\displaystyle \textbf{Question 1: }\text{If in a }\triangle ABC,\ \angle A=45^\circ,\ \angle B=60^\circ\text{ and}
\displaystyle \angle C=75^\circ,\text{ find the ratio of its sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,\ b\text{ and }c\text{ be the sides opposite }\angle A,\ \angle B\text{ and }\angle C\text{ respectively.}
\displaystyle \text{By the Sine Rule,}
\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k
\displaystyle \therefore \frac{a}{\sin45^\circ}=\frac{b}{\sin60^\circ}=\frac{c}{\sin75^\circ}=k
\displaystyle \sin45^\circ=\frac{1}{\sqrt{2}},\qquad \sin60^\circ=\frac{\sqrt{3}}{2}
\displaystyle \sin75^\circ=\sin(45^\circ+30^\circ)
\displaystyle =\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ
\displaystyle =\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac{1}{2}
\displaystyle =\frac{\sqrt{3}+1}{2\sqrt{2}}
\displaystyle \therefore \frac{a}{\frac{1}{\sqrt{2}}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{c}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=k
\displaystyle \therefore a=\frac{k}{\sqrt{2}},\qquad b=\frac{\sqrt{3}}{2}k,\qquad c=\frac{\sqrt{3}+1}{2\sqrt{2}}k
\displaystyle \therefore a:b:c=\frac{1}{\sqrt{2}}:\frac{\sqrt{3}}{2}:\frac{\sqrt{3}+1}{2\sqrt{2}}
\displaystyle \text{Multiplying each term by }2\sqrt{2},
\displaystyle \therefore a:b:c=2:\sqrt{6}:(\sqrt{3}+1)
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If in }\triangle ABC,\ \angle C=105^\circ,\ \angle B=45^\circ\text{ and }a=2,   \text{find }b.
\displaystyle \text{Answer:}
\displaystyle \angle A=180^\circ-(105^\circ+45^\circ)=30^\circ
\displaystyle \text{By the Sine Rule,}
\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}
\displaystyle \therefore \frac{2}{\sin30^\circ}=\frac{b}{\sin45^\circ}
\displaystyle \therefore b=2\times\frac{\sin45^\circ}{\sin30^\circ}
\displaystyle =2\times\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=2\sqrt{2}
\displaystyle \therefore b=2\sqrt{2}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In }\triangle ABC,\text{ if }a=18,\ b=24,\ c=30\text{ and }\angle C=90^\circ,
\displaystyle \text{find }\sin A,\ \sin B\text{ and }\sin C.
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule,}
\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
\displaystyle \frac{18}{\sin A}=\frac{24}{\sin B}=\frac{30}{1}
\displaystyle \therefore \sin A=\frac{18}{30}=\frac{3}{5}
\displaystyle \therefore \sin B=\frac{24}{30}=\frac{4}{5}
\displaystyle \therefore \sin C=\sin90^\circ=1
\displaystyle \\

\displaystyle \text{In any triangle }ABC,\text{ prove the following:}
\displaystyle \textbf{Question 4: }\frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A\text{ and }b=k\sin B.
