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
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