Consider $\tan 15^{\circ}$ tan $195^{\circ}$ $=\tan 15^{\circ} \tan \left(180+15^{\circ}\right)$
$=\tan 15^{\circ} \tan 15^{\circ} \quad(\because \tan (180+\theta)=\tan \theta)$
$\begin{aligned} &=\left(\tan 15^{\circ}\right)^{2} \\ &=(2-\sqrt{3})^{2} \quad\left(\because \tan 15^{\circ}=2-\sqrt{3}\right) \\ &=4+3-4 \sqrt{3}=7-4 \sqrt{3} \\ \text { 109. (b) } & \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{\sin ^{2} x+1+\cos ^{2} x+2 \cos x}{(1+\cos x)(\sin x)} \\ &=\frac{2+2 \cos x}{(1+\cos x)(\sin x)}=\frac{2(1+\cos x)}{(1+\cos x)(\sin x)} \\ &=\frac{2}{\sin x}=2 \operatorname{cosec} x \\ & \Rightarrow 3 \sin A-4 \sin ^{3} A=1 \\ & \Rightarrow 4 \sin ^{3} A-3 \sin A+1=0 \\ & \Rightarrow(\sin A+1)\left(4 \sin ^{2} A-4 \sin A+1\right)=0 \\ & \Rightarrow(\sin A+1)(2 \sin A-1)^{2}=0 \\ & \Rightarrow \sin A=-1 \text { or } \frac{1}{2} \end{aligned}$
Hence, $\sin A$ can take two distinct values.