Question

Prove that 1/tanA-secA=secA/sinA-1

Answer 1

Heya..!!!


1/(tanA - secA)

⇒1/[(sinA/cosA - 1/cosA)]

⇒1/[ (sinA - 1)/cosA]

⇒cosA/sinA - 1

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Answer 2

LHS

$secA(1-sinA)(secA+tanA) \\ \\ = \frac{1}{cosA} (1-sinA)(secA+tanA) \\ \\ = ( \frac{1}{cosA} - \frac{ sinA}{cosA} )(secA+tanA) \\ \\ = (secA-tanA)(secA+tanA) \\ \\ = {sec}^{2} A - {tan}^{2} A\\ \\ = 1$

RHS

=1

LHS=RHS