Question

Sec4A(1-Sin4A) - 2tan2A=1

prove the following ​

Answer 1

Answer:

Step-by-step explanation:

sec⁴A(1-sin⁴A)-2tan²A

=sec⁴A-sec⁴Asin⁴A-2tan²A

=sec⁴A-sin⁴A(1/cos⁴A)-2tan²A

=sec⁴A-tan⁴A-2tan²A

={(sec²A)²-(tan²A)²}-2tan²A

=(sec²A+tan²A)(sec²A-tan²A)-2tan²A

=sec²A+tan²A-2tan²A [∵, sec²A-tan²A=1]

=sec²A-tan²A

=1 (Proved)

Answer 2

Answer :

sec4A ( 1 - sin4A) - 2tan2A =

1/ COS4A (1 - sin4A) - 2tan2A

= (1/ COS4A - sin4A/ COS4A) - 2tan2A

= { ( 1 -S IN4A)/ COS4A }- 2tan2A

= { (1 - SIN2 A ) (1+SIN2 A)/ COS4A }- 2tan2A

={COS2A (1+SIN2 A)/ COS4A }- 2tan2A

= (1+SIN2 A)/ COS2A - 2S IN2A / COS2 A

=( (1+SIN2 A) - 2S IN2A) / COS2 A

= (1 - SIN2 A )/ COS2 A = COS2 A / COS2 A = 1

(1 - SIN2 A ) = COS2 A

So Cos'2A = Cos'2A

So Proved 1=1