question_answer

Prove that:  \[{{\sec }^{2}}\theta -{{\cot }^{2}}(90{}^\circ -\theta )=co{{s}^{2}}(90{}^\circ -\theta )+co{{s}^{2}}\theta .\]

Answer:

To prove: \[{{\sec }^{2}}\theta -{{\cot }^{2}}(90{}^\circ -\theta )=co{{s}^{2}}(90{}^\circ -\theta )+co{{s}^{2}}\theta .\]

\[L.H.S.=se{{c}^{2}}\theta -co{{t}^{2}}(90{}^\circ -\theta )\]

\[=se{{c}^{2}}\theta -\text{  }\!\![\!\!\text{ cot}\,\text{(90-}\theta \text{)}{{\text{ }\!\!]\!\!\text{ }}^{2}}\]

\[=se{{c}^{2}}\theta -{{(tan\,\,\theta )}^{2}}\]

\[=se{{c}^{2}}\theta -ta{{n}^{2}}\theta \]

\[=1\]

\[R.H.S.={{\cos }^{2}}(90-\theta )+co{{s}^{2}}\theta \]

\[=[cos{{(90-\theta )}^{2}}]+co{{s}^{2}}\theta \]

\[={{(sin\,\theta )}^{2}}+{{\cos }^{2}}\theta \]

\[={{\sin }^{2}}\theta +{{\cos }^{2}}\theta \]

\[=1\]

Hence,                L.H.S. = R.H.S.                                              Hence Proved