Question

prove that

(sec theta- tan theta)2=1- sin theta/1+ sin theta

Answer 1

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Answer 2

Hey there !!

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

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