Question

cosec²A- tan( 90-A) = sin²A + sin²( 90-A) cosec²A- cotA

Answer 1

$$\begin{gathered} \mathrm{LHS}={\mathrm{cosec}}^2\mathrm{A}-\tan \left(90°-\mathrm{A}\right) \\ ={\mathrm{cosec}}^2\mathrm{A}-\cot \mathrm{A} \\ \mathrm{RHS}=\left[{\sin }^2\mathrm{A}+{\sin }^2\left(90°-\mathrm{A}\right)\right]\left({\mathrm{cosec}}^2\mathrm{A}-\cot \mathrm{A}\right) \\ =\left[{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}\right]\left({\mathrm{cosec}}^2\mathrm{A}-\cot \mathrm{A}\right) \\ =1\times \left({\mathrm{cosec}}^2\mathrm{A}-\cot \mathrm{A}\right) \\ ={\mathrm{cosec}}^2\mathrm{A}-\cot \mathrm{A} \\ \mathrm{So},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$