Prove Identity: $\frac{\cos A - \sin A}{\cos A + \sin A}$

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SUMMARY

The discussion centers on proving the trigonometric identity $\frac{\cos A - \sin A}{\cos A + \sin A} = \frac{1 - \tan(A)}{1 + \tan(A)}$. Participants demonstrate that the left-hand side simplifies to $\frac{\cos(A) - \sin(A)}{\cos(A) + \sin(A)}$ using the tangent and cotangent functions. The conversation suggests dividing both the numerator and denominator by $\sin A$ to facilitate further simplification. This method effectively leads to the desired identity.

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$\frac{1-\tan\left({A}\right)}{1+\tan\left({A}\right)}=\frac{\cot\left({A}\right)-1}{\cot\left({A}\right)+1}$

$L..H.S=\frac{1-\frac{\sin\left({A}\right)}{\cos\left({A}\right)}}{1+\frac{\sin\left({A}\right)}{\cos\left({A}\right)}}$

$=\frac{\cos\left({A}\right)-\sin\left({A}\right)}{\cos\left({A}\right)+\sin\left({A}\right)}$

What should be done from here?
 
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Silver Bolt said:
$\frac{1-\tan\left({A}\right)}{1+\tan\left({A}\right)}=\frac{\cot\left({A}\right)-1}{\cot\left({A}\right)+1}$

$L..H.S=\frac{1-\frac{\sin\left({A}\right)}{\cos\left({A}\right)}}{1+\frac{\sin\left({A}\right)}{\cos\left({A}\right)}}$

$=\frac{\cos\left({A}\right)-\sin\left({A}\right)}{\cos\left({A}\right)+\sin\left({A}\right)}$

What should be done from here?

Hi Silver Bolt! ;)

Can we divide both the numerator and the denominator by $\sin A$?
 
$$\frac{1-\tan(A)}{1+\tan(A)}\cdot\frac{\cot(A)}{\cot(A)}$$
 

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