Question

If $E.tan(x-{30}^{\circ })=j.tan(x+{120}^{\circ })$, then $\frac{E+J}{E-J}=$

• A. $sin2x$
• B. $2cos2x$
• C. $tan2x$
• D. None of these.

Answer 1

Answer: (B)

$2cos2x$

Given $E\tan (x-{30}^{\circ })=J\tan (x+{120}^{\circ })$

$⟹\frac{E}{J}=\frac{\tan (x+{120}^{\circ })}{\tan (x-{30}^{\circ })}$

Applying compoundo and dividendo rule

$\frac{E+J}{E-J}=\frac{\tan (x-{30}^{\circ })+\tan (x+{120}^{\circ })}{\tan (x+{120}^{\circ })-\tan (x-{30}^{\circ })}=\frac{\frac{\sin (x-{30}^{\circ })}{\cos (x-{30}^{\circ })}+\frac{\sin (x+{120}^{\circ })}{\cos (x+{120}^{\circ })}}{\frac{\sin (x+{120}^{\circ })}{\cos (x+{120}^{\circ })}-\frac{\sin (x-{30}^{\circ })}{\cos (x-{30}^{\circ })}}$

$=\frac{\sin (x-{30}^{\circ })\cos (x+{120}^{\circ })+\sin (x+{120}^{\circ })\cos (x-{30}^{\circ })}{\sin (x+{120}^{\circ })\cos (x-{30}^{\circ })-\sin (x-{30}^{\circ })\cos (x+{120}^{\circ })}$

$=\frac{\sin (x-{30}^{\circ }+x+{120}^{\circ })}{\sin (x+{120}^{\circ }-x+{30}^{\circ })}$

$=\frac{\sin ({90}^{\circ }+2x)}{\sin {150}^{\circ }}$

$=2\cos 2x$