Question

Prove each of the following identities:

$$\begin{gathered} \left(\mathrm{i}\right){\cot }^2\theta -\frac{1}{{\sin }^2\theta }=-1 \\ \left(\mathrm{ii}\right){\tan }^2\theta -\frac{1}{{\cos }^2\theta }=-1 \\ \left(\mathrm{iii}\right){\cos }^2\theta +\frac{1}{\left(1+{\cot }^2\theta \right)}=1 \end{gathered}$$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{LHS}={\cot }^2\theta -\frac{1}{{\sin }^2\theta } \\ =\frac{{\cos }^2\theta }{{\sin }^2\theta }-\frac{1}{{\sin }^2\theta } \\ =\frac{{\cos }^2\theta -1}{{\sin }^2\theta } \\ =\frac{-{\sin }^2\theta }{{\sin }^2\theta } \\ =-1 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{LHS}={\tan }^2\theta -\frac{1}{{\cos }^2\theta } \\ =\frac{{\sin }^2\theta }{{\cos }^2\theta }-\frac{1}{{\cos }^2\theta } \\ =\frac{{\sin }^2\theta -1}{{\cos }^2\theta } \\ =\frac{-{\cos }^2\theta }{{\cos }^2\theta } \\ =-1 \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{LHS}={\cos }^2\theta +\frac{1}{\left(1+{\cot }^2\theta \right)} \\ ={\cos }^2\theta +\frac{1}{{\mathrm{cosec}}^2\theta } \\ ={\cos }^2\theta +{\sin }^2\theta \\ =1 \\ =\mathrm{RHS} \end{gathered}$$