Question

1/cotA. tanA=sinA. cosA​

Answer 1

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1/(cotA +tanA)

=1/[(cosA/sinA) + (sinA/cosA)]

=1/[(cos² +sin²A)/sinAcosA]

=sinAcosA

Answer 2

$\frac{1}{cotA + tanA} \\ \\ = \frac{1}{ \frac{cosA}{sinA} + \frac{sinA}{cosA} } \\ \\ = \frac{1}{ \frac{ {cos}^{2}A + {sin}^{2} A}{sinAcosA} } \\ \\ = \frac{1}{ \frac{1}{sinAcosA} } \\ \\ = sinAcosA$