Question

$\frac{ \tan{}^{3} theta - 1}{ \tan \: theta - 1 } = \sec^{2} theta \: + \tan \: theta$

pls solve
no fraud answers pls​

Answer 1

Answer:

Put 1=sec

2

θ−tan

2

θ=(secθ−tanθ)(secθ+tanθ)

⟹

tanθ−secθ+1

secθ+tanθ−(secθ−tanθ)(secθ+tanθ)

⟹

(1−secθ+tanθ)

secθ+tanθ(1−secθ+tanθ)

⟹secθ+tanθ

Put secθ=

cosθ

1

and tanθ=

cosθ

sinθ

⟹

cosθ

1

+

cosθ

sinθ

Multiply and divide by 1−sinθ

⟹

cosθ(1−sinθ)

1−sin

2

θ

⟹

1−sinθ

cosθ