If A B and tan A - tan B = x and cot B - co | vedclass

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

Question

(C) Given,$	an A - 	an B = x$ $\cot B - \cot A = y$ $\Rightarrow \frac{1}{	an B} - \frac{1}{	an A} = y$ $\Rightarrow \frac{	an A - 	an B}{	an A

Vedclass · Accepted Answer

$\frac{1}{x} + \frac{1}{y}$

Vedclass · Answer

(C) Given,$	an A - 	an B = x$ $\cot B - \cot A = y$ $\Rightarrow \frac{1}{	an B} - \frac{1}{	an A} = y$ $\Rightarrow \frac{	an A - 	an B}{	an A 	an B} = y$ $\Rightarrow \frac{x}{	an A 	an B} = y$ $\Rightarrow 	an A 	an B = \frac{x}{y}$ Now,$\cot (A - B) = \frac{1}{	an (A - B)} = \frac{1 + 	an A 	an B}{	an A - 	an B}$ Substituting the values,we get: $\cot (A - B) = \frac{1 + \frac{x}{y}}{x} = \frac{\frac{y + x}{y}}{x} = \frac{x + y}{xy} = \frac{x}{xy} + \frac{y}{xy} = \frac{1}{y} + \frac{1}{x}$

If $A > B$ and $\tan A - \tan B = x$ and $\cot B - \cot A = y$,then $\cot (A - B) = $

Similar Questions

$\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ} = $

Let $\cos (\alpha+\beta)=\frac{4}{5}$ and $\sin (\alpha-\beta)=\frac{5}{13}$,where $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,then $\tan 2 \alpha=$

In an isosceles right-angled triangle,a straight line is drawn from the midpoint of one of the equal sides to the opposite vertex. Then a pair of possible values of the cotangents of the two angles so formed at that vertex are

Prove that $\sin^{2} 6x - \sin^{2} 4x = \sin 2x \sin 10x$.

$\cos \alpha \sin (\beta-\gamma) + \cos \beta \sin (\gamma-\alpha) + \cos \gamma \sin (\alpha-\beta)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module