Question

"Question6
If cot θ=15/8,Then evaluate [2+2sin θ][1-sinθ]/[1+cos θ][2-2 cos θ].
Chapter 5 , Trigonometry, Exercise -5 , Page number - 273"

Answer 1

Hey there!
Given cotθ = 15/8
Adjacent side = 15 , Opposite side = 8

By Pythagoras theorem, Hypotenuse = 17

Now, sinθ = 8/17 ( Opposite side to theta / Hypotenuse
cosθ = 15/17 ( Adjacent side to theta / Hypotenuse)

Now,
$= \frac{ (2 +2 sin \theta) ( 1 - sin\theta ) } { (1 + cos \theta ) ( 2 - 2 cos \theta) }$

= $\frac { 2 ( 1 + sin \theta) ( 1 - sin \theta) }{ 2( 1- cos \theta ) ( 1 + cos \theta ) }$

= $\frac{ 1 - sin^2 \theta } { [ 1 - cos^2 \theta ] }$

Now, We can either directly insert sinθ and cosθ or use formula. I am doing it by formula

We know that, sin²θ + cos²θ = 1 , From this, 1 - cos²θ = sin²θ , 1 - sin²θ = cos²θ

So, $\frac{ 1 - sin^2 \theta } { [ 1 - cos^2 \theta ] }$

= cos²θ / sin²θ

= cot²θ .

= 15²/8²

= 225/64

[ Note : There was no need to find sinθ and cosθ values. But if we were to substitute them, we should find them ]

Therefore, Value of $\frac{ (2 +2 sin \theta ) ( 1 - sin \theta ) } { (1 + cos \theta ) ( 2 - 2 cos \theta ) }$ is 225/64 if cotθ = 15/8

Answer 2

HELLO DEAR,

GIVEN THAT:-

cot∅ = 15/8

on squaring both side,

we get,

cot²∅ = 225/64--------(1)

now,

[2+2sin θ][1-sinθ]/[1+cos θ][2-2 cos θ]

= [2(1 + sin∅)(1 - sin∅)] /[(1 + cos∅)(1 - cos∅)2]

= 2(1 - sin²∅)/2(1 - cos²∅)

=cos²∅/sin²∅
[ sin²∅ + cos²∅ = 1]

= cot²∅ = 225/64 ------from-----(1)

I HOPE ITS HELP YOU DEAR,
THANKS