If angle theta is divided into two parts such | vedclass

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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 = $

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