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Cos squaretheta +cos squarethetacotsquretheta=cot squaretheta |
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Answer» Here, I replaced the Theta "θ"\xa0with "x".Please take θ in your solution. To prove : cos2x + (cos2x X cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx) = cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx ?Solution : Taking the LHS => cos2x + (cos2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx X cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx)\xa0 (Taking\xa0cos2x common from both\xa0the terms) =>\xa0cos2x (1 +\xa0cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx)\xa0 (Since 1 +\xa0cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx = cosec2x ) =>\xa0cos2x (cosec2x\xa0)\xa0 (Since cosec2x = 1/ sin2x ) =>\xa0 cos2x / sin2x => (cos x / sin x )2 (Since cosx / sinx = cotx ) => (cot\xa0x )2=> cot2xwhich is equals to RHSThus, LHS = RHS\u200b\u200b\u200b\u200b\u200b\u200b\u200bHence proved. |
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