Question

If cot theta =7/8
evaluate
(1 + sin 6)(1-sino)
(1 + cos ) (1 - cos )​

Answer 1

Answer:

49 /64

Step-by-step explanation:

cotA = 7/8 = base /perpendicular

h = √(7^2 + 8^2) = √113

sinA = 8/√113 & cosA = 7/√113

(1+ sinA)(1- sinA)/(1 + cosA)(1 - cosA )

= (1^2- sin^2A)/(1^2 - cos^2A)

=[1- (8/√113)^2]/[1 - (7/√113)^2]

= 49/113/64/113 = 49/64

Answer 2

Answer:

Step-by-step explanation:

given is cotθ=7/8 

opposite side = 8 and adjacent side=7

so, hypotenuse will be square root 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