Question

If cotθ = 7/8, evaluate
(i)(1+sinθ)(1-sinθ)/(1+cosθ)(1-cosθ)
(ii)cot2θ

Answer 1

Answer:

Refer attachment for your answer

Answer 2

(i) Required to evaluate:= cot θ = 7/8

Taking the numerator, we have

(1+sin θ)(1–sin θ) = 1 – sin2 θ [Since, (a+b)(a-b) = a2 – b2]

Similarly,

(1+cos θ)(1–cos θ) = 1 – cos2 θ

We know that,

sin2 θ + cos2 θ = 1

⇒ 1 – cos2 θ = sin2 θ

And,

1 – sin2 θ = cos2 θ

Thus,

(1+sin θ)(1 –sin θ) = 1 – sin2 θ = cos2 θ

(1+cos θ)(1–cos θ) = 1 – cos2 θ = sin2 θ

⇒

= cos2 θ/ sin2 θ

= (cos θ/sin θ)2

And, we know that (cos θ/sin θ) = cot θ

⇒

= (cot θ)2

= (7/8)2

= 49/ 64

(ii) Given,

cot θ = 7/8

So, by squaring on both sides we get

(cot θ)2 = (7/8)2

∴ cot θ2 = 49/64