Question

Prove that ( cosecA-sinA ) ( secA-cosA ) = 1÷tanA+cotA

Answer 1

QUESTION➡


Prove that ( cosecA-sinA ) ( secA-cosA ) = 1÷tanA+cotA



SOLUTION➡



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

hey mate here is your answer


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)

..I hope it helps you....