Question

sec^6A-tan^6A-3sec^2A*tan^2A​

Answer 1

Answer:

sec

6

A−tan

6

A=1+3tan

2

A+3tan

4

A

We know that a

3

−b

3

=(a−b)(a

2

+ab+b

2

)

∴sec

6

A−tan

6

A=(sec

2

A−tan

2

A)((sec

2

A)

2

+sec

2

Atan

2

A+(tan

2

A)

2

)

sec

6

A−tan

6

A=((1+tan

2

A)−tan

2

A)((1+tan

2

A)

2

+(1+tan

2

A)tan

2

A+tan

4

A)[∵sec

2

A=1+tan

2

A]

sec

6

A−tan

6

A=(1+tan

2

A−tan

2

A)(1+tan

4

A+2tan

2

A+tan

2

A+tan

4

A+tan

4

A)

sec

6

A−tan

6

A=1+3tan

2

A+3tan

4

A