In ABC prove that b-ca-tan B-2-tan C-2tan B-2- tan C-2

From sine rulea = 2 R sin Ab = 2 R sin BC = 2 R sin C ⇒ #160; b ca = 2R sin B 2R sin C2R sin A = sin B sinCsinA #160; = 2sin ( B C2 ) cos ...

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Standard XII

Mathematics

Trigonometric Ratios of Half Angles

In ABC prov...

Question

In $△ A B C$ prove that $\frac{b - c}{a} = \frac{tan (B / 2) - tan (C / 2)}{tan (B / 2) + tan (C / 2)}$

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Solution

From sine rule
a = 2 R sin A
b = 2 R sin B
C = 2 R sin C
$\Rightarrow \frac{b - c}{a} = \frac{2 R s i n B - 2 R s i n C}{2 R s i n A}$
$= \frac{s i n B - s i n C}{s i n A}$
$= \frac{2 s i n (\frac{B - C}{2}) c o s (\frac{B + C}{2})}{2 s i n (\frac{B + C}{2}) c o s (\frac{B + C}{2})}$
$= \frac{\frac{s i n (\frac{B - C}{2})}{\frac{c o s B}{2} c o s \frac{C}{2}}}{\frac{s i n (\frac{B + C}{2})}{c o s \frac{B}{2} c o s \frac{C}{2}}}$
$= \frac{\frac{s i n \frac{B}{2} c o s \frac{C}{2}}{c o s \frac{B}{2} c o s \frac{C}{2}} - \frac{c o s \frac{B}{2} s i n \frac{C}{2}}{c o s \frac{B}{2} c o s \frac{C}{2}}}{\frac{s i n \frac{B}{2} c o s \frac{C}{2}}{c o s \frac{B}{2} c o s \frac{C}{2}} + \frac{c o s \frac{B}{2} s i n \frac{C}{2}}{c o s \frac{B}{2} c o s \frac{C}{2}}}$
$= \frac{t a n \frac{B}{2} - t a n \frac{C}{2}}{t a n \frac{B}{2} + t a n \frac{C}{2}}$

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