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Third question

Dear student,

W e h a v e, \cos θ + \cos^{2} θ = 1 \cos θ = 1 - \cos^{2} θ \cos θ = \sin^{2} θ . . . (1) T o p r o v e : \sin^{12} θ + 3 \sin^{10} θ + 3 \sin^{8} θ + \sin^{6} θ + 2 \sin^{4} θ + 2 \sin^{2} θ - 2 = 1 T a k i n g L H S \sin^{12} θ + 3 \sin^{10} θ + 3 \sin^{8} θ + \sin^{6} θ + 2 \sin^{4} θ + 2 \sin^{2} θ - 2 = {(\sin^{4} θ)}^{3} + 3 {(\sin^{4} θ)}^{2} (\sin^{2} θ) + 3 (\sin^{4} θ) {(\sin^{2} θ)}^{2} + {(\sin^{2} θ)}^{3} + 2 (\sin^{4} θ + \sin^{2} θ) - 2 . . . [u \sin g e x p o n e n t i a l l a w s] = \{{(\sin^{4} θ)}^{3} + {(\sin^{2} θ)}^{3} + 3 {(\sin^{4} θ)}^{2} (\sin^{2} θ) + 3 (\sin^{4} θ) {(\sin^{2} θ)}^{2}\} + 2 (\sin^{4} θ + \sin^{2} θ) - 2 = {(\sin^{4} θ + \sin^{2} θ)}^{3} + 2 (\sin^{4} θ + \sin^{2} θ) - 2 . . . [u \sin g {(a + b)}^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}] = {(\cos^{2} θ + \sin^{2} θ)}^{3} + 2 (\cos^{2} θ + \sin^{2} θ) - 2 [u \sin g (1)] = 1^{3} + 2 \times 1 - 2 . . . [\sin c e \cos^{2} θ + \sin^{2} θ = 1] = 1 + 2 - 2 = 1 ∴ L H S = R H S

Hence proved

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