Question

(cot^2x/(cscx + 1)) = cos^2x/(sin x(1+sin x)) help me prove the identity simplifying the right side

Answer 1

you have \[\frac{\cos^2 x}{\sin x (1 + \sin x)}\]
first express cos^2 x in terms of sin..you know the pythagorean identity yes?

Answer 2

not well

Answer 3

here's a hint:
\[\sin^2 x + \cos^2 x = 1\]

Answer 4

ok so cos^2x = 1 - sin^2x

Answer 5

right. so if you put that in your equation \[\frac{1 - \sin^2 x}{\sin x (1 + \sin x)}\]

do you agree that the numerator is a difference of squares?

Answer 6

yes

Answer 7

so if you expand the numerator, what would it be?

Answer 8

(1 + sin x)( 1 - sin x)

Answer 9

right.
\[\frac{(1 + \sin x)(1 - \sin x)}{\sin x (1 + \sin x)}\]
something can be cancelled here

Answer 10

(1- sin x)/ sin x

Answer 11

correct?

Answer 12

right. and since the numerator is subtraction, you can separate the terms \[\frac 1{\sin x} - \frac{\sin x}{\sin x}\]
simplify

Answer 13

wait how can you simplify that?

Answer 14

do you mean how i got there? or how to go from there?

Answer 15

how to go from there

Answer 16

it helps if you know the trigonometric identity \(\csc x = \frac 1{\sin x}\)
the second term...i assume you know how to simplify that?

Answer 17

ok so csc x - 1

Answer 18

then what?

Answer 19

now... a trigonometry magic.... multiply numerator and denominator by csc x + 1

Answer 20

\[\csc x - 1 \times \frac{\csc x + 1}{\csc x + 1}\]

Answer 21

so csc^2x + 1/ csc x + 1

Answer 22

i mean csc^2x - 1 which equals tan^2x

Answer 23

csc^2 x - 1 is cot^2 x not tan^2 x

Answer 24

tan^2 x is sec^2 x - 1

Answer 25

so your equation has been proven. QED

Answer 26

ok thankyou

Answer 27

LHS, numerator \[cot^2 x= \frac{cos ^2x}{sin^2x}\]
denominator \[csc x +1= \frac{1}{sin x}+1= \frac{1+sinx}{sinx}\]combine them
\[\huge\frac{\frac{cos^2x}{sin^2x}}{\frac{1+sinx}{sinx}}\]
\[= \frac{cos^2x}{sin^2x}*\frac{sinx}{1+sinx}\]
cancel out sinx from the terms
\[\frac{cos^2x}{sinx(1+sinx)}\] and that's it