Prove that (sec A + tan A)^2 = cosec A + 1 / cosec A - 1
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To Prove :
⇒ (sec A + tan A )² = cosec A + 1 / cosec A - 1
Proof :
On solving L.H.S
⇒ ( sec A + tan A )²
⇒ ( 1 / cos A + sin A / cos A )²
⇒ [ (1 + sin A ) / cos A ]²
⇒ ( 1 + sin A)² / ( 1 - sin² A )
⇒ ( 1 + sin A ) ( 1 + sin A ) / ( 1 + sin A ) ( 1 - sin A )
⇒ ( 1 + sin A ) / ( 1 - sin A )
⇒ ( 1 + 1 / cosec A) / ( 1 - 1 / cosec A )
⇒ ( 1 + cosec A ) / ( 1- cosec A )
L.H.S = R.H.S
Hence proved
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