Question

if cot θ = 7/8, evaluate: (i) (1+ sin θ )(1- sin θ)/ (1+ cos θ)(1- cos θ) (ii) cot^6 θ

Answer 1

I will use q symbol as theta below
cotq=7/8 ie b/p
b=7,p=8
h²=b²+p²
h² =7²+8²
=49+64
=113
h²=113
h=√113
(1)(1+sinq)(1-sinq) /(1+cosq)(1-cosq)
(1-sin²q) / (1-cos²q)
{1-(8/√113)²} / {1-(7/√113)²}
1-64/113 / 1-49/113
113-64/113 / 113-49/113
49/113 / 64/113
49/64
(2)
${ cotq}^{6}$
(7/8)^6
117649/262144

Answer 2

Given:

$\tt \: cot \theta = \dfrac{7}{8} \\ \\ \tt \: tan \theta = \dfrac{1}{cot \theta} = \dfrac{8}{7}$

We know that,

$\tt tan \theta = \dfrac{opposite \: side}{adjacent \: side}$

From Pythagoras theorem,

hypotenuse² = opposite side² + adjacent side²

hypotenuse² = 8² + 7²

hypotenuse² = 64 + 49 = 113

hypotenuse² = ⬇️

$\tt \sqrt{113}$

Solution:

$\tt \dfrac{(1 + \sin\theta)(1 - \sin\theta )}{(1 + \ \cos \theta)(1 - \cos \theta )}$

We have,

a² - b² = (a+b)(a-b)

Similarly,

$\tt (1 + \ {sin}^{2} \theta) = (1 + { sin \theta \: } )(1 - sin \theta)$

$\tt (1 - \ {cos}^{2} \theta) = (1 + { cos \theta \: } )(1 - cos \theta)$

$\tt = \dfrac{{1 -( \frac{8}{ \sqrt{113} }^){2} } }{1 - \frac{7}{{ \sqrt{113} }^{2} } } \\ \\ = \dfrac{49}{64}$