# Trig Identities: a) sin^4 x - cos^4 x / sin^2 x - cos^2 x = 1 b) 1-2 cos^2 x = tan^2 x-1/tan^2 x+1 c) secx + tanx = cosx/1 - sinx Most of these identities hold true but some of them are...

Trig Identities:

a) sin^4 x - cos^4 x / sin^2 x - cos^2 x = 1

b) 1-2 cos^2 x = tan^2 x-1/tan^2 x+1

c) secx + tanx = cosx/1 - sinx

Most of these identities hold true but some of them are false. Prove LS = RS or show why the rule does not hold

### 1 Answer | Add Yours

a) `(sin^4(x) - cos^4(x))/(sin^2(x)-cos^2(x)) = 1`

LHS:

`((sin^2(x) - cos^2(x))(sin^2(x)+ cos^2(x)))/(sin^2(x)-cos^2(x))`

We know, from basic trignometry,

sin^2(x)+ cos^2(x) = 1

Therefore,

`LHS=((sin^2(x) - cos^2(x))(1))/(sin^2(x)-cos^2(x))`

LHS = 1.

Therefore LHS = RHS and

`(sin^4(x) - cos^4(x))/(sin^2(x)-cos^2(x)) = 1`

b) `(1-2cos^2(x)) = (tan^2(x)-1)/(tan^2(x)+1)`

For this I will start with RHS.

`RHS = (tan^2(x)-1)/(tan^2(x)+1)`

`= ((sin^2(x))/(cos^2(x))-1)/((sin^2(x))/(cos^2(x))+1)`

`= (sin^2(x)-cos^2(x))/(sin^2(x)+cos^2(x))`

`= (sin^2(x)-cos^2(x))/1`

`= (1-cos^2(x))-cos^2(x)`

`= 1-2cos^2(x)`

Therefore LHS = RHS and

`(1-2cos^2(x)) = (tan^2(x)-1)/(tan^2(x)+1)`

c) `sec(x)+tan(x) = cos(x)/(1-sin(x))`

`LHS = sec(x)+tan(x)`

`= 1/cos(x) + sin(x)/cos(x)`

`=(1+sin(x))/cos(x)`

Multiplying both numerator and denominator by `(1-sin(x))`

`= (1+sin(x))(1-sin(x))/(cos(x)(1-sin(x)))`

`= (1-sin^2(x))/(cos(x)(1-sin(x)))`

`= (cos^2(x))/(cos(x)(1-sin(x)))`

`= cos(x)/(1-sin(x))`

LHS = RHS and

`sec(x)+tan(x) = cos(x)/(1-sin(x))`

All the identities are true.