Verifying trig identities specific problem

Question

$$\frac1{1-\cos y} + \frac1{1+\cos y} = 2\csc^2y $$ My attempt was me trying to find a common denominator on the left side but I don't know what to do after that.

score 5 · Answer 1 · answered Nov 25 '14 at 02:22

5

Your idea is correct. So we have

$$\frac{1}{1-\cos y}+\frac{1}{1+\cos y} = \frac{1+\cos y+ 1-\cos y}{(1+\cos y)(1-\cos y)} = \frac{2}{1-\cos^2 y}.$$

What do you know about $1-\cos^2 y$? Can you take it from here?

answered Nov 25 '14 at 02:22

Cameron Williams

29,432

score 2 · Answer 2 · answered Nov 25 '14 at 02:22

2

$$\frac{1}{1-\cos y} + \frac1{1+\cos y} =\frac{1}{1-\cos y}\cdot\frac{1+\cos y}{1+\cos y}+\frac{1}{1+\cos y}\cdot\frac{1-\cos y}{1-\cos y}$$$$=\frac{1+\cos y+1-\cos y}{1-\cos^2y}=\frac{2}{1-\cos^2y}=\frac{2}{\sin^2y}=2\csc^2 y$$

answered Nov 25 '14 at 02:22

Aditya Hase

8,851

score 0 · Answer 3 · answered Nov 25 '14 at 19:11

This is an another "lengthy" method...

$\frac1{1-\cos y} + \frac1{1+\cos y}$ = $\frac1{1-\frac {1}{\sec y}} + \frac1{1+\frac {1}{\sec y}}$

=$\frac{1}{\frac {\sec y - 1}{\sec y}}$ + $\frac{1}{\frac {\sec y + 1}{\sec y}}$

=$\frac{\sec y}{\sec y -1}$+$\frac{\sec y}{\sec y +1}$

=$\sec y[\frac{1}{\sec y -1}+\frac{1}{\sec y +1}]$

=$\sec y[\frac{2 \sec y}{\tan^2 y}]$

=$2 \frac{\sec^2 y}{\tan^2 y}$