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.
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