Question

sec A - tan A / sec A + tan A = 1+2 tan^2 A - 2 sec A tanA

Answer 1

LHS:
=(secA − tanA) / (secA + tanA) 
= (1/cosA − sinA/cosA) / (1/cosA + sinA/cosA) 
= (1 − sinA) / (1 + sinA) 
= {(1 − sinA)(1 − sinA)} / {(1 + sinA)(1 − sinA)}
= (1 − 2sinA + sin²A) / (1 − sin²A) 
= (cos²A + sin²A − 2sinA + sin²A) / cos²A 
= (cos²A − 2sinA + 2sin²A) / cos²A 
= cos²A/cos²A − 2 (1/cosA)( sinA/cosA) + 2(sin²A/cos²A )
= 1 − 2secAtanA + 2tan²A
= RHS
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I hope this helps you.

Answer 2

Hi ,

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We know the trigonometric identity

Sec² A - tan² A = 1

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LHS = ( secA - tanA )/( secA + tanA )

= (SecA-tanA)(secA-tanA)/(secA+tanA)(secA-tanA)

= ( Sec A- tanA)² / ( sec² A - tan² A )

= ( SecA - tanA )²

= Sec² A + tan² A - 2tanAsecA

= 1 + tan² A + tan² A - 2tanAsecA

= 1 + 2tan² A - 2tanAsecA

= RHS

I hope this helps you.

:)