If angle theta is divided into two parts such | vedclass

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

Question

(A) Let the two parts be $A$ and $B$ such that $A + B = 	heta$ and $A - B = \phi$.Given that $	an A = k 	an B$,we have $\frac{	an A}{	an B} = k$.

Vedclass · Accepted Answer

$\frac{k + 1}{k - 1} \sin \phi$

Vedclass · Answer

(A) Let the two parts be $A$ and $B$ such that $A + B = 	heta$ and $A - B = \phi$.Given that $	an A = k 	an B$,we have $\frac{	an A}{	an B} = k$.Expanding the tangents in terms of sine and cosine:$\frac{\sin A \cos B}{\cos A \sin B} = \frac{k}{1}$.Applying the componendo and dividendo rule:$\frac{	an A + 	an B}{	an A - 	an B} = \frac{k + 1}{k - 1}$.Using the identity $\frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B} = \frac{\sin(A + B)}{\sin(A - B)}$,we get:$\frac{\sin(A + B)}{\sin(A - B)} = \frac{k + 1}{k - 1}$.Substituting $A + B = 	heta$ and $A - B = \phi$:$\frac{\sin 	heta}{\sin \phi} = \frac{k + 1}{k - 1}$.Therefore,$\sin 	heta = \frac{k + 1}{k - 1} \sin \phi$.

If angle $\theta$ is divided into two parts such that the tangent of one part is $k$ times the tangent of the other and $\phi$ is their difference,then $\sin \theta = $

Similar Questions

Let $S = \{x \in R : \cos(x) + \cos(\sqrt{2}x) < 2\}$,then

Evaluate: $(\cos 252^{\circ} - \sin 126^{\circ})(\cos 252^{\circ} + \sin 126^{\circ})(\sin^2 126^{\circ} + \sin^2 186^{\circ} + \sin^2 66^{\circ})$

The value of $\tan 40^{\circ} + \tan 11^{\circ} + \tan 20^{\circ} - \tan 56^{\circ} + \tan 56^{\circ} \tan 11^{\circ} + \sqrt{3} \tan 40^{\circ} \tan 20^{\circ}$ is

$\frac{\cot A}{1-\tan A} + \frac{\tan A}{1-\cot A} = ?$

The value of $\frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\sin \theta + \cos \theta}{\tan^2 \theta - 1}$ for all permissible values of $\theta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module