If the sides a, b, c of a are in H.P., prove that ^2 A 2 , ^2 B 2 , ^2 C 2 are in H.P.

If the sides a, b, c of a are in H.P. 1 a ,1 b and 1 c are in AP 1 b -1 a =1 c -1 b a-b ba = b-c ca - = - ...[Using sine rule] 2 A-B 2 A + B 2 = 2 B-C 2 B + C 2 But A + B + C = A + B= -C A + B 2 = ( 2 - C 2 )= C 2 ^2 C 2 C 2 A-B 2 = B-C 2 A 2 ^2 A 2 ^2 C 2 A + B 2 A-B 2 = B-C 2 B + C 2 ^2 A 2 ^2 C 2 [ ^2 A 2 - ^2 B 2 ]= ^2 A 2 [ ^2 B 2 - ^2 C 2 ] ^2 C 2 ^2 A 2 - ^2 C 2 ^2 B 2 = ^2 A 2 ^2 B 2 - ^2 A 2 ^2 C 2 1 ^2 B 2 -1 ^2 A 2 =1 ^2 C 2 -1 ^2 B 2 Hence 1 ^2 A 2 ,1 B 2 ,1 ^2 C 2 are in A.P. ^2 A 2 , ^2 B 2 , ^2 C 2 are in H.P.

If the sides a, b, c of a are in H.P., prove that ^2 A 2 , ^2 B 2…

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If the sides a, b, c of a $\triangle\text{ABC}$ are in H.P., prove that $\sin^2\frac{\text{A}}{2},\sin^2\frac{\text{B}}{2},\sin^2\frac{\text{C}}{2}$ are in H.P.

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If the sides a, b, c of a $\triangle\text{ABC}$ are in H.P. $\therefore\frac{1}{\text{a}},\frac{1}{\text{b}}$ and $\frac{1}{\text{c}}$ are in AP $\therefore\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$ $\Rightarrow\frac{\text{a}-\text{b}}{\text{ba}}=\frac{\text{b}-\text{c}}{\text{ca}}$ $\Rightarrow\frac{\sin\text{A}-\sin\text{B}}{\sin\text{B}\sin\text{A}}=\frac{\sin\text{B}-\sin\text{C}}{\sin\text{C}\sin\text{B}}$ ...[Using sine rule] $\Rightarrow\frac{2\sin\frac{\text{A}-\text{B}}{2}\cos\frac{\text{A + B}}{2}}{\sin\text{A}}=\frac{2\sin\frac{\text{B}-\text{C}}{2}\cos\frac{\text{B + C}}{2}}{\sin\text{C}}$ But $\text{A + B + C = } \pi$ $\text{A + B}=\pi-\text{C}$ $\cos\frac{\text{A + B}}{2}=\cos\Big(\frac{\pi}{2}-\frac{\text{C}}{2}\Big)=\sin\frac{\text{C}}{2}$ $\sin^2\frac{\text{C}}{2}\cos\frac{\text{C}}{2}\sin\frac{\text{A}-\text{B}}{2}=\sin\frac{\text{B}-\text{C}}{2}\cos\frac{\text{A}}{2}\sin^2\frac{\text{A}}{2}$ $\sin^2\frac{\text{C}}{2}\sin\frac{\text{A + B}}{2}\sin\frac{\text{A}-\text{B}}{2}=\sin\frac{\text{B}-\text{C}}{2}\cos\frac{\text{B + C}}{2}\sin^2\frac{\text{A}}{2}$ $\sin^2\frac{\text{C}}{2}\Big[\sin^2\frac{\text{A}}{2}-\sin^2\frac{\text{B}}{2}\Big]=\sin^2\frac{\text{A}}{2}\Big[\sin^2\frac{\text{B}}{2}-\sin^2\frac{\text{C}}{2}\Big]$ $\sin^2\frac{\text{C}}{2}\sin^2\frac{\text{A}}{2}-\sin^2\frac{\text{C}}{2}\sin^2\frac{\text{B}}{2}=\sin^2\frac{\text{A}}{2}\sin^2\frac{\text{B}}{2}-\sin^2\frac{\text{A}}{2}\sin^2\frac{\text{C}}{2}$ $\frac{1}{\sin^2\frac{\text{B}}{2}}-\frac{1}{\sin^2\frac{\text{A}}{2}}=\frac{1}{\sin^2\frac{\text{C}}{2}}-\frac{1}{\sin^2\frac{\text{B}}{2}}$ Hence $\frac{1}{\sin^2\frac{\text{A}}{2}},\frac{1}{\sin\frac{\text{B}}{2}},\frac{1}{\sin^2\frac{\text{C}}{2}}$ are in A.P. $\therefore\sin^2\frac{\text{A}}{2},\sin^2\frac{\text{B}}{2},\sin^2\frac{\text{C}}{2}$ are in H.P.

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