Question

The solution of $3\tan \left(A-{15}^{\circ }\right)=\tan \left(A+{15}^{\circ }\right)$ is

• A. $n\pi +\frac{\pi }{4}$
• B. $2n\pi +\frac{\pi }{4}$
• C. $2n\pi -\frac{\pi }{4}$
• D. $\frac{n\pi }{2}+{\left(-1\right)}^n\frac{\pi }{2}$

Answer 1

The correct option is B $2n\pi +\frac{\pi }{4}$

$3\tan (a-15=\tan (a+15)$

$\frac{3}{1}=\frac{\tan (a-15)}{\tan (a+15)}$

Use components and devedendo rule

$\frac{3+1}{3-1}=\frac{\tan (a+15)+\tan (a-15)}{\tan (a+15)-\tan (a-15)}$

$2=\frac{dfrac\sin (a+15)\cos (a+15)+\sin (a-15)\cos (a+15)\cos (a+15)\cos (a+15)}{\frac{\sin (a+15)\cos (a-15)-\sin (a+15)\cos (a+15)}{\cos (a+15)\cos (a-15)}}$

$2=\frac{\sin (a+15+a-15)}{\sin (a+15-a+15)}$

$2=\frac{\sin 2a}{\sin 30}$

$\sin 2a=1=\sin 90=\sin (2n\pi +90)$

$2a=90$ or $2a=2m\pi +90$

$a=45$ $a=n\pi +45$