Question

if cot A = 7/8 , evaluate ( 1+sinA) (1-sinA) / (1+cos A) (1-cos A)​

Answer 1

Answer:

Heya..

Step-by-step explanation:

Let AB = 7k and BC = 8k, where k is a positive integer.

By applying Pythagoras theorem in Δ ABC, we get

AC2 = AB2 + BC2

= (7k)2 + (8k)2

= 49k2 + 64k2

= 113k2

AC = √113k2

= √113k

Therefore, sin θ = side opposite to θ / hypotenuse = BC/AC = 8k/√113k = 8/√113

cos θ = side adjacent to θ / hypotenuse = AB/AC = 7k/√113k = 7/√113

(i) (1 + sin θ) (1 - sin θ) / (1 + cos θ) (1 - cos θ) = 1 - sin2θ / 1 - cos2θ  [Since, (a + b)(a - b) = (a2 - b2)]

= [1 - (8/√113)2] / [1 - (7/√113)2]

= (1 - 64/113) / (1 - 49/113)

= (49/113) / (64/113)

= 49/64

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Answer 2

Answer:

cotθ= 87

tanθ= cotθ

1

= 78

We know that,

tanθ= adjacentSide

oppositeSide

From Pythagoras theorem,

Hypotenuse 2

=OppositeSide 2

+AdjacentSide 2

Hypotenuse 2 =8

2

+7

2

Hypotenuse 2

=64+49=113

Hypotenuse=

113

sinθ=

Hypotenuse

oppositeSide = 113 8

cosθ=

Hypotenuse

AdjacentSide = 113

Step-by-step explanation:

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