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If cos theta + Sin theta =m and sec theta + Coses theta =n prove n(m²-1)= 2m |
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Answer» Given that, sin theta + COS theta= m, SEC theta + COSEC theta = n.Consider LHSn(m^2 - 1)n(m 2 −1)(sec\theta + cosec\theta)[(cos\theta + sin\theta)^2 - 1](secθ+cosecθ)[(cosθ+sinθ) 2 −1](sec\theta + cosec\theta)[cos^2 \theta + sin^2 \theta + 2sin\theta cos\theta - 1](secθ+cosecθ)[cos 2 θ+sin 2 θ+2sinθcosθ−1](sec\theta + cosec\theta)[2sin\theta cos\theta] since {cos^2 \theta + sin^2 \theta = 1}(secθ+cosecθ)[2sinθcosθ]since cos 2 θ+sin 2 θ=1sec\theta . 2sin\theta cos\theta + cosec\theta . 2sin\theta cos\thetasecθ.2sinθcosθ+cosecθ.2sinθcosθ2sin\theta + 2 cos\theta2sinθ+2cosθ2[sin\theta + cos\theta]2[sinθ+cosθ]= 2mStep-by-step explanation:hope it helps youplease give me LIKES to support my answers.... |
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