Question

(cosA - SinA) (SecA - cosA ) = 1/TanA+CotA​

Answer 1

Answer:

1/(tanA+cotA)

Step-by-step explanation:

The given question should be as follows:-

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

L.H.S.

=(cosecA-sinA) (secA-cosA).

=(1/sinA-sinA)(1/cosA-cosA).

=(1-sin^2A) (1-cos^2A)/sinA.cosA.

=cos^2A.sin^2A/sinA.còsA.

=sinA.cosA/1.

=sinA.cosA/(sin^2A+cos^2A).

On dividing above and below by sinA.cosA.

=(sinA.cosA/sinA.cosA)/(sin^2A/sinA.cosA+cos^2A/sinA.cosA).

= 1/(tanA+cotA).

Answer 2

$(coecA - sinA)(secA - cosA) \\ \\ = \frac{1 - {sin}^{2}A }{sinA} \times \frac{1 - {cos}^{2}A }{cosA} \\ \\ = \frac{ {cos}^{2} A \times {sin}^{2} A}{sinAcosA} \\ \\ = sinAcosA \\ \\ = \frac{1}{ \frac{1}{sinAcosA} } \\ \\ = \frac{1}{ \frac{ {sin}^{2}A + {cos}^{2} A }{sinAcosA} } \\ \\ = \frac{1}{ \frac{sinA}{cosA} + \frac{cosA}{sinA} } \\ \\ = \frac{1}{tanA + cotA}$