Question

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

Answer 1

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.

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