Question

if cotO= 7/8, evaluate (1+sinO)(1-sinO)/(1+cosO)(1-cosO)​

Answer 1

Answer: 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

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Answer 2

Answer:

49/64

Step-by-step explanation:

1^2 - sin^2O/1^2 - cos^2O

1 - sin^2O/1 - cos^2O

cos^2O/sin^2O

cot^2O

(7/8)^2

49/64