\( \begin{array}{ll}\text { (g) } \cos \theta(1+\tan \theta)=\cos \theta+\sin \theta & \text { (h) } \frac{1-\cos ^{2} \theta}{\cos ^{2} \theta+2 \cos \theta+1}=\frac{1-\cos \theta}{1+\cos \theta} \\ \text { (i) } \frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=\frac{2}{\sin ^{2} \theta} & \text { (j) } \frac{1}{\sin \theta}+\frac{1}{\tan \theta}=\frac{1+\cos \theta}{\sin \left(180^{\circ}-\theta\right)}\end{array} \)

**(g)** Starting with the left-hand side, \[ \cos \theta(1+\tan \theta)=\cos \theta+\cos \theta \tan \theta. \] Recall that \[ \tan \theta=\frac{\sin \theta}{\cos \theta}, \] so \[ \cos \theta\tan \theta=\cos \theta\left(\frac{\sin \theta}{\cos \theta}\right)=\sin \theta. \] Thus, \[ \cos \theta(1+\tan \theta)=\cos \theta+\sin \theta, \] which equals the right-hand side. --- **(h)** Consider the expression \[ \frac{1-\cos ^{2} \theta}{\cos ^{2} \theta+2 \cos \theta+1}. \] Notice that the numerator is a difference of squares: \[ 1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta). \] The denominator can be written as a perfect square: \[ \cos^2\theta+2\cos\theta+1=(\cos \theta+1)^2. \] Thus, the expression becomes \[ \frac{(1-\cos\theta)(1+\cos\theta)}{(\cos\theta+1)^2}=\frac{1-\cos\theta}{1+\cos\theta}, \] assuming \(\cos\theta\neq -1\). This matches the right-hand side. --- **(i)** Examine the left-hand side: \[ \frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}. \] To combine the fractions, find a common denominator: \[ (1-\cos\theta)(1+\cos\theta)=1-\cos^2\theta. \] Thus, \[ \frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=\frac{(1+\cos\theta)+(1-\cos\theta)}{1-\cos^2\theta}. \] Simplify the numerator: \[ (1+\cos\theta)+(1-\cos\theta)=2. \] Also, use the Pythagorean identity \[ 1-\cos^2\theta=\sin^2\theta. \] Therefore, \[ \frac{2}{\sin^2\theta}, \] which is the simplified form on the right-hand side. --- **(j)** For the left-hand side, we have \[ \frac{1}{\sin \theta}+\frac{1}{\tan \theta}. \] Write \(\frac{1}{\tan \theta}\) in terms of sine and cosine: \[ \frac{1}{\tan \theta}=\frac{\cos \theta}{\sin \theta}. \] Now, \[ \frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\frac{1+\cos \theta}{\sin \theta}. \] On the right-hand side, we have \[ \frac{1+\cos \theta}{\sin (180^\circ-\theta)}. \] Recall that \[ \sin (180^\circ-\theta)=\sin \theta. \] Thus, \[ \frac{1+\cos \theta}{\sin (180^\circ-\theta)}=\frac{1+\cos \theta}{\sin \theta}, \] which is identical to the expression obtained for the left-hand side. --- Each identity is verified as true.