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JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

In a triangle

A B C, \tan \frac{A-B}{2} \cot \frac{A+B}{2}

is equal to

View answer and explanation

Correct option: C

\frac{a-b}{a+b}

Option C (

\frac{a-b}{a+b}

) is correct because using Napier's analogy,

\tan \frac{A-B}{2} = \frac{a-b}{a+b} \cot \frac{C}{2}

, and since

A+B+C = \pi

,

\cot \frac{A+B}{2} = \tan \frac{C}{2}

.

More Questions from PROPERTIES OF TRIANGLES

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{p_1},{p_2},{p_3}

are altitudes of a

\Delta \;ABC

drawn from the vertices

A,B,C

\Delta

the area of the triangle, then

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

is equal to If the sides of the triangles are

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

and radius of incircle is 6, the value of

k

is equal to In a

\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

is the area and

2 s

is the sum of the sides of a triangle, then