prove that √(1−sinA/1+sinA) = secA tanA.
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Answer:
The given statement is proved
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Answered by
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Answer:
sry I got secA-tanA
Step-by-step explanation:
first consider cosA=√(sin^2A)
=√((1+sinA)(1-sinA))
therefore √1+sinA=cosA/√(1-sinA)
and √1-sinA=cosA/√(1+sinA)
and also, 1+sinA=cos^2A/1-sinA
√1-sinA/√1+sinA
=(cosA/√1+sinA)/√1+sinA
=cosA/1+sinA
substituting values
=cosA(1-sinA)/cos^2A
=1-sinA/cosA
=(1/cosA)-(sinA/cosA)
=secA-tanA
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