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1-sec-sec-sin-2-1-cos-




Question Number 19659 by thukada last updated on 14/Aug/17
((1+secθ)/(secθ))=((sin^(2 ) θ)/(1−cosθ))
1+secθsecθ=sin2θ1cosθ
Answered by prakash jain last updated on 14/Aug/17
((1+sec θ)/(sec θ))=1+(1/(sec θ))  =1+cos θ=(((1+cos θ)(1−cos θ))/((1−cos θ)))  =((1−cos^2 θ)/(1−cos θ))=((sin^2 θ)/(1−cos θ))■
1+secθsecθ=1+1secθ=1+cosθ=(1+cosθ)(1cosθ)(1cosθ)=1cos2θ1cosθ=sin2θ1cosθ◼

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