Question

1+tanA/sinA + 1+cotA/cosA = 2 (secA+cosecA)

Answer 1

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Answer 2

$\frac{1 + cotA}{cosA} + \frac{1 + tanA}{sinA} \\ \\ = \frac{sinA + cosA}{sinA \: cosA} + \frac{cosA + sinA}{sinA \: cosA} \\ \\ = 2( \frac{sinA + cosA}{sinA \: cosA} ) \\ \\ = 2(secA + cosecA)$