Question

sec⁶θ= tan⁶θ + 3 tan²θ sec²θ +1​

Answer 1

Step-by-step explanation:

sec

6

θ−tan

6

θ−3sec

2

θtan

2

θ=1

Taking LHS

sec

6

θ−tan

6

θ−3sec

2

θtan

2

θ

=(sec

2

θ)

3

−(tan

2

θ)

3

−3sec

2

θtan

2

θ a

3

−b

3

=(a−b)(a

2

+b

2

+ab)

=(sec

2

θ−tan

2

θ)(sec

4

θ+tan

4

θ+sec

2

θtan

2

θ)−3sec

2

θtan

2

θ {sec

2

θ−tan

2

θ=1}

=1(sec

4

θ+tan

4

θ−2sec

2

θtan

2

θ)

(sec

2

θ−tan

2

θ)

2

or (tan

2

θ−sec

2

θ)

2

(1)

2

or (−1)

2

=1

Answer 2

plz find the attachment…hope this hlp