The solution of 3tan (A - {15^o}) = tan (A + | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

$n\pi + \frac{\pi }{4}$

Vedclass · Answer

(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 $ $A = n\pi + \frac{\pi }{4}$.

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

Similar Questions

If$\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0$, where $0 < \theta < {180^o}$, then $\theta =$

The number of solutions to $\sin \left(\pi \sin ^2 \theta\right)+\sin \left(\pi \cos ^2 \theta\right)=2 \cos \left(\frac{\pi}{2} \cos \theta\right)$ satisfying $0 \leq \theta \leq 2 \pi$ is

Find the principal solutions of the equation $\tan x=-\frac{1}{\sqrt{3}}.$

One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval

If $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x)2\cos x = 0$ then