# Proving Trigonometric Equality

I have this trigonometric equality to prove:

$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$ I started with the left hand side, reducing the fractions to common denominator and got this: $$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos x+\cos x(\frac{\sin x}{\cos x})-\sin x+\sin x(\frac{\sin x}{\cos x})}{1-\left(\frac{\sin^2x}{\cos^2x}\right)}\\=\frac{\cos x+\left(\frac{\sin^2x}{\cos x}\right)}{1-\frac{\sin^2}{\cos^2x}}$$and by finding common denominator top and bottom and then multiplying the fractions i got: $$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}$$ which is far from the right hand side and I don't know what am I doing wrong.
What is the correct way to prove this equality?

• Cancel one $\cos x$ and use $\sin^2x=1-\cos^2x$. – A.Γ. May 27 '17 at 16:08
• You can as well remember that $\sin^2 x=1-\cos^2x$ – kingW3 May 27 '17 at 16:10
• It's usually better to use the least common denominator. Multiplying your two denominators of $\cos x$ and $\cos^2x$ gives a common denominator, but the fact that it wasn't the least common meant you couldn't see your next step. – Teepeemm May 27 '17 at 16:37

From the last line of your working, and using $\sin^2x+\cos^2x=1,$ $$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}=\frac{\cos x}{\cos^2x-\sin^2x}=\frac{\cos x}{\cos^2x-(1-\cos^2x)}=\frac{\cos x}{2\cos^2x-1}$$

Multiplying numerator/denominator by $\cos x$,

$$\frac{c}{1-t}-\frac{s}{1+t}=c\left(\frac c{c-s}-\frac s{c+s}\right)=c\frac{c^2+cs-cs+s^2}{c^2-s^2}=\frac c{2c^2-1}.$$

• +1 for nice notation – Khosrotash May 27 '17 at 16:39

$$\frac { cosx }{ 1-tanx } -\frac { sinx }{ 1+tanx } =\frac { cosx }{ 2cos^{ 2 }x-1 } \\ \frac { \cos { x } \left( 1+tanx \right) }{ \left( 1-tanx \right) \left( 1+tanx \right) } -\frac { \sin { x } \left( 1-tanx \right) }{ \left( 1-tanx \right) \left( 1-tanx \right) } =\\ =\frac { \cos { x } +\sin { x } -\sin { x } +\frac { \sin ^{ 2 }{ x } }{ \cos { x } } }{ 1-\tan ^{ 2 }{ x } } =\frac { \frac { \cos ^{ 2 }{ x+\sin ^{ 2 }{ x } } }{ \cos { x } } }{ \frac { \cos ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } } =\\ =\frac { \cos { x } }{ \cos ^{ 2 }{ x-\sin ^{ 2 }{ x } } } =\frac { \cos { x } }{ 2\cos ^{ 2 }{ x-1 } }$$

Factorise cosx out of the bottom line of the last equation, then you have

$\cos \left( x \right)\left( 2\cos \left( x \right)^{2}-1 \right)$

as your bottom line. The rest is fine

Here's a hint: $$\frac{\cos x + \frac{\sin x}{\cos x}^2}{1 - \left(\frac{\sin x}{\cos x}\right)^2} = \frac{\frac{\sin x^2 + \cos x^2}{\cos x}}{\frac{\cos x^2 - \sin x^2}{\cos x^2}}.$$

$$\frac{c^2}{c-s}-\frac{cs}{c+s}=\frac{\cos x}{2\cos^2x-1}$$ $$\frac{c^3+sc^2-c^2s+cs^2}{c^2-s^2} =\frac{\cos x}{2\cos^2x-1}$$ $$\frac{ c }{c^2-s^2}=\frac{\cos x}{2\cos^2x-1}$$ $$\frac{ c }{2c^2-1}=\frac{\cos x}{2\cos^2x-1}$$

The steps for proving a trigonometric identity are:

1. Take the more complicated side and simplify, writing all other trig functions in terms of sines and cosines. (You did this. By the way, you're missing an x, and you still have a common factor of cos x.)

2. Try working with the other side. (I don't see too much to do here.)

3. Subtract the results from 1 and 2 to see if you can get 0.

You statement that you result is far from the right side is wrong. Once you remove the cos x factor, the numerators will both be cos x, and the dominators should be recognized as equivalent forms from the double angle formulas for cosine.