{A}}{2{\cos }^3\mathrm{A}-\mathrm{cosA}}=\mathrm{tanA}$

$\begin{array}{rcl}& & \\ \mathrm{LHS}& =& \frac{\mathrm{sinA}-2{\sin }^3\mathrm{A}}{2{\cos }^3\mathrm{A}-\mathrm{cosA}}\\ & =& \frac{\mathrm{sinA}\left(1-2{\sin }^2\mathrm{A}\right)}{\mathrm{cosA}\left(2{\cos }^2\mathrm{A}-1\right)}\\ & =& \mathrm{tanA}\frac{\left({\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}-2{\sin }^2\mathrm{A}\right)}{\left(2{\cos }^2\mathrm{A}-{\sin }^2\mathrm{A}-{\cos }^2\mathrm{A}\right)}\left(\because {\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}=1\right)\\ & =& \mathrm{tanA}\left(\frac{{\cos }^2\mathrm{A}-{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}-{\sin }^2\mathrm{A}}\right)\\ & =& \mathrm{tanA}\\ & =& \mathrm{RHS}\\ & & \end{array}$
Hence proved.