Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$\frac{(\cos A–\sin A+1)}{(\cos A+\sin A–1)}=\mathrm{cosec}A+\cot A$, using the identity ${\mathrm{cosec}}^{2}A=1+{\cot }^{2}A$.

Answer 1

Proof:

Consider the LHS of the given expression.

$$\begin{gathered} \mathrm{LHS}=\frac{(\cos A–\sin A+1)}{(\cos A+\sin A–1)} \\ =\frac{{\displaystyle \frac{(\cos A–\sin A+1)}{\sin A}}}{{\displaystyle \frac{(\cos A+\sin A–1)}{\sin A}}}\left[\mathrm{Divide}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\right] \\ =\frac{cotA-1+\cos ecA}{cotA+1-\cos ecA}\left[\mathrm{Use}\cos e{c}^{2}A-co{t}^{2}A=1\right] \\ =\frac{cotA-\cos e{c}^{2}A+co{t}^2+\cos ecA}{cotA+1-\cos ecA} \\ =\frac{cotA+\cos ecA+(cotA-\cos ecA)(cotA+\cos ecA)}{cotA+1-\cos ecA} \\ =\frac{cotA+\cos ecA(1+cotA-\cos ecA)}{cotA+1-\cos ecA} \\ =\cot A+\mathrm{cosec}A \\ =\mathrm{RHS} \end{gathered}$$

Hence, proved.