if cot theta=7/8 evaluate (1+sin theta)(1 - sin theta)/ (1+cot theta)(1-cos theta)
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Answered by
818
the attachment is the solution
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TrapNation:
brainliest
but excuse me which standard you are in cause so easily you solved it
10
in question in denominator it is (1-cot theta)(1+cos theta) u took it as 1-cos^2theta
she is in 10th now :)
the question is wrng =_=
typos error
it is understood btw -_-
thanks.. :D
Answered by
231
Hi friend,
given is cotθ=7/8 ⇒ opposite side = 8 and adjacent side=7 so, hypotenuse will be squareroot of sum of sqaures of 7 and 8 i.e sqrt7^2+8^2=√113 so now sinθ=8/√113 and cosθ=7/√113
(1+sinθ)(1-sinθ)/(1+cotθ)(1-cosθ)=1-sin^2θ/1+cotθ(1-cosθ)
= 1-64/113/1+7/8(1-7/√113)
on solving we get 49/(√113-7)√113
given is cotθ=7/8 ⇒ opposite side = 8 and adjacent side=7 so, hypotenuse will be squareroot of sum of sqaures of 7 and 8 i.e sqrt7^2+8^2=√113 so now sinθ=8/√113 and cosθ=7/√113
(1+sinθ)(1-sinθ)/(1+cotθ)(1-cosθ)=1-sin^2θ/1+cotθ(1-cosθ)
= 1-64/113/1+7/8(1-7/√113)
on solving we get 49/(√113-7)√113
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