\displaystyle \text{LHS}=\frac{a-b}{a+b}
\displaystyle =\frac{k\sin A-k\sin B}{k\sin A+k\sin B}
\displaystyle =\frac{\sin A-\sin B}{\sin A+\sin B}
\displaystyle =\frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}
\displaystyle =\frac{\sin\frac{A-B}{2}}{\cos\frac{A-B}{2}}\times\frac{\cos\frac{A+B}{2}}{\sin\frac{A+B}{2}}
\displaystyle =\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)}=\text{RHS}
\displaystyle \therefore \frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }(a-b)\cos\frac{C}{2}=c\sin\left(\frac{A-B}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=(a-b)\cos\frac{C}{2}
\displaystyle =k(\sin A-\sin B)\cos\frac{C}{2}
\displaystyle =2k\cos\frac{A+B}{2}\sin\frac{A-B}{2}\cos\frac{C}{2}
\displaystyle =2k\cos\frac{180^\circ-C}{2}\sin\frac{A-B}{2}\cos\frac{C}{2}
\displaystyle =2k\sin\frac{C}{2}\sin\frac{A-B}{2}\cos\frac{C}{2}
\displaystyle =k\left(2\sin\frac{C}{2}\cos\frac{C}{2}\right)\sin\frac{A-B}{2}
\displaystyle =k\sin C\sin\frac{A-B}{2}
\displaystyle =c\sin\frac{A-B}{2}=\text{RHS}
\displaystyle \therefore (a-b)\cos\frac{C}{2}=c\sin\left(\frac{A-B}{2}\right).\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\frac{c}{a-b}=\frac{\tan\frac{A}{2}+\tan\frac{B}{2}}{\tan\frac{A}{2}-\tan\frac{B}{2}}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=\frac{c}{a-b}
\displaystyle =\frac{k\sin C}{k\sin A-k\sin B}
\displaystyle =\frac{\sin C}{\sin A-\sin B}
\displaystyle =\frac{2\sin\frac{C}{2}\cos\frac{C}{2}}{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}
\displaystyle \text{Since }A+B+C=180^\circ,
\displaystyle \cos\frac{C}{2}=\sin\frac{A+B}{2}\text{ and }\cos\frac{A+B}{2}=\sin\frac{C}{2}.
\displaystyle \therefore \text{LHS}=\frac{\sin\frac{C}{2}\sin\frac{A+B}{2}}{\sin\frac{C}{2}\sin\frac{A-B}{2}}
\displaystyle =\frac{\sin\frac{A+B}{2}}{\sin\frac{A-B}{2}}
\displaystyle =\frac{\sin\frac{A}{2}\cos\frac{B}{2}+\cos\frac{A}{2}\sin\frac{B}{2}}{\sin\frac{A}{2}\cos\frac{B}{2}-\cos\frac{A}{2}\sin\frac{B}{2}}
\displaystyle \text{Dividing the numerator and denominator by }\cos\frac{A}{2}\cos\frac{B}{2},
\displaystyle \text{LHS}=\frac{\tan\frac{A}{2}+\tan\frac{B}{2}}{\tan\frac{A}{2}-\tan\frac{B}{2}}=\text{RHS}
\displaystyle \therefore \frac{c}{a-b}=\frac{\tan\frac{A}{2}+\tan\frac{B}{2}}{\tan\frac{A}{2}-\tan\frac{B}{2}}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\frac{c}{a+b}=\frac{1-\tan\frac{A}{2}\tan\frac{B}{2}}{1+\tan\frac{A}{2}\tan\frac{B}{2}}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=\frac{c}{a+b}
\displaystyle =\frac{k\sin C}{k\sin A+k\sin B}
\displaystyle =\frac{\sin C}{\sin A+\sin B}
\displaystyle =\frac{2\sin\frac{C}{2}\cos\frac{C}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}
\displaystyle \text{Since }A+B+C=180^\circ,
\displaystyle \sin\frac{C}{2}=\cos\frac{A+B}{2}\text{ and }\sin\frac{A+B}{2}=\cos\frac{C}{2}.
\displaystyle \therefore \text{LHS}=\frac{\cos\frac{A+B}{2}\cos\frac{C}{2}}{\cos\frac{C}{2}\cos\frac{A-B}{2}}
\displaystyle =\frac{\cos\frac{A+B}{2}}{\cos\frac{A-B}{2}}
\displaystyle =\frac{\cos\frac{A}{2}\cos\frac{B}{2}-\sin\frac{A}{2}\sin\frac{B}{2}}{\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{A}{2}\sin\frac{B}{2}}
\displaystyle \text{Dividing the numerator and denominator by }\cos\frac{A}{2}\cos\frac{B}{2},
\displaystyle \text{LHS}=\frac{1-\tan\frac{A}{2}\tan\frac{B}{2}}{1+\tan\frac{A}{2}\tan\frac{B}{2}}=\text{RHS}
\displaystyle \therefore \frac{c}{a+b}=\frac{1-\tan\frac{A}{2}\tan\frac{B}{2}}{1+\tan\frac{A}{2}\tan\frac{B}{2}}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\frac{a+b}{c}=\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\frac{C}{2}}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=\frac{a+b}{c}
\displaystyle =\frac{k\sin A+k\sin B}{k\sin C}
\displaystyle =\frac{\sin A+\sin B}{\sin C}
\displaystyle =\frac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{2\sin\frac{C}{2}\cos\frac{C}{2}}
\displaystyle \text{Since }A+B=180^\circ-C,
\displaystyle \sin\frac{A+B}{2}=\sin\frac{180^\circ-C}{2}=\cos\frac{C}{2}.
