If m=sin +cos theta and n = sec theta+cosec theta then prove that n(m 2 -1)=2m
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m=sinθ+cosθ and n=secθ+cosecθ
n(m²-1)
=(secθ+cosecθ)[(sinθ+cosθ)²-1]
=(1/cosθ+1/sinθ)(sin²θ+2sinθcosθ+cos²θ-1)
={(sinθ+cosθ)/sinθcosθ}(2sinθcosθ) [ Since, sin²θ+cos²θ=1]
=2(sinθ+cosθ)
=2m (Proved)
n(m²-1)
=(secθ+cosecθ)[(sinθ+cosθ)²-1]
=(1/cosθ+1/sinθ)(sin²θ+2sinθcosθ+cos²θ-1)
={(sinθ+cosθ)/sinθcosθ}(2sinθcosθ) [ Since, sin²θ+cos²θ=1]
=2(sinθ+cosθ)
=2m (Proved)
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