The trigonometric identity `cos^4x+sin^2x*cos^2x + sin^2x + tan^2x = 1/(cos^2 x)` has to be proved.
Start with the left hand side:
`cos^4x+sin^2x*cos^2x + sin^2x + tan^2x`
= `cos^4x+sin^2x*cos^2x + sin^2x + (sin^2x)/(cos^2x)`
= `cos^4x+(1 - cos^2x)*cos^2x + 1 - cos^2x + (1 - cos^2x)/(cos^2x)`
= `cos^4x+cos^2x - cos^4x + 1...
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The trigonometric identity `cos^4x+sin^2x*cos^2x + sin^2x + tan^2x = 1/(cos^2 x)` has to be proved.
Start with the left hand side:
`cos^4x+sin^2x*cos^2x + sin^2x + tan^2x`
= `cos^4x+sin^2x*cos^2x + sin^2x + (sin^2x)/(cos^2x)`
= `cos^4x+(1 - cos^2x)*cos^2x + 1 - cos^2x + (1 - cos^2x)/(cos^2x)`
= `cos^4x+cos^2x - cos^4x + 1 - cos^2x + (1 - cos^2x)/(cos^2x)`
= `1 + (1 - cos^2x)/(cos^2x)`
= `(cos^2x)/(cos^2x) + (1 - cos^2x)/(cos^2x)`
` `=> `1/(cos^2x)`
This proves that ` cos^4x+sin^2x*cos^2x + sin^2x + tan^2x = 1/(cos^2 x)`