(a)OR(b)

Prove that : $\displaystyle \frac{\cos \theta-2 \cos ^{3} \theta}{\sin \theta-2 \sin ^{3} \theta}+\cot \theta=0$.

\[\begin{aligned} \text { LHS } & =\frac{\cos \theta-2 \cos ^{3} \theta}{\sin \theta-2 \sin ^{3} \theta}+\cot \theta \\ & =\frac{\cos \theta\left(1-2 \cos ^{2} \theta\right)}{\sin \theta\left(1-2 \sin ^{2} \theta\right)}+\cot \theta \\ & =\frac{\cos \theta}{\sin \theta}\left[\frac{\sin ^{2} \theta+\cos ^{2} \theta-2 \cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta-2 \sin ^{2} \theta}\right]+\cot \theta \\ & =\frac{\cot \theta\left(\sin ^{2} \theta-\cos ^{2} \theta\right)}{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}+\cot \theta \\ & =-\cot \theta+\cot \theta \\ & =0=\text { RHS } \end{aligned} \]

Given: \(\displaystyle \sin \theta+\cos \theta=x\)

Squaring both sides

\[\begin{aligned} & \sin ^{2} \theta+\cos ^{2} \theta+2 \cos \theta \sin \theta=x^{2} \\ & 2 \sin \theta \cos \theta=x^{2}-1 \end{aligned} \]

\[\begin{aligned} \text { RHS } & =\frac{2-(2 \sin \theta \cos \theta)^{2}}{2} \\ & =\frac{2-4 \sin ^{2} \theta \cos ^{2} \theta}{2} \\ & =1-2 \sin ^{2} \theta \cos ^{2} \theta \\ & =\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cos ^{2} \theta \\ & =\left(\sin ^{4} \theta+\cos ^{4} \theta\right)=\mathrm{LHS} \end{aligned} \]