MCQ
The solution of $3\tan (A - {15^o}) = \tan (A + {15^o})$ is
- ✓$n\pi + \frac{\pi }{4}$
- B$2n\pi + \frac{\pi }{4}$
- C$2n\pi - \frac{\pi }{4}$
- D$\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{2}$
✓
Answer
Correct option: A.
$n\pi + \frac{\pi }{4}$
a
(a) $\frac{{3\sin (A - {{15}^o})}}{{\cos (A - {{15}^o})}} = \frac{{\sin (A + {{15}^o})}}{{\cos (A + {{15}^o})}}$
(a) $\frac{{3\sin (A - {{15}^o})}}{{\cos (A - {{15}^o})}} = \frac{{\sin (A + {{15}^o})}}{{\cos (A + {{15}^o})}}$
==>$3\sin (A - {15^o})\cos (A + {15^o})$$ = \cos (A - {15^o})\sin (A + {15^o})$
$ \Rightarrow $ $2\sin (A - {15^o})\cos (A + {15^o}) = \frac{1}{2}$
$ \Rightarrow $ $\sin 2A - \sin {30^o} = \frac{1}{2}$
$ \Rightarrow $$2A = 2n\pi + \frac{\pi }{2}$
$ \Rightarrow $ $A = n\pi + \frac{\pi }{4}$.
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