\displaystyle \therefore \text{LHS}=\frac{2\cos\frac{C}{2}\cos\frac{A-B}{2}}{2\sin\frac{C}{2}\cos\frac{C}{2}}
\displaystyle =\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\text{RHS}
\displaystyle \therefore \frac{a+b}{c}=\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\frac{C}{2}}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\sin\left(\frac{B-C}{2}\right)=\frac{b-c}{a}\cos\frac{A}{2}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{RHS}=\frac{b-c}{a}\cos\frac{A}{2}
\displaystyle =\frac{k\sin B-k\sin C}{k\sin A}\cos\frac{A}{2}
\displaystyle =\frac{\sin B-\sin C}{\sin A}\cos\frac{A}{2}
\displaystyle =\frac{2\cos\frac{B+C}{2}\sin\frac{B-C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}\cos\frac{A}{2}
\displaystyle =\frac{\cos\frac{B+C}{2}\sin\frac{B-C}{2}}{\sin\frac{A}{2}}
\displaystyle \text{Since }B+C=180^\circ-A,
\displaystyle \cos\frac{B+C}{2}=\cos\frac{180^\circ-A}{2}=\sin\frac{A}{2}.
\displaystyle \therefore \text{RHS}=\frac{\sin\frac{A}{2}\sin\frac{B-C}{2}}{\sin\frac{A}{2}}
\displaystyle =\sin\frac{B-C}{2}=\text{LHS}
\displaystyle \therefore \sin\left(\frac{B-C}{2}\right)=\frac{b-c}{a}\cos\frac{A}{2}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\frac{a^2-c^2}{b^2}=\frac{\sin(A-C)}{\sin(A+C)}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=\frac{a^2-c^2}{b^2}
\displaystyle =\frac{k^2\sin^2A-k^2\sin^2C}{k^2\sin^2B}
\displaystyle =\frac{\sin^2A-\sin^2C}{\sin^2B}
\displaystyle \text{Since }B=180^\circ-(A+C),
\displaystyle \sin B=\sin\left(180^\circ-(A+C)\right)=\sin(A+C).
\displaystyle \therefore \text{LHS}=\frac{\sin^2A-\sin^2C}{\sin^2(A+C)}
\displaystyle =\frac{\sin(A+C)\sin(A-C)}{\sin^2(A+C)}
\displaystyle =\frac{\sin(A-C)}{\sin(A+C)}=\text{RHS}
\displaystyle \therefore \frac{a^2-c^2}{b^2}=\frac{\sin(A-C)}{\sin(A+C)}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }b\sin B-c\sin C=a\sin(B-C)
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=b\sin B-c\sin C
\displaystyle =k\sin^2B-k\sin^2C
\displaystyle =k\left(\sin^2B-\sin^2C\right)
\displaystyle =k\sin(B+C)\sin(B-C)
\displaystyle \text{Since }B+C=180^\circ-A,
\displaystyle \sin(B+C)=\sin(180^\circ-A)=\sin A.
\displaystyle \therefore \text{LHS}=k\sin A\sin(B-C)
\displaystyle =a\sin(B-C)=\text{RHS}
\displaystyle \therefore b\sin B-c\sin C=a\sin(B-C).\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }a^2\sin(B-C)=(b^2-c^2)\sin A
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{RHS}=(b^2-c^2)\sin A
\displaystyle =k^2(\sin^2B-\sin^2C)\sin A
\displaystyle =k^2\sin(B+C)\sin(B-C)\sin A
\displaystyle \text{Since }B+C=180^\circ-A,
\displaystyle \sin(B+C)=\sin(180^\circ-A)=\sin A.
