[Math] Prove $\frac{1-\sin(2A)}{\cos(2A)}=\frac{1-\tan A}{1+\tan A}$

trigonometry

How would I prove the following double angle identity?

$$\frac{1-\sin(2A)}{\cos(2A)}=\frac{1-\tan A}{1+\tan A}$$

My work thus far is

$$\frac{1-2\sin A\cos A}{\cos^2A-\sin^2A}$$

$$\frac{1-2\sin A\cos A}{(\cos A+\sin A)(\cos A-\sin A)}$$

Sadly I am stuck.

Best Answer

$$ \begin{align} \frac{1-\sin(2A)}{\cos(2A)} &=\frac{1-2\sin(A)\cos(A)}{\cos^2(A)-\sin^2(A)}\tag{1}\\ &=\frac{\sec^2(A)-2\tan(A)}{1-\tan^2(A)}\tag{2}\\ &=\frac{1+\tan^2(A)-2\tan(A)}{1-\tan^2(A)}\tag{3}\\ &=\frac{(1-\tan(A))^2}{1-\tan^2(A)}\tag{4}\\ &=\frac{1-\tan(A)}{1+\tan(A)}\tag{5} \end{align} $$

  1. double angle formulas

  2. multiply numerator and denominator by $\sec^2(A)$

  3. $\sec^2(A)=1+\tan^2(A)$

  4. collect square of a difference

  5. cancel $1-\tan(A)$ from numerator and denominator

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