Question

Prove that:

\[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A = 1\]

Answer 1

L.H.S = \[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A\]

\[ = \sec^4 A - \sec^4 A \sin^4 A - 2 \tan^2 A\]

= `(sec^2A)^2 - 1/(cos^4A). sin^4A - 2tan^2A   ...[secθ = 1/cosθ]`

= `(1 + tan^2A)^2 - (sin^4A)/(cos^4A) - 2 tan^2A  ...[1 + tan^2θ = sec^2θ]`

= `1^2 + 2 xx 1 xx  tan^2A + (tan^2A)^2 - tan^4A - 2tan^2A   ...[(a + b)^2 = a^2 + 2ab + b^2 tanθ = (sinθ)/(cosθ)]`

= `1 + cancel(2tan^2A) + cancel(tan^4A) - cancel(tan^4A) - cancel(2tan^2A)`

= 1

= R.H.S