R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: Exercise 13A

Prove the following identity:

$\frac{1}{\left(1+{\tan }^2\theta \right)}+\frac{1}{\left(1+{\cot }^2\theta \right)}=1$

Prove: $\frac{{\cot }^2\theta \left(\sec \theta -1\right)}{\left(1+\sin \theta \right)}+\frac{{\sec }^2\theta \left(\sin \theta -1\right)}{\left(1+\sec \theta \right)}=0$

Prove: $\left\{\frac{1}{\left({\sec }^2\theta -{\cos }^2\theta \right)}+\frac{1}{\left({\mathrm{cosec}}^2\theta -{\sin }^2\theta \right)}\right\}\left({\sin }^2\theta {\cos }^2\theta \right)=\frac{1-{\sin }^2\theta {\cos }^2\theta }{2+{\sin }^2\theta {\cos }^2\theta }$

Prove each of the following Identities:
$\frac{(\sin A-\sin B)}{(\cos A+\cos B)}+\frac{(\cos A-\cos B)}{(\sin A+\sin B)}=0$

Show that the following is not an identity:
${\cos }^2\theta +\cos \theta =1$

Show that the following expression is not an identity:
${\sin }^2\theta +\sin \theta =2$

Prove that $\left(\sin \theta -2{\sin }^3\theta \right)=\left(2{\cos }^3\theta -\cos \theta \right)\tan \theta$

If $1+{\sin }^2\theta =3 \sin \theta \cos \theta$ then prove that $\tan \theta =1$ or $\frac{1}{2}$.