If 0<theta<90°, sin theta+ cos theta=P and q=2p/p^2=1
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Answered by
1
Answer:
p=sinθ+cosθ and q=secθ+cosecθ
q(p²-1)
=(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)
=2p (Proved)
Answered by
0
p=sinθ+cosθ and q=secθ+cosecθ
q(p²-1)
=(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)
=2p (Proved)
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