Prove that sec thetha + tan thetha = cos thetha/1-sin thetha
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Step-by-step explanation:
SecA + TanA = CosA/(1-SinA)
Taking LHS,
=> 1/CosA + SinA/CosA
Taking LCM,
=> (1+SinA)/CosA
By Rationalising denominator,
=> [(1+SinA)(CosA)]/cos²A
Using, 1 - sin²A = Cos²A
=> [(1+SinA)(CosA)]/[(1-Sin²A)]
Using a² - b² = (a+b)(a-b)
=> [(1+SinA)(CosA)]/[(1+SinA)(1-SinA)]
Cancelling (1+SinA) from both numerator and denominator,
=> (CosA)/(1-SinA)
=> RHS
.°. LHS = RHS
Hence proved!
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