Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta$
$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos \theta+\cos \theta}=\tan \theta$
$L.H.S.=\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}$
$=\frac{\sin \theta\left(1-2 \sin ^{2} \theta\right)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$
$=\frac{\sin \theta \times\left(1-2 \sin ^{2} \theta\right)}{\cos \theta \times\left\{2\left(1-\sin ^{2} \theta\right)-1\right\}}$
$=\frac{\sin \theta \times\left(1-2 \sin ^{2} \theta\right)}{\cos \theta \times\left(1-2 \sin ^{2} \theta\right)}$
$=\tan \theta= R \cdot H.S.$
Similar Questions
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=........$
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=........$
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
Evaluate $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
Evaluate $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
If $\angle A$ and $\angle B$ are acute angles such that $\cos A =\cos B ,$ then show that $\angle A =\angle B$.
If $\angle A$ and $\angle B$ are acute angles such that $\cos A =\cos B ,$ then show that $\angle A =\angle B$.