Question

(Sec theta - tan theta)2 (1 + sin theta)= 1 - sin theta

Answer 1

Answer:

Prove that :-

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

Solution :-

→ (sec θ - tan θ )².

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

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

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

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

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

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

Hence, it is proved.

THANKS

Answer 2

Step-by-step explanation:

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