In $\Delta \;ABC,\frac{{b + c}}{a}$ is equal to

$\frac{{\sin \frac{1}{2}\left( {B - C} \right)}}{{\cos \frac{1}{2}\;A}}$

$\frac{{\cos \frac{1}{2}\left( {B + C} \right)}}{{\cos \frac{1}{2}\;A}}$

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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Application +4 | -1

\Delta \;ABC,\frac{{b + c}}{a}

is equal to

View answer and explanation

Correct option: A

\frac{{\cos \frac{1}{2}\left( {B - C} \right)}}{{\sin \frac{1}{2}\;A}}

Option A (

\frac{{\cos \frac{1}{2}({B - C})}}{{\sin \frac{1}{2}\,A}}

) is correct because by using the sine rule and sum-to-product trigonometric identities, the expression

\frac{{b + c}}{a}

can be simplified to

\frac{{\cos \frac{1}{2}({B - C})}}{{\sin \frac{1}{2}\,A}}