Originally Posted by

**ticbol** :

:

sinA in terms of tanA:

tanA = sinA/cosA ----identity

Multiply both sides by cosA,

cosA*tanA = sinA

Express the cosA in terms of sinA,

[sqrt(1 -sin^2(A)]tanA = sinA ------------(i)

There, only sinA and tanA are involved.

Solve for sinA in terms of tanA and you're done.

Here is one way of doing that,

Square both sides of (i),

(1 -sin^2(A)[tan^2(A)] = sin^2(A)

Expand,

tan^2(A) -[sin^2(A)][tan^2(A)] = sin^2(A)

Collect the sinA terms,

tan^2(A) = sin^2(A) +[sin^2(A)][tan^2(A)]

Factor out the sin^2(A),

tan^2(A) = [sin^2(A)][1 +tan^2(A)]

Isolate the sin^2(A),

[tan^2(A)] /[1 +tan^2(A)] = sin^2(A)

So, take the square roots of both sides,

sinA = (tanA) /[sqrt(1 +tan^2(A))] -------------answer.