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NIMCET 2021 Question Paper with solutions

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Question : 7 of 120

Marks: +1, -0

In

\Delta ABC

\tan^2 \frac{A}{2} + \tan^2 \frac{B}{2} + \tan^2 \frac{C}{2} = k

k

is always

$>1$
$\geq 1$
$=2$
$=1$

Solution:

A+B+C=180^{\circ}

C=180^{\circ}-(A+B)

\Rightarrow \frac{C}{2} = \frac{\pi}{2} - \frac{A+B}{2}

\tan\left(\frac{C}{2}\right) = \tan\left(\frac{\pi}{2} - \frac{A+B}{2}\right)

= \frac{1}{\tan\left(\frac{A+B}{2}\right)} = \frac{1-\tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right)}{\tan\left(\frac{A}{2}\right)+\tan\left(\frac{B}{2}\right)}

Let

a=\tan\left(\frac{A}{2}\right), b=\tan\left(\frac{B}{2}\right), c=\tan\left(\frac{C}{2}\right)

all positive, the constraint becomes

c=\frac{1-ab}{a+b}

Which is equivalent to

ab+bc+ca=1

And we know that,

a^{2}+b^{2}+c^{2} \geq ab+bc+ca

\Rightarrow a^{2}+b^{2}+c^{2} \geq 1

\Rightarrow \tan^{2}\left(\frac{A}{2}\right)+\tan^{2}\left(\frac{B}{2}\right)+\tan^{2}\left(\frac{C}{2}\right) \geq 1

Therefore, Answer is

k \geq 1

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