If $\frac{{\tan \,\,3A}}{{\tan A}}\,\, = k$, then $\frac{{\sin \,\,3A}}{{\sin A}}$ is equal to

$\frac{{2k}}{{k - 1}},\;\,k \in R$

$\frac{{2k}}{{k - 1}},\;\,k \in \left[ {1/3,\;\,3} \right]$

$\frac{{k - 1}}{{2k}},\,\;k \notin \left[ {1/3,\;\,3} \right]\,$

JEE MATHEMATICS MCQ: TRANSFORMATIONS

1 Hard

1 of 1

Analyzing +4 | -1

\frac{{\tan \,\,3A}}{{\tan A}}\,\, = k

, then

\frac{{\sin \,\,3A}}{{\sin A}}

is equal to

View answer and explanation

Correct option: C

\frac{{2k}}{{k - 1}},\;\,k \notin \left[ {1/3,\,\;3} \right]

Option C (

\frac{{2k}}{{k - 1}},\;k \notin \left[ {1/3,\;3} \right]

) is correct because by using trigonometric identities and the given condition, we can express

\frac{{\sin \,3A}}{{\sin A}}

in terms of k, and determine the range of k.

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