NIMCET 2021 Question Paper with solutions

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Question : 7 of 120

Marks: +1, -0

In

\ΔA B C

if

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

, then

k

is always

$>1$
$\≥ 1$
$=2$
$=1$

Solution:

A+B+C=180°

C=180°-(A+B) \

⇒{C}/{2}={\π}/{2}-{A+B}/{2}

\tan ({C}/{2})=\tan ({\π}/{2}-{(A+B)}/{2})

={1}/{\tan (A+B) \/ 2)}={1-\tan (A \/ 2) \tan (B \/ 2)}/{\tan (A \/ 2)+\tan (B \/ 2)}

Let

a=\tan (A \/ 2), b=\tan (B \/ 2), c=\tan (C \/ 2)

all positive, the constraint becomes

c=(1-a b) \/(a+b)

Which is equivalent to

a b+b c+c a=1

And we know that,

a^{2}+b^{2}+c^{2} \≥ a b+b c+c a

⇒ a^{2}+b^{2}+c^{2} \≥1

\⇒\tan ^{2}(A \/ 2)+\tan ^{2}(B \/ 2)+\tan ^{2}(C \/ 2) \≥1

Therefore, Answer is

k \≥1

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