question_answer

Prove that: \[\frac{\sec \,A-1}{\sec \,A+1}={{\left( \frac{\sin \,A}{1+\cos \,A} \right)}^{2}}={{(cot\,A-cosec\,A)}^{2}}\]

Answer:

\[L.H.S.=\frac{\sec \,A-1}{\sec \,A+1}\]

\[=\frac{\frac{1}{\cos \,A}-1}{\frac{1}{\cos \,A}+1}=\frac{\frac{1-\cos \,A}{\cos \,A}}{\frac{1+\cos \,A}{\cos \,A}}\]

\[=\frac{1-\cos \,A}{1+\cos \,A}\]

\[=\frac{(1-cos\,A)(1+cos\,A)}{(1+cos\,A)(1+cos\,A)}\]

\[=\frac{1-{{\cos }^{2}}A}{{{(1+cos\,A)}^{2}}}\]

\[=\frac{{{\sin }^{2}}A}{{{(1+cos\,A)}^{2}}}\]

\[={{\left( \frac{\sin \,A}{1+\cos \,A} \right)}^{2}}\]                         Hence Proved.

And,     \[{{\left( \frac{\sin \,A}{1+\cos \,A} \right)}^{2}}=\left[ \left( \frac{\sin \,A}{1+\cos \,A} \right)\times \frac{(1-cos\,A)}{(1-\cos \,A)} \right]\]

\[={{\left[ \frac{\sin \,A(1-cos\,A)}{1-{{\cos }^{2}}A} \right]}^{2}}\]

\[={{\left[ \frac{\sin \,A(1-cos\,A)}{si{{n}^{2}}A} \right]}^{2}}\]

\[={{\left[ \frac{1-cos\,A}{sinA} \right]}^{2}}\]

\[={{(cosec\,A-cot\,A)}^{2}}\]

\[={{(-1)}^{2}}{{[cot\,A-cosec\,A]}^{2}}\]

\[={{[cot\,A-cosec\,A]}^{2}}=R.H.S.\]                      Hence Proved.