Question

Prove sin^2A+cot^2A=cosec^2A-cos^2A

Answer 1

Dear student

$$\begin{gathered} \mathrm{LHS}={\sin }^2\mathrm{A}+{\cot }^2\mathrm{A} \\ ={\sin }^2\mathrm{A}+\frac{{\cos }^2\mathrm{A}}{{\sin }^2\mathrm{A}} \\ =\frac{{\sin }^4\mathrm{A}+{\cos }^2\mathrm{A}}{{\sin }^2\mathrm{A}} \\ =\frac{{\left({\sin }^2\mathrm{A}\right)}^2+{\cos }^2\mathrm{A}}{{\sin }^2\mathrm{A}} \\ =\frac{{\left(1-{\cos }^2\mathrm{A}\right)}^2+{\cos }^2\mathrm{A}}{{\sin }^2\mathrm{A}} \\ =\frac{1+{\cos }^4\mathrm{A}-2{\cos }^2\mathrm{A}+{\cos }^2\mathrm{A}}{{\sin }^2\mathrm{A}} \\ =\frac{{\cos }^4\mathrm{A}-{\cos }^2\mathrm{A}+1}{{\sin }^2\mathrm{A}} \\ =\frac{{\cos }^2\mathrm{A}\left({\cos }^2\mathrm{A}-1\right)+1}{{\sin }^2\mathrm{A}} \\ =\frac{{\cos }^2\mathrm{A}\left(-{\sin }^2\mathrm{A}\right)+1}{{\sin }^2\mathrm{A}} \\ =-{\cos }^2\mathrm{A}+{\mathrm{cosec}}^2\mathrm{A} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved} \end{gathered}$$
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