tan-cot/tan+cot=2sin^2-1

Question

prove the following is an identity please thank you.

Answer 1

I'm going to guess your equation is (tan(x) - cot(x)) / (tan(x) + cot(x)) = 2sin²(x) - 1.

 

 

We'll first work with tan(x) - cot(x), the numerator on the left-hand side. Let's try to combine those terms into one term that's comprised of just sines and cosines. Remember, tan = sin/cos and cot = cos/sin.

 

tan(x) - cot(x)

sin(x)/cos(x) - cos(x)/sin(x)

sin²(x)/(sin(x)cos(x)) - cos²(x)/(sin(x)cos(x))

(sin²(x) - cos²(x))/(sin(x)cos(x))

 

 

Now, let's do the same to the denominator:

 

tan(x) + cot(x)

sin(x)/cos(x) + cos(x)/sin(x)

sin²(x)/(sin(x)cos(x)) + cos²(x)/(sin(x)cos(x))

(sin²(x) + cos²(x))/(sin(x)cos(x))

 

 

Since sin²(x) + cos²(x) = 1, the denominator is equal to:

 

1/(sin(x)cos(x))

 

 

 

Now, on the left-hand side, we've got:

 

((sin²(x) - cos²(x))/(sin(x)cos(x))) / (1/(sin(x)cos(x)))

 

That's the same as:

 

((sin²(x) - cos²(x))/(sin(x)cos(x))) × ((sin(x)cos(x))/1)

(sin²(x) - cos²(x)) / 1

sin²(x) - cos²(x)

 

 

 

So, overall, the equation is:

 

sin²(x) - cos²(x) = 2sin²(x) - 1

sin²(x) - cos²(x) = sin²(x) + sin²(x) - 1

 

Since sin²(x) + cos²(x) = 1, it's also true that sin²(x) - 1 = -cos²(x). So, substitute that in for sin²(x) - 1 on the right-hand side:

 

sin²(x) - cos²(x) = sin²(x) + (-cos²(x))

sin²(x) - cos²(x) = sin²(x) - cos²(x)

 

The two sides are equal.