# Show that sec^4 x - sec^2 x = tan^4x + tan^2x

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### 2 Answers

The trigonometric identity `sec^4 x - sec^2 x = tan^4x + tan^2x ` has to be proved.

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

Start from the left hand side

`sec^4 x - sec^2 x `

As shown earlier `sec^2x = 1 + tan^2x` . Make the appropriate substitution.

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

= `1 + tan^4x + 2*tan^2x - 1 - tan^2x`

= `tan^4x + tan^2x`

**This proves that `sec^4 x - sec^2 x = tan^4x + tan^2x` **

Problem is to prove

`sec^4 (x)-sec^2(x)=tan^4(x)+tan^2(x)`

Before proving let us recall

`1+tan^2(x)=sec^2(x) ` ,

Thus

`LHS=sec^4(x)-sec^2(x)`

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

`=(1+tan^2(x))(tan^2(x)`

`=tan^2(x)+tan^4(x)`

`=tan^4(x)+tan^2(x)=RHS`

Hence proved.