Question

sin^2x+tan^2x= sec^2x - cos^2x

How to prove this identity

Answer 1

I have to prove that LS = RS?

Answer 2

You usually start with the most complex side. Either side seems complex.

Answer 3

he needs to know Left side and make it right side.

Answer 4

The easiest way to do it is to break down the tan^2x and sec^x into sin and cos,

tan^2x = sin^x/cos^x and sec^x = 1/cos^x

Answer 5

I got one side to equal

(1-cos^4x)/cos^2x the left side but how do i do this to the right side ?

Answer 6

wolfram alpha solution, with steps and all

Answer 7

You usually do one side, and leave the other side alone

Answer 8

u can zoom in

Answer 9

wow ill take a look at that :)

Answer 10

i know another way to prove it
use 30 degrees and if its equal, it works

Answer 11

sec^2x = 1/cos^2x and cos^2x = cos^2x /1

(1/cos^2x) - (cos^2x /1)

Get them both over the same denominator by multiplying the second term (cos^2x /1) by (cos^2x/cos^2x)

You now have (1/cox^2x) - (cos^4/cos^2x), which equals (1-cos^4x)/cos^2x

Answer 12

left side
Sin^2x + tan^2x = 1
sin^2x + sin^2/cos^2x
sin^2xcos^2x + sin^2x all over cos^2x
sin^2xcos^2x + (1-cos^2x) /cos^2x
sin^2xCos^2x + 1 - cos^2x /cos^2x
1 + cos^2x (sin^2x -1) /cos^2x
1/cos^2x - cos^2x
sec^2x- cos^2x = right side