(1+sintheta)(1-sintheta)/(1+costheta)(1-costheta) evaluate if cot theta=7/8
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Answered by
6
Answer:
(1-sinthita)(1+sinthita)/(1-costhita)(1+costhita)=1-sin2thita/ 1-cos2thita
=cos2thita/sin2thita ( :1-sin2thita=cos2thita and 1-cos2thita= sin2thita)
=cot2thita
And cotthita= 7/8
So cot2thita= 49/64
Step-by-step explanation:
domnalnishra:
thnku
Welcome
cant we do 1=sin*1-sin=
We can
1-sin*1+sin=1^2+sin^
2+2sintheta and make it cos^2theta
then what happens to sin a
plz tl
I think I should think it first
Answered by
4
Given cotθ = 7/8
Now {(1 + sinθ)*(1 - sinθ)}/{(1 + cosθ)*(1 - cosθ)}
= (1 - sin2 θ)/(1 - cos2 θ)
= cos2 θ/sin2 θ {since sin2 θ + cos2 θ = 1 }
= cot2 θ
= (7/8)2
= 49/64
Now {(1 + sinθ)*(1 - sinθ)}/{(1 + cosθ)*(1 - cosθ)}
= (1 - sin2 θ)/(1 - cos2 θ)
= cos2 θ/sin2 θ {since sin2 θ + cos2 θ = 1 }
= cot2 θ
= (7/8)2
= 49/64
thnku
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