Question

prove that :
(sec theta - tan theta)² = 1-sin theta/1+sin theta​

Answer 1

Answer:

Hey there !!

Prove that :-

→ (secθ - tanθ)² = \bf \frac{ 1-sin \theta }{1+sin \theta }

1+sinθ

1−sinθ

.

Solution :-

→ (sec θ - tan θ )².

⇒ (\bf \frac{1}{ cos \theta } - \frac{ sin \theta }{ cos \theta }

cosθ

1

−

cosθ

sinθ

)² .

⇒ ( \bf \frac{ 1 - sin \theta }{ cos \theta }

cosθ

1−sinθ

)² .

⇒ \bf\frac{{(1 - sin \theta })^{2}} {{cos}^{2} \theta} .

cos

2

θ

(1−sinθ)

2

.

⇒ \bf \frac{ ( 1 - sin \theta )(1 - sin \theta ) }{1 - {sin}^{2} \theta }

1−sin

2

θ

(1−sinθ)(1−sinθ)

⇒ \bf \frac{ \cancel{ ( 1 - sin \theta )} (1 - sin \theta ) }{ \cancel{ ( 1 - sin \theta ) } ( 1 + sin \theta ) }

(1−sinθ)

(1+sinθ)

(1−sinθ)

(1−sinθ)

⇒ (secθ - tanθ)² = \bf \frac{ 1-sin \theta }{1+sin \theta }

1+sinθ

1−sinθ

.

Answer 2

Answer:

hey I think this identity is invalid, but there's a similar id with the lhs. Please check it in the attachment