\displaystyle \therefore \text{RHS}=k^2\sin^2A\sin(B-C)
\displaystyle =a^2\sin(B-C)=\text{LHS}
\displaystyle \therefore a^2\sin(B-C)=(b^2-c^2)\sin A.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}   =\frac{a+b-2\sqrt{ab}}{a-b},\qquad a\ne b
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\frac{a+b-2\sqrt{ab}}{a-b}
\displaystyle =\frac{(\sqrt a-\sqrt b)^2}{(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)}
\displaystyle =\frac{\sqrt a-\sqrt b}{\sqrt a+\sqrt b}
\displaystyle \text{By the Sine Rule, let }a=k\sin A\text{ and }b=k\sin B.
\displaystyle \therefore \text{RHS}=\frac{\sqrt{k\sin A}-\sqrt{k\sin B}}{\sqrt{k\sin A}+\sqrt{k\sin B}}
\displaystyle =\frac{\sqrt{k}\sqrt{\sin A}-\sqrt{k}\sqrt{\sin B}}{\sqrt{k}\sqrt{\sin A}+\sqrt{k}\sqrt{\sin B}}
\displaystyle =\frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}=\text{LHS}
\displaystyle \therefore \frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}
\displaystyle =\frac{a+b-2\sqrt{ab}}{a-b}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }a(\sin B-\sin C)+b(\sin C-\sin A)+c(\sin A-\sin B)=0
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \therefore \sin A=\frac{a}{k},\qquad \sin B=\frac{b}{k},\qquad \sin C=\frac{c}{k}
\displaystyle \text{LHS}=a(\sin B-\sin C)+b(\sin C-\sin A)+c(\sin A-\sin B)
\displaystyle =a\left(\frac{b}{k}-\frac{c}{k}\right)+b\left(\frac{c}{k}-\frac{a}{k}\right)+c\left(\frac{a}{k}-\frac{b}{k}\right)
\displaystyle =\frac{1}{k}(ab-ac+bc-ab+ac-bc)
\displaystyle =\frac{1}{k}(0)=0=\text{RHS}
\displaystyle \therefore a(\sin B-\sin C)+b(\sin C-\sin A)+c(\sin A-\sin B)=0.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\frac{a^2\sin(B-C)}{\sin A}+\frac{b^2\sin(C-A)}{\sin B}   +\frac{c^2\sin(A-B)}{\sin C}=0
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=\frac{a^2\sin(B-C)}{\sin A}+\frac{b^2\sin(C-A)}{\sin B}
\displaystyle \qquad+\frac{c^2\sin(A-B)}{\sin C}
\displaystyle =ka\sin(B-C)+kb\sin(C-A)+kc\sin(A-B)
\displaystyle =k\left[a\sin(B-C)+b\sin(C-A)+c\sin(A-B)\right]
\displaystyle =k^2\left[\sin A\sin(B-C)+\sin B\sin(C-A)\right.
\displaystyle \left.\qquad+\sin C\sin(A-B)\right]
\displaystyle =k^2\left[\sin(B+C)\sin(B-C)+\sin(C+A)\sin(C-A)\right.
\displaystyle \left.\qquad+\sin(A+B)\sin(A-B)\right]
\displaystyle =k^2\left[(\sin^2B-\sin^2C)+(\sin^2C-\sin^2A)\right.
