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| 1. |
1+cot^2/1+cosec=1/sin |
| Answer» LHS= \u200b{tex}\\frac{1}{\\cot ^{2} \\theta}+\\frac{1}{1+\\tan ^{2} \\theta}{/tex}= tan2\xa0{tex}\\theta{/tex}\xa0+\xa0{tex}\\frac{1}{\\sec ^{2} \\theta}{/tex}= tan2\xa0{tex}\\theta{/tex}\xa0+ cos2\xa0{tex}\\theta{/tex}= (sec2{tex}\\theta{/tex}\xa0- 1) +cos2{tex}\\theta{/tex}= sec2{tex}\\theta{/tex}\xa0- ( 1 - cos2{tex}\\theta{/tex}\xa0)= sec2{tex}\\theta{/tex}\xa0- sin2{tex}\\theta{/tex}{tex}=\\frac{1}{\\cos ^{2} \\theta}{/tex}\xa0- sin2{tex}\\theta{/tex}{tex}= \\frac{1}{1-\\sin ^{2} \\theta}-\\frac{1}{\\ cosec ^{2} \\theta}{/tex}\u200b\u200b\u200b\u200b\u200b= R.H.S\xa0Hence proved. | |