In a $\Delta \;ABC,\angle {\rm{ }}C = 60^\circ $ then $\frac{1}{{a + c}} + \frac{1}{{b + c}}$ is equal to

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

In a

\Delta \;ABC,\angle \mathrm{ }C = 60^\circ

then

\frac{1}{{a + c}} + \frac{1}{{b + c}}

is equal to

View answer and explanation

Correct option: C

\frac{3}{{a + b + c}}

Option C (

\frac{3}{{a + b + c}}

) is correct because by using the cosine rule and algebraic manipulation, the given expression simplifies to this form.

Practice All Chapter MCQs →

{p_1},{p_2},{p_3}

\Delta \;ABC

A,B,C

\Delta

p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2}

5k,\;6k,\;5k

k

\Delta \,ABC

a = \left( {b - c} \right)\sec \theta

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =

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

\Delta = {a^2} - {\left( {b - c} \right)^2}

\Delta

\Delta \;ABC

\tan A

A

2 s