prove that √secA-tanA/secA +tanA= 1-sinA/cosA
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Answer:
Step-by-step explanation:
LHS=(secA-tanA+sec^2A-tan^2A)/(secA+tanA+1)
=[secA-tanA+(secA-tanA)(secA+tanA)/(secA+tanA+1) [since, a^2 - b^2 = (a-b)(a+b)]
Taking (secA-tanA) common from the terms in the numerator
=(secA-tanA)(1+secA+tanA)/(secA+tanA+1)
= secA-tanA
= 1/cosA - sinA/cosA
=(1-sinA)/cosA
Hence, proved
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