If tan B = frac{2 sin A sin C}{sin (A+C)},the | vedclass

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

Question

(B) Given $	an B = \frac{2 \sin A \sin C}{\sin (A+C)}$.Taking the reciprocal on both sides,we get $\frac{1}{	an B} = \frac{\sin (A+C)}{2 \sin A \sin

Vedclass · Accepted Answer

Harmonic progression

Vedclass · Answer

(B) Given $	an B = \frac{2 \sin A \sin C}{\sin (A+C)}$.Taking the reciprocal on both sides,we get $\frac{1}{	an B} = \frac{\sin (A+C)}{2 \sin A \sin C}$.Using the expansion $\sin (A+C) = \sin A \cos C + \cos A \sin C$,we have $\frac{1}{	an B} = \frac{\sin A \cos C + \cos A \sin C}{2 \sin A \sin C}$.This simplifies to $\frac{1}{	an B} = \frac{1}{2} (\cot A + \cot C) = \frac{1}{2} (\frac{1}{	an A} + \frac{1}{	an C})$.Multiplying by $2$,we get $\frac{2}{	an B} = \frac{1}{	an A} + \frac{1}{	an C}$.This is the condition for $	an A, 	an B, 	an C$ to be in Harmonic Progression.

If $\tan B = \frac{2 \sin A \sin C}{\sin (A+C)}$,then $\tan A, \tan B$ and $\tan C$ are in

Similar Questions

Which number should be added to the numbers $13, 15, 19$ so that the resulting numbers are the consecutive terms of a $H.P.$?

Five numbers are in $HP$. The middle term is $1$ and the ratio of the second and the fourth terms is $2:1$. Then, the sum of the first three terms is

If $a, b, c$ are in harmonic progression,then the straight line $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ always passes through a fixed point. That point is:

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ are in $H.P.$,then the expression ${a_1}{a_2} + {a_2}{a_3} + \dots + {a_{n - 1}}{a_n}$ is equal to:

If $a, b, c$ are in $G.P.$,then $\log_a x, \log_b x, \log_c x$ are in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module