Question

please anyone please solve it... If u get the answer I will make it brainliest answer...... ; )

Answer 1

$<h4>SOLUTION :$

L. H. S. = Sec⁴ θ ( 1 - Sin ² θ )² - 2 tan² θ

=> Sec⁴ θ ( cos² θ )² - 2 tan²θ

=> sec ⁴ θ × cos⁴ θ - 2 tan² θ

=> ( 1 / Cos⁴ θ ) × cos⁴ θ - 2 tan² θ

=> 1 - 2tan ² θ

=> 1 - 2 sin² θ/ Cos² θ

=> ( cos² θ - 2 sin² θ ) / Cos² θ

=> [cos ² θ - 2 ( 1 - cos² θ)] /Cos ² θ

=> ( cos² θ - 2 + 2 cos² θ ) /cos² θ

=>[ cos² θ ( 1 - 2 +2 )] /Cos ² θ

=> 1

= R. H. S.

Answer 2

Answer:

L. H. S. = Sec⁴ θ ( 1 - Sin ² θ )² - 2 tan² θ

=> Sec⁴ θ ( cos² θ )² - 2 tan²θ

=> sec ⁴ θ × cos⁴ θ - 2 tan² θ

=> ( 1 / Cos⁴ θ ) × cos⁴ θ - 2 tan² θ

=> 1 - 2tan ² θ

=> 1 - 2 sin² θ/ Cos² θ

=> ( cos² θ - 2 sin² θ ) / Cos² θ

=> [cos ² θ - 2 ( 1 - cos² θ)] /Cos ² θ

=> ( cos² θ - 2 + 2 cos² θ ) /cos² θ

=>[ cos² θ ( 1 - 2 +2 )] /Cos ² θ

=> 1

= R. H. S.

its your answer ☺️

nice to help you