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TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Mathematics

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Question : 28 of 160

Marks: +1, -0

In a triangle

ABC

\;\frac{r}{r_1}=\;\frac{1}{2}

4 \tan \;\frac{A}{2}\left(\tan \;\frac{B}{2}+\tan \;\frac{C}{2}\right)=

1
2
3
4

Solution:

Given,

\;\frac{r}{r_1}=\;\frac{1}{2}

\;\frac{4 R \sin \;\frac{A}{2} \sin \;\frac{B}{2} \sin \;\frac{C}{2}}{4 R \sin \;\frac{A}{2} \cos \;\frac{B}{2} \cos \;\frac{C}{2}}=\;\frac{1}{2}

\tan \;\frac{B}{2} \tan \;\frac{C}{2}=\;\frac{1}{2}

.....(i)

A+B+C=\pi

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

\Rightarrow \;\frac{\tan \frac{B}{2}+\tan \frac{C}{2}}{1-\tan \;\frac{B}{2} \tan \;\frac{C}{2}}=\cot \;\frac{A}{2}

\Rightarrow \;\frac{\tan \frac{B}{2}+\tan \frac{C}{2}}{1-\;\frac{1}{2}}=\;\frac{1}{\tan \;\frac{A}{2}}

[

\because

From Eq. (i)]

\Rightarrow 2 \tan \;\frac{A}{2} \left[\tan \;\frac{B}{2}+\tan \;\frac{C}{2}\right] =1

Multiply by 2 on both sides,

4 \tan \;\frac{A}{2} \left[\tan \;\frac{B}{2} \tan \;\frac{C}{2}\right] =2

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