C-a-b-tanA-2-tanB-2-tan1-2-tan1-2

LHS c/a b =k sin C/k sin A k sin B =sin C/sin A sin B =2sinC/2cosC/2/sin A sin B =2sinC/2cosC/2/2cos(A+B/2 )sin(A B/2 ) =sinC/2cos(π (A+B ) )/cos(π C/2 ) ...

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Byju's Answer

Standard XII

Mathematics

Sine Rule

c/a-b=tanA/2+...

Question

$\frac{c}{a - b} = \frac{t a n (\frac{A}{2}) + t a n (\frac{B}{2})}{t a n (\frac{A}{2}) - t a n (\frac{B}{2})}$

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Solution

$LHS \frac{c}{a - b} = \frac{k s i n C}{k s i n A - k s i n B} = \frac{s i n C}{s i n A - s i n B} = \frac{2 s i n \frac{C}{2} c o s \frac{C}{2}}{s i n A - s i n B} = \frac{2 s i n \frac{C}{2} c o s \frac{C}{2}}{2 c o s (\frac{A + B}{2}) s i n (\frac{A - B}{2})} = \frac{s i n \frac{C}{2} c o s (π - (A + B))}{c o s (\frac{π - C}{2}) s i n (\frac{A - B}{2})} = \frac{s i n \frac{C}{2} s i n \frac{(A - B)}{2}}{s i n \frac{C}{2} s i n \frac{(A - B)}{2}} = \frac{s i n \frac{(A + B)}{2}}{s i n \frac{(A - B)}{2}} = \frac{s i n (\frac{A}{2}) . c o s (\frac{B}{2}) + s i n (\frac{B}{2}) . c o s (\frac{A}{2})}{s i n (\frac{A}{2}) c o s (\frac{B}{2}) - s i n (\frac{B}{2}) c o s (\frac{A}{2})} = \frac{t a n (\frac{A}{2}) + t a n (\frac{B}{2})}{t a n (\frac{A}{2}) - t a n (\frac{B}{2})} (Dividing both Numerator and denominator by cos) (\frac{A}{2}) . c o s (\frac{B}{2})$

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ca−b=tan(A2) + tan(B2)tan(A2) − tan(B2)

$\frac{c}{a - b} = \frac{t a n (\frac{A}{2}) + t a n (\frac{B}{2})}{t a n (\frac{A}{2}) - t a n (\frac{B}{2})}$