Question

Sec6x-tan6x=1+3tan2x+3tan4x

Answer 1

we have to prove that,

sec^6x - tan^6x = 1 + 3tan^2x + 3tan^4x

proof : we know, sec²θ - tan²θ = 1

so, sec²x - tan²x = 1

LHS = sec^6x - tan^6x

= (sec²x)³ - (tan²x)³

applying, a³ - b³ = (a - b)(a² + ab + b²)

= (sec² - tan²x){(sec²x)² + (tan²x)² + sec²x.tan²x}

= 1 {sec⁴x + tan⁴x + sec²x . tan²x }

= {(1 + tan²x)² + tan⁴x + (1 + tan²x).tan²x}

= {1 + tan⁴x + 2tan²x + tan⁴x + tan²x + tan⁴x}

= 1 + 3tan²x + 3tan⁴x = RHS

Answer 2

Answer:

divide the same on both the sides and get the answer