1+sin/1-sin=(sec+tan)²
Answers
Step-by-step explanation:
Can you prove the identity: (sec A - tan A) ^2=1-sin A/1+sin A?
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Prove that (sec A - tan A)² = (1 - sin A)/(1 + sin A)
Solution
To prove choose the complex side. In this equation both are equally complex. You can choose either side and methodically simplify moving towards the goal side. Apply the formulae correctly.
RHS
= (1-sinA)/(1+sinA)
= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization
= (1-sinA)²/(1+sinA)(1-sinA)
= (1-sinA)/(1+sinA) = LHS
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Alternatively,
LHS
= (sec A - tan A)²; express all as sine and cosine
= (1/cos A - sin A/cos A)²
= [(1 - sin A)/cos A)]²
= (1 - sin A)²/cos² A
= (1 - sin A)²/(1 - sin² A)
= (1 - sin
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(secA−tanA)2=1−sinA1+sinA
Considering LHS,
(secA−tanA)2=(1cosA−sinAcosA)2
=(1−sinAcosA)2
=((1−sinA)2cos2A)
=((1−sinA)21−sin2A)
=((1−sinA)2(1−sinA)(1+sinA))
=1−sinA1+sinA=RHS