Question

(cosA-tanA)²+1/cosecA(secA-tanA)²=2tanA

Answer 1

you did mistake in typing above question.

question should be ----> {(secA - tanA)² + 1 }/{cosecA(secA - tanA)}

LHS = {(secA - tanA)² + 1}/{cosecA(secA - tanA)}

= {sec²A + tan²A - 2secA.tanA + 1}/{cosecA(secA - tanA)}

= {sec²A - 2secA.tanA + (1 + tan²A)}/{cosecA(secA - tanA)}

[ we know, sec²x - tan²x = 1 so, 1 + tan²A = sec²A ]

= {sec²A - 2secA.tanA + sec²A}/{cosecA(secA - tanA)}

= {2sec²A - 2secA.tanA}/{cosecA(secA - tanA)}

= {2secA(secA - tanA)}/{cosecA(secA- tanA)}

= 2 (secA/cosecA)

[ as we know, secA = 1/cosA and cosecA = 1/sinA ]

= 2(sinA/cosA) = 2tanA = RHS