InterviewSolution
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InterviewSolution
| 1. |
1+cot square alpha/1+cosec alpha= cosec alpha |
| Answer» LHS\xa0{tex} = 1 + \\frac{{{{\\cot }^2}\\theta }}{{1 + \\cos ec\\theta }}{/tex}{tex} = 1 + \\frac{{\\cos e{c^2}\\theta - 1}}{{1 + \\cos ec\\;\\theta }}{/tex}\xa0{tex}\\left[ {\\because {{\\cot }^2}\\theta = \\cos e{c^2}\\theta - 1} \\right]{/tex}{tex} = 1 + \\frac{{(\\cos ec\\theta + 1)(\\cos ec\\theta - 1)}}{{(1 + \\cos ec\\theta )}}{/tex}\xa0{tex}\\left[ {\\because {a^2} - {b^2} = (a + b)(a - b)} \\right]{/tex}{tex} = 1 + \\cos ec\\;\\theta - 1{/tex}{tex} = \\cos ec\\theta {/tex}= RHSHence proved. | |