C -a- b -1-tan A - L tan B -2-1-tan A -2 tan B -2

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

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Byju's Answer

Standard XII

Mathematics

Sine Rule

c /a+ b =1-ta...

Question

$\frac{c}{a + b} = \frac{1 - t a n (\frac{A}{2}) t a n (\frac{B}{2})}{1 + t a n (\frac{A}{2}) t a n (\frac{B}{2})}$

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Solution

$LHS \frac{c}{a + b} = \frac{k s i n C}{k s i n A + k s i n B} = \frac{2 s i n \frac{C}{2} c o s \frac{C}{2}}{2 s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2})} = \frac{s i n \frac{C}{2} c o s \frac{C}{2}}{s i n (\frac{π - C}{2}) . c o s (\frac{A - B}{2})} = \frac{s i n (\frac{π - (A + B)}{2}) c o s \frac{C}{2}}{c o s (\frac{C}{2}) . c o s (\frac{A - B}{2})} = \frac{c o s (\frac{A + B}{2})}{c o s (\frac{A - B}{2})} = \frac{c o s \frac{A}{2} . c o s \frac{B}{2} - s i n \frac{A}{2} s i n \frac{B}{2}}{c o s \frac{A}{2} . c o s \frac{B}{2} + s i n \frac{A}{2} s i n \frac{B}{2}} Dividing both Numerator and denominator by (c o s \frac{A}{2} c o s \frac{B}{2}) = \frac{1 - t a n \frac{A}{2} t a n \frac{B}{2}}{1 + t a n \frac{A}{2} t a n \frac{B}{2}}$

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Similar questions

Q. If

A + B + C = π

, prove that

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

If A+B +C = $180^{\circ}$ , then $\sum t a n \frac{A}{2} t a n \frac{B}{2}$ is equal

Q.
In any triangle

A B C, 1 - tan (\frac{A}{2}) tan (\frac{B}{2}) =

Q. If

A + B + C = π

sin (A + \frac{C}{2}) = k sin \frac{C}{2}

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tan \frac{A}{2} tan \frac{B}{2} =

prove

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

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ca+b=1−tan(A2)tan(B2)1+tan(A2)tan(B2)

LHSca+b=k sin Ck sin A+k sin B=2sinC2cosC22sin(A+B2)cos(A−B2)=sinC2cosC2sin(π−C2).cos(A−B2)=sin(π−(A+B)2)cosC2cos(C2).cos(A−B2)=cos(A+B2)cos(A−B2)=cosA2.cosB2−sinA2sinB2cosA2.cosB2+sinA2sinB2Dividing both Numerator and denominator by (cosA2cosB2)=1−tanA2tanB21+tanA2tanB2

$\frac{c}{a + b} = \frac{1 - t a n (\frac{A}{2}) t a n (\frac{B}{2})}{1 + t a n (\frac{A}{2}) t a n (\frac{B}{2})}$