Question

Solve this please....​

Answer 1

Hey there !!

Prove that :-

→ (secθ - tanθ)² = \bf \frac{ 1-sin \theta }{1+sin \theta }1+sinθ1−sinθ .

Solution :-

→ (sec θ - tan θ )².

⇒ (\bf \frac{1}{ cos \theta } - \frac{ sin \theta }{ cos \theta }cosθ1−cosθsinθ )² .

⇒ ( \bf \frac{ 1 - sin \theta }{ cos \theta }cosθ1−sinθ )² .

⇒ \bf\frac{{(1 - sin \theta })^{2}} {{cos}^{2} \theta} .cos2θ(1−sinθ)2.

⇒ \bf \frac{ ( 1 - sin \theta )(1 - sin \theta ) }{1 - {sin}^{2} \theta }1−sin2θ(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+sinθ1−sinθ .

Answer 2

Answer:

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