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If cosecθ-sinθ =m^3 and secθ-cosθ=n^3 then prove that n^2 m^4 +n^4 m^2 = 1 |
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Answer» cosec∅ - sin∅ = m -----(1) sec∅ - cos∅ = N --------(2) from EQUATION ,(1) cosec∅ - sin∅ = m 1/sin∅ - sin∅ = m 1 - sin²∅ = m.sin∅ sin²∅ + m.sin∅ -1 = 0 m.sin∅ = 1 - sin²∅ = cos²∅ m = cos²∅/sin∅ ---------(3) again, similarly , from equation ,(2) sec∅ - cos∅ = n 1/cos∅ - cos∅ = n 1 - cos²∅ = n.cos∅ cos²∅ + n.cos∅ -1 = 0 n.cos∅ = 1 - cos²∅ = sin²∅ n = sin²∅/cos∅ -------(4) put equation (3) and (4) in (m²n)⅔ + (mn²)⅔ LHS = {(cos²∅/sin∅)²(sin²∅/cos∅)}⅔ + {(cos²∅/sin∅)(sin²∅/cos∅)²}⅔ = { cos⁴∅ × sin²∅/sin²∅ × cos∅}⅔ + { cos²∅×sin⁴∅/sin∅×cos²∅}⅔ = { cos³∅}⅔ + {sin³∅}⅔ = cos²∅ + sin²∅ = 1 = RHS [ we KNOW , sin²x + cos²x = 1 ] mark me as BRAINLIEST
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