\displaystyle \left.\qquad+(\sin^2A-\sin^2B)\right]
\displaystyle =k^2(0)=0=\text{RHS}
\displaystyle \therefore \frac{a^2\sin(B-C)}{\sin A}+\frac{b^2\sin(C-A)}{\sin B}
\displaystyle \qquad+\frac{c^2\sin(A-B)}{\sin C}=0.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }a^2(\cos^2B-\cos^2C)+b^2(\cos^2C-\cos^2A)   +c^2(\cos^2A-\cos^2B)=0
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \therefore \sin A=\frac{a}{k},\qquad\sin B=\frac{b}{k},\qquad\sin C=\frac{c}{k}
\displaystyle \text{LHS}=a^2(\cos^2B-\cos^2C)+b^2(\cos^2C-\cos^2A)
\displaystyle \qquad+c^2(\cos^2A-\cos^2B)
\displaystyle =a^2\left[(1-\sin^2B)-(1-\sin^2C)\right]
\displaystyle \qquad+b^2\left[(1-\sin^2C)-(1-\sin^2A)\right]
\displaystyle \qquad+c^2\left[(1-\sin^2A)-(1-\sin^2B)\right]
\displaystyle =a^2(\sin^2C-\sin^2B)+b^2(\sin^2A-\sin^2C)
\displaystyle \qquad+c^2(\sin^2B-\sin^2A)
\displaystyle =\frac{a^2}{k^2}(c^2-b^2)+\frac{b^2}{k^2}(a^2-c^2)
\displaystyle \qquad+\frac{c^2}{k^2}(b^2-a^2)
\displaystyle =\frac{1}{k^2}\left(a^2c^2-a^2b^2+a^2b^2-b^2c^2+b^2c^2-a^2c^2\right)
\displaystyle =\frac{1}{k^2}(0)=0=\text{RHS}
\displaystyle \therefore a^2(\cos^2B-\cos^2C)+b^2(\cos^2C-\cos^2A)
\displaystyle \qquad+c^2(\cos^2A-\cos^2B)=0.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }b\cos B+c\cos C=a\cos(B-C)
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=b\cos B+c\cos C
\displaystyle =k\sin B\cos B+k\sin C\cos C
\displaystyle =\frac{k}{2}\left(2\sin B\cos B+2\sin C\cos C\right)
\displaystyle =\frac{k}{2}(\sin2B+\sin2C)
\displaystyle =\frac{k}{2}\left[2\sin(B+C)\cos(B-C)\right]
\displaystyle =k\sin(B+C)\cos(B-C)
\displaystyle \text{Since }B+C=180^\circ-A,
\displaystyle \sin(B+C)=\sin(180^\circ-A)=\sin A.
\displaystyle \therefore \text{LHS}=k\sin A\cos(B-C)
\displaystyle =a\cos(B-C)=\text{RHS}
\displaystyle \therefore b\cos B+c\cos C=a\cos(B-C).\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\frac{\cos2A}{a^2}-\frac{\cos2B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }\sin A=ka\text{ and }\sin B=kb.
\displaystyle \text{LHS}=\frac{\cos2A}{a^2}-\frac{\cos2B}{b^2}
\displaystyle =\frac{1-2\sin^2A}{a^2}-\frac{1-2\sin^2B}{b^2}
\displaystyle =\left(\frac{1}{a^2}-\frac{1}{b^2}\right)-2\left(\frac{\sin^2A}{a^2}-\frac{\sin^2B}{b^2}\right)
\displaystyle =\left(\frac{1}{a^2}-\frac{1}{b^2}\right)-2\left(\frac{k^2a^2}{a^2}-\frac{k^2b^2}{b^2}\right)
\displaystyle =\left(\frac{1}{a^2}-\frac{1}{b^2}\right)-2(k^2-k^2)
\displaystyle =\frac{1}{a^2}-\frac{1}{b^2}=\text{RHS}
\displaystyle \therefore \frac{\cos2A}{a^2}-\frac{\cos2B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\frac{\cos^2B-\cos^2C}{b+c}+\frac{\cos^2C-\cos^2A}{c+a}   +\frac{\cos^2A-\cos^2B}{a+b}=0
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }\sin A=ka,\ \sin B=kb\text{ and }\sin C=kc.
\displaystyle \text{LHS}=\frac{\cos^2B-\cos^2C}{b+c}+\frac{\cos^2C-\cos^2A}{c+a}
\displaystyle \qquad+\frac{\cos^2A-\cos^2B}{a+b}
\displaystyle =\frac{(1-\sin^2B)-(1-\sin^2C)}{b+c}
\displaystyle \qquad+\frac{(1-\sin^2C)-(1-\sin^2A)}{c+a}
\displaystyle \qquad+\frac{(1-\sin^2A)-(1-\sin^2B)}{a+b}
\displaystyle =\frac{\sin^2C-\sin^2B}{b+c}+\frac{\sin^2A-\sin^2C}{c+a}
\displaystyle \qquad+\frac{\sin^2B-\sin^2A}{a+b}
\displaystyle =k^2\left[\frac{c^2-b^2}{b+c}+\frac{a^2-c^2}{c+a}+\frac{b^2-a^2}{a+b}\right]
\displaystyle =k^2\left[\frac{(c-b)(c+b)}{b+c}+\frac{(a-c)(a+c)}{c+a}\right.
