5) $$ \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}= an 2 x $$

Question

5) $$ \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=	an 2 x $$

UpStudy AI · Accepted Answer

Solve the equation $$ \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=	an 2x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{\cos\left(xight)}{\cos\left(xight)-\sin\left(xight)}-\frac{\cos\left(xight)}{\cos\left(xight)+\sin\left(xight)}=	an\left(2xight)$$
- step1: Find the domain:
  $$\frac{\cos\left(xight)}{\cos\left(xight)-\sin\left(xight)}-\frac{\cos\left(xight)}{\cos\left(xight)+\sin\left(xight)}=	an\left(2xight),x
eq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step2: Rewrite the expression:
  $$\frac{\cos\left(xight)}{\cos\left(xight)-\sin\left(xight)}-\frac{\cos\left(xight)}{\cos\left(xight)+\sin\left(xight)}=\frac{2	an\left(xight)}{1-	an^{2}\left(xight)}$$
- step3: Rewrite the expression:
  $$\frac{\cos\left(xight)}{\cos\left(xight)-\sin\left(xight)}-\frac{\cos\left(xight)}{\cos\left(xight)+\sin\left(xight)}=\frac{2	imes \frac{\sin\left(xight)}{\cos\left(xight)}}{1-\left(\frac{\sin\left(xight)}{\cos\left(xight)}ight)^{2}}$$
- step4: Expand the expression:
  $$\frac{\cos\left(xight)}{\cos\left(xight)-\sin\left(xight)}-\frac{\cos\left(xight)}{\cos\left(xight)+\sin\left(xight)}=\frac{\frac{2\sin\left(xight)}{\cos\left(xight)}}{1-\frac{\sin^{2}\left(xight)}{\cos^{2}\left(xight)}}$$
- step5: Calculate:
  $$\frac{\cos\left(xight)}{\cos\left(xight)-\sin\left(xight)}-\frac{\cos\left(xight)}{\cos\left(xight)+\sin\left(xight)}=\frac{2\sin\left(xight)\cos\left(xight)}{\cos^{2}\left(xight)-\sin^{2}\left(xight)}$$
- step6: Multiply both sides of the equation by LCD:
  $$\left(\frac{\cos\left(xight)}{\cos\left(xight)-\sin\left(xight)}-\frac{\cos\left(xight)}{\cos\left(xight)+\sin\left(xight)}ight)\left(\cos\left(xight)-\sin\left(xight)ight)\left(\cos\left(xight)+\sin\left(xight)ight)=\frac{2\sin\left(xight)\cos\left(xight)}{\cos^{2}\left(xight)-\sin^{2}\left(xight)}	imes \left(\cos\left(xight)-\sin\left(xight)ight)\left(\cos\left(xight)+\sin\left(xight)ight)$$
- step7: Simplify the equation:
  $$2\sin\left(xight)\cos\left(xight)=2\sin\left(xight)\cos\left(xight)$$
- step8: Move the expression to the left side:
  $$2\sin\left(xight)\cos\left(xight)-2\sin\left(xight)\cos\left(xight)=0$$
- step9: Calculate:
  $$0=0$$
- step10: The statement is true:
  $$x \in \mathbb{R}$$
- step11: Check if the solution is in the defined range:
  $$x \in \mathbb{R},x
eq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step12: Find the intersection:
  $$x
eq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
The solution to the equation $$ \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=	an 2x $$ is $$ x 
eq \frac{\pi}{4} + \frac{k\pi}{2}, k \in \mathbb{Z} $$.
The solution to the equation is all real numbers except $$ x = \frac{\pi}{4} + \frac{k\pi}{2} $$ for any integer $$ k $$.

5) \( \frac{\cos x}{\cos x-\sin x}-\frac{\cos x}{\cos x+\sin x}=\tan 2 x \)

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