question_answer

Prove that: \[{{(cot\text{ }A+sec\text{ }B)}^{2}}-{{(tan\text{ }B-cosec\text{ }A)}^{2}}=2(cot\text{ }A.sec\text{ }B+tan\text{ }B.cosec\text{ }A).\]

Answer:

L.H.S.

\[={{(cot\text{ }A+sec\text{ }B)}^{2}}-{{(tan\text{ }B-cosec\text{ }A)}^{2}}\]

\[=(co{{t}^{2}}A+se{{c}^{2}}B+2\,cot\,A\,sec\,B)-(ta{{n}^{2}}B+\cos e{{c}^{2}}A-2\,tan\,B\,cosec\,A)\]

\[=co{{t}^{2}}A+se{{c}^{2}}B+2\,cot\,A\,sec\,B-ta{{n}^{2}}B\,-cose{{c}^{2}}A+2\,tan\,B\,cosec\,A\]

\[=(se{{c}^{2}}B-ta{{n}^{2}}B)-(cose{{c}^{2}}A-co{{t}^{2}}A)+2(cot\,A\,sec\,B+tan\,B\,cosec\,A)\]

\[=1-1+2(cot\text{ }A\text{ }sec\text{ }B+tan\text{ }B\text{ }cosec\text{ }A)\]

\[\left[ \begin{align}   & \because \,{{\sec }^{2}}B-{{\tan }^{2}}B=1 \\  & \cos e{{c}^{2}}A-{{\cot }^{2}}A=1 \\ \end{align} \right]\]

\[=2(cot\text{ }A\text{ }sec\text{ }B+tan\text{ }B\text{ }cosec\text{ }A)=R.H.S.\]             Hence Proved.