\displaystyle \left.\qquad+\frac{(b-a)(b+a)}{a+b}\right]
\displaystyle =k^2[(c-b)+(a-c)+(b-a)]
\displaystyle =k^2(0)=0=\text{RHS}
\displaystyle \therefore \frac{\cos^2B-\cos^2C}{b+c}+\frac{\cos^2C-\cos^2A}{c+a}
\displaystyle \qquad+\frac{\cos^2A-\cos^2B}{a+b}=0.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }a\sin\frac{A}{2}\sin\left(\frac{B-C}{2}\right)+b\sin\frac{B}{2}\sin\left(\frac{C-A}{2}\right)   +c\sin\frac{C}{2}\sin\left(\frac{A-B}{2}\right)=0
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=a\sin\frac{A}{2}\sin\frac{B-C}{2}+b\sin\frac{B}{2}\sin\frac{C-A}{2}
\displaystyle \qquad+c\sin\frac{C}{2}\sin\frac{A-B}{2}
\displaystyle \text{Since }A+B+C=180^\circ,
\displaystyle \sin\frac{A}{2}=\cos\frac{B+C}{2},\qquad\sin\frac{B}{2}=\cos\frac{C+A}{2},
\displaystyle \sin\frac{C}{2}=\cos\frac{A+B}{2}.
\displaystyle \therefore \text{LHS}=a\cos\frac{B+C}{2}\sin\frac{B-C}{2}
\displaystyle \qquad+b\cos\frac{C+A}{2}\sin\frac{C-A}{2}
\displaystyle \qquad+c\cos\frac{A+B}{2}\sin\frac{A-B}{2}
\displaystyle =\frac{a}{2}(\sin B-\sin C)+\frac{b}{2}(\sin C-\sin A)
\displaystyle \qquad+\frac{c}{2}(\sin A-\sin B)
\displaystyle =\frac{1}{2}\left[a(\sin B-\sin C)+b(\sin C-\sin A)\right.
\displaystyle \left.\qquad+c(\sin A-\sin B)\right]
\displaystyle \text{By the Sine Rule, let }\sin A=ka,\ \sin B=kb\text{ and }\sin C=kc.
\displaystyle \therefore \text{LHS}=\frac{k}{2}\left[a(b-c)+b(c-a)+c(a-b)\right]
\displaystyle =\frac{k}{2}(ab-ac+bc-ab+ac-bc)
\displaystyle =\frac{k}{2}(0)=0=\text{RHS}
\displaystyle \therefore a\sin\frac{A}{2}\sin\left(\frac{B-C}{2}\right)+b\sin\frac{B}{2}\sin\left(\frac{C-A}{2}\right)
\displaystyle \qquad+c\sin\frac{C}{2}\sin\left(\frac{A-B}{2}\right)=0.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\frac{b\sec B+c\sec C}{\tan B+\tan C}=\frac{c\sec C+a\sec A}{\tan C+\tan A}   =\frac{a\sec A+b\sec B}{\tan A+\tan B}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \frac{b\sec B+c\sec C}{\tan B+\tan C}
\displaystyle =\frac{k\sin B\sec B+k\sin C\sec C}{\tan B+\tan C}
\displaystyle =\frac{k\tan B+k\tan C}{\tan B+\tan C}
\displaystyle =k
\displaystyle \text{Similarly,}
\displaystyle \frac{c\sec C+a\sec A}{\tan C+\tan A}
\displaystyle =\frac{k\tan C+k\tan A}{\tan C+\tan A}=k
\displaystyle \text{and}
\displaystyle \frac{a\sec A+b\sec B}{\tan A+\tan B}
\displaystyle =\frac{k\tan A+k\tan B}{\tan A+\tan B}=k
\displaystyle \therefore \frac{b\sec B+c\sec C}{\tan B+\tan C}=\frac{c\sec C+a\sec A}{\tan C+\tan A}
\displaystyle =\frac{a\sec A+b\sec B}{\tan A+\tan B}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }a\cos A+b\cos B+c\cos C=2a\sin B\sin C
\displaystyle =2b\sin A\sin C=2c\sin A\sin B
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=a\cos A+b\cos B+c\cos C
\displaystyle =