If tan A = frac{n}{n+1} and tan B = frac{1}{2 | vedclass

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

Question

(B) We use the trigonometric identity: $	an (A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substitute the given values: $	an (A+B) = \frac{\frac

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We use the trigonometric identity: $	an (A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substitute the given values: $	an (A+B) = \frac{\frac{n}{n+1} + \frac{1}{2n+1}}{1 - (\frac{n}{n+1} 	imes \frac{1}{2n+1})}$.Simplify the numerator: $\frac{n(2n+1) + 1(n+1)}{(n+1)(2n+1)} = \frac{2n^2 + n + n + 1}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)}$.Simplify the denominator: $1 - \frac{n}{(n+1)(2n+1)} = \frac{(n+1)(2n+1) - n}{(n+1)(2n+1)} = \frac{2n^2 + 3n + 1 - n}{(n+1)(2n+1)} = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)}$.Dividing the numerator by the denominator: $	an (A+B) = \frac{2n^2 + 2n + 1}{(n+1)(2n+1)} 	imes \frac{(n+1)(2n+1)}{2n^2 + 2n + 1} = 1$.

If $\tan A = \frac{n}{n+1}$ and $\tan B = \frac{1}{2n+1}$,the value of $\tan (A+B) =$

Similar Questions

The shadow of a tower standing on a level plane is found to be $60\, m$ longer when the angle of elevation of the sun is $30^{\circ}$ than when it is $45^{\circ}$. Find the height of the tower.

If $\tan A = \frac{1 - \cos B}{\sin B}$,find $\tan 2A$ in terms of $\tan B$ and show that:

The value of $\sum_{r=1}^{18} \cos^2(5r)^\circ$,where $x^\circ$ denotes the $x$ degree,is equal to

If $\cos \theta = \frac{1}{2}\left( a + \frac{1}{a} \right),$ then the value of $\cos 3\theta$ is

The value of $\sqrt{2+\sqrt{2(1+\cos 4A)}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module