Question

cos(90-A) sec(90-A)tanA/ cosec(90-A) sin(90-A) cot(90-A)+ tan(90-A)/ cot A=2

Answer 1

this is the best way to solve it

Answer 2

$LHS = \red{\frac{cos(90-A)sec(90-A)tanA}{Cosec(90-A)sin(90-A)cot(90-A)} + \frac{tan(90-A)}{cotA}}$

$= \frac{sinAcosecAtanA}{secAcosAtanA} + \frac{cotA}{cotA} \\= \frac{sinAcosecA}{secAcosA} + 1 \\= \frac{sinA\times \frac{1}{sinA}}{\frac{1}{cosA} \times cosA } + 1\\= 1 + 1$

$= 2 \\= RHS$

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