(a) Prove that: √(sec^2 θ+ cosec^2 θ)= tan θ+ cot θ

LHS \(\displaystyle =\sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}\)

\[\begin{aligned} & =\frac{1}{\sin \theta \cdot \cos \theta} \\ & =\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta} \\ & =\frac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta} \\ & =\tan \theta+\cot \theta=\mathrm{RHS} \end{aligned} \]

\[\begin{aligned} & \begin{aligned} \cot ^{2} \theta=\operatorname{cosec}^{2} \theta-1 & =\left(x+\frac{1}{4 x}\right)^{2}-1 \\ & =\left(x-\frac{1}{4 x}\right)^{2} \end{aligned} \\ & \Rightarrow \cot \theta=\left(x-\frac{1}{4 x}\right) \text { or }\left(\frac{1}{4 x}-x\right) \\ & \operatorname{cosec} \theta+\cot \theta=\left(x+\frac{1}{4 x}\right)+\left(x-\frac{1}{4 x}\right) \text { or }\left(x+\frac{1}{4 x}\right)+\left(\frac{1}{4 x}-x\right) \\ & \quad=2 x \text { or } \frac{1}{2 x} \end{aligned} \]