Question

If cot $A=\frac{4}{5}$, prove that $\frac{(\sin A+\cos A)}{(\sin A-\cos A)}=9.$

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{cotA}=\frac{4}{5} \\ \mathrm{Writing}\cot \mathrm{A}=\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\mathrm{and}\mathrm{sqauring}\mathrm{the}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{}\frac{{\cos }^2\mathrm{A}}{{\sin }^2\mathrm{A}}=\frac{16}{25} \\ \mathrm{=>25}{\cos }^2\mathrm{A}=16{\sin }^2\mathrm{A} \\ \mathrm{=>25}{\cos }^2\mathrm{A}=16-16{\cos }^2\mathrm{A} \\ \mathrm{=>}{\cos }^2\mathrm{A}=\frac{16}{41} \\ \mathrm{=>}\mathrm{cosA}=\frac{4}{\sqrt{41}} \\ \therefore {\sin }^2\mathrm{A}=1-{\cos }^2\mathrm{A} \\ \mathrm{=1}-\frac{16}{41} \\ \mathrm{Now},\sin \mathrm{A}=\sqrt{\frac{25}{41}} \\ \mathrm{=>sin}\mathrm{A}=\frac{5}{\sqrt{41}} \\ \therefore \mathrm{LHS=}\frac{\sin \mathrm{A}+\mathrm{cosA}}{\sin \mathrm{A}-\mathrm{cosA}} \\ \mathrm{=}\frac{\frac{5}{\sqrt{41}}+\frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}} \\ \mathrm{=}\frac{9}{1} \\ \mathrm{=9}=\mathrm{RHS} \end{gathered}$$