In ABC - 1-tan A -2tan B -2-Roorkee 1973-A- 2 c-a-b-cB- d-a-k-cC- z2-a-p-cD- 4 a-a-b-c

The correct option is A 2c/a+b+c 1 tan A/2 tan B/2 = cosA/2cosB/2 sinA/2sinB/2sinB/2/cosA/2cosB/2 = cos ( A/2+B/2 )/cos A/2 cos B/2 = sin C/2/cos A/2 cos B/ ...

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Standard XII

Mathematics

Trigonometric Ratios of Half Angles

In ABC , 1-ta...

Question

In $△ A B C, 1 - t a n \frac{A}{2} t a n \frac{B}{2} =$
[Roorkee 1973]

$\frac{2 c}{a + b + c}$

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$\frac{a}{a + b + c}$

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$\frac{2}{a + b + c}$

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$\frac{4 a}{a + b + c}$

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Solution

The correct option is A
$\frac{2 c}{a + b + c}$

$1 - t a n \frac{A}{2} t a n \frac{B}{2} = \frac{c o s \frac{A}{2} c o s \frac{B}{2} - s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{B}{2}}{c o s \frac{A}{2} c o s \frac{B}{2}}$
$= \frac{c o s (\frac{A}{2} + \frac{B}{2})}{c o s \frac{A}{2} c o s \frac{B}{2}} = \frac{s i n \frac{C}{2}}{c o s \frac{A}{2} c o s \frac{B}{2}}$
$= {[\frac{(s - a) (s - b) b c . a c}{a b . s (s - a) s (s - b)}]}^{1 / 2} = \frac{c}{s} = \frac{2 c}{a + b + c} .$

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In △ABC,1−tan A2 tan B2= [Roorkee 1973]

The correct option is A 2ca+b+c 1−tanA2 tan B2=cosA2cosB2−sinA2sinB2sinB2cosA2cosB2 =cos(A2+B2)cosA2cosB2=sinC2cosA2cosB2 =[(s−a)(s−b)bc.acab.s(s−a)s(s−b)]1/2=cs=2ca+b+c.

In $△ A B C, 1 - t a n \frac{A}{2} t a n \frac{B}{2} =$
[Roorkee 1973]