Prove that secAcosecA-cosA=tanA

Question

It is solved using trigonometric identities.it is a difficult question.

Answer 1

If this is an identity it's true for all angles A. So let A=45 degrees, so that tanA=1, sinA=cosA=1/√2 and secAcosecA=2.

Therefore 2-1/√2≠1=tanA. The given is not an identity but it could be an equation. Let's assume it is an equation:

secA=1/cosA and cosecA=1/sinA so we have 1/(sinAcosA)-cosA=tanA.

(1-sinAcos^2)/sinAcosA=tanA;

(1-sinA(1-sin^2(A))/sinAcosA=

(1-sinA+sin^3(A))/sinAcosA=tanA=sinA/cosA;

(1-sinA+sin^3(A))/sinA=sinA.

Therefore, 1-sinA+sin^3(A)=sin^2(A).

Let x=sinA then we have x^3-x^2-x+1=0, which factorises:

(x+1)(x-1)^2=0 from which x=-1 or 1; so sinA=±1 and A=90 or -90, or more generally 90(2n-1) where n is an integer. This "solution" however cannot be tried out in the original equation because it generates infinite values. The problem is that cos90=0. That makes secA and tanA infinite.