Prove the following trig identity

Question

sin^2x-cos^2x/sinxcosx = tanx-cotx

Answer 1

Do you mean (sin^2x-cos^2x)/sinxcosx?

 

Assuming you do...

 

By the double angle formula for cosine, sin²x - cos²x = -(cos²x - sin²x) = - cos 2x

 

And by the double angle formula for sine, sinx cosx = 1/2 (2 sinx cosx) = 1/2 sin 2x

 

So the left side is equal to -cos 2x / (1/2 sin 2x) = -2 cos 2x / sin 2x = -2 cot 2x (Remember this, call it fact #1)

 

The right side, taxx - cotxis equal to tanx - 1/tanx which is equal to (tan²x - 1) / tan x. (Remember this too, call it fact #2)

 

That looks familiar doesn't it?  It looks a bit like the double angle formula for tangent, but not quite:

 

tan 2x = 2 tanx / (1 − tan²x)

 

Let's see what we can make of that.

 

tan 2x = 2 tanx / (1 − tan²x)

 

Divide both sides by 2:

1/2 tan 2x = tanx / (1 - tan²x)

 

Take the reciprocal of both sides:

1 / (1/2 tan 2x) = (1 - tan²x) / tanx

2 cot 2x = (1 - tan²x) / tanx

 

Multiply both sides by -1

-2 cot 2x = (tan²x - 1) / tanx

 

The right side of this equation is exactly what we said earlier was the right side of the original equation, in fact #2

 

So that means that the right side of the original equation is also equal to -2 cot 2x.

 

But in fact #1, that's exactly the same as what the left side of the original equation is.

 

So the left and right side of the original equation are equal, which is what was needed to be proved.