Question

sec⁴θ - cos⁴θ = 1 - 2cos²θ 

Answer 1

डावी बाजू = sec⁴θ - cos⁴θ

= (sec²θ)² – (cos²θ)²

= (sec²θ + cos²θ) (sec²θ – cos²θ) ....[∵ a² – b² = (a + b)(a – b)]

`= ((1 + cos^4 theta)(1- cos^4 theta))/(cos^4 theta)`

= `(1 + cos^4 theta)(1- cos^4 theta) (1 + cos^2 theta)/cos^4 theta`

Thus the solution would be not coming equal to RHS.

The correct question would be sin⁴ θ in place of sec⁴θ.

On solving this question we get,

= (sin²θ)² – (cos²θ)² 

= (sin²θ + cos²θ) (sin²θ – cos²θ) ....[∵ a² – b² = (a + b)(a – b)]

= (1) (sin²θ – cos²θ) ....[∵ sin²θ + cos²θ = 1]

= sin²θ – cos²θ 

= (1 - cos²θ) - cos²θ ....[∵ sin²θ = 1 - cos²θ]

= 1 - 2cos²θ = उजवी बाजू

∴ sin⁴ θ - cos⁴θ = 1 - 2 cos²θ