If $\sin A = nsin\;B$, then $\frac{{n - 1}}{{n + 1}}\,\,\tan \,\,\frac{{A + B}}{2}$ is equal to

JEE MATHEMATICS MCQ: TRANSFORMATIONS

1 Easy

1 of 1

Application +4 | -1

\sin A = nsin\;B

, then

\frac{{n - 1}}{{n + 1}}\,\,\tan \,\,\frac{{A + B}}{2}

is equal to

View answer and explanation

Correct option: B

\tan \,\,\frac{{A - B}}{2}

Option B

\tan \,\,\frac{{A - B}}{2}

is correct because applying componendo and dividendo to

\frac{n}{1} = \frac{{\sin A}}{{\sin B}}

and simplifying leads to the given expression being equal to

\tan \,\,\frac{{A - B}}{2}

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