# Another prove: double angles

• May 15th 2010, 11:17 AM
tallguywhoplaysguitar
Another prove: double angles
So here's another problem that's been puzzling me for a couple hours. I got my answer as

---1---- does not equal--- 1--- but I am very unsure whether this is right
4cos^2x---------------cos^2x
Here's the question
Prove that:
2csc2xtanx=sec^2x
• May 15th 2010, 02:11 PM
skeeter
Quote:

Originally Posted by tallguywhoplaysguitar
So here's another problem that's been puzzling me for a couple hours. I got my answer as

---1---- does not equal--- 1--- but I am very unsure whether this is right
4cos^2x---------------cos^2x
Here's the question
Prove that:
2csc2xtanx=sec^2x

$\displaystyle 2\csc(2x)\tan{x} =$

$\displaystyle \frac{2}{\sin(2x)} \cdot \frac{\sin{x}}{\cos{x}} =$

$\displaystyle \frac{2}{2\sin{x}\cos{x}} \cdot \frac{\sin{x}}{\cos{x}}$

finish it ...
• May 16th 2010, 11:48 AM
tallguywhoplaysguitar
thanks, I was getting it wrong in the first step by placing the 2 in the denominator