The area of a $\Delta \,ABC$ is ${b^2} - {\left( {c - a} \right)^2}$. Then, $\tan \,\,B = $

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

The area of a

\Delta \,ABC

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

. Then,

\tan \,\,B =

View answer and explanation

Correct option: C

\frac{8}{15}

Option C (

9\frac{1}{2}

) is correct because by pairing terms using the identity sin²θ + cos²θ = 1 and accounting for the remaining terms, the sum equals 9.5.

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