/sides a, b, c of saec are in Hp, p...
Question

# /sides a, b, c of saec are in Hp, prove that sri (4) (9) and (9) are in HP Sides a b c of AABC are in H.P. prove that in * Sinta 1 Sine are in HP

JEE/Engineering Exams
Maths
Solution
95
4.0 (1 ratings)
As a , b and c are in ( mathrm{HP} ). ( S o frac{1}{a}, frac{1}{b} ) and ( frac{1}{c} ) are in ( A P ) Hence( frac{1}{b}-frac{1}{a}=frac{1}{c}-frac{1}{b} ) ( Rightarrow frac{sin A-sin B}{sin A cdot sin B}=frac{sin B-sin C}{sin C sin B} ) ( Rightarrow frac{2 sin frac{A-B}{2} cos frac{A+B}{2}}{sin A}=frac{2 sin frac{B C}{2} cos frac{B_{t} C}{2}}{sin C} ) ( Rightarrow frac{2 sin frac{A-B}{2} cos frac{A+B}{2}}{2 sin frac{A}{2} cos frac{A}{2}}=frac{2 sin frac{B C}{2} cos frac{B+C}{2}}{2 sin frac{C}{2} cos frac{C}{2}} ) ( B u t A+B+C=pi ) So ( A+B=pi-C ) So ( cos frac{A+B}{2}=cos left(frac{pi}{2}-frac{C}{2}right)=sin frac{C}{2} ) So ( sin ^{2} frac{C}{2} cos frac{C}{2} sin frac{A-B}{2}=sin frac{B-C}{2} cos frac{A}{2} sin ^{2} frac{A}{2} ) ( sin ^{2} frac{mathrm{C}}{2} sin frac{mathrm{A}+mathrm{B}}{2} sin frac{mathrm{A}-mathrm{B}}{2}=sin frac{mathrm{B}-mathrm{C}}{2} cos frac{mathrm{B}+mathrm{C}}{2} sin ^{2} frac{mathrm{A}}{2} ) Or ( sin ^{2} frac{mathrm{C}}{2}left[sin ^{2} frac{mathrm{A}}{2}-sin ^{2} frac{mathrm{B}}{2}right]=sin ^{2} frac{mathrm{A}}{2}left[sin ^{2} frac{mathrm{B}}{2}-sin ^{2} frac{mathrm{C}}{2}right] ) ( sin ^{2} frac{C}{2} sin ^{2} frac{A}{2}-sin ^{2} frac{C}{2} sin ^{2} frac{B}{2}=sin ^{2} frac{A}{2} sin ^{2} frac{B}{2}-sin ^{2} frac{A}{2} sin ^{2} frac{C}{2} ) Divide both sides by ( sin ^{2} frac{mathrm{C}}{2} sin ^{2} frac{mathrm{A}}{2} sin ^{2} frac{mathrm{B}}{2} ) ( frac{1}{sin ^{2} frac{1}{2}}-frac{1}{sin ^{2} frac{Lambda}{2}}=frac{1}{sin ^{2} frac{C}{2}}-frac{1}{sin ^{2} frac{1}{2}} ) Hence ( frac{1}{sin ^{2} frac{Lambda}{2}}, frac{1}{sin ^{2} frac{pi}{2}}, frac{1}{sin ^{2} frac{C}{2}} ) are is ( mathrm{AP} ) So ( sin ^{2} frac{A}{2}, sin ^{2} frac{B}{2} ) and ( sin ^{2} frac{C}{2} ) Are in ( mathrm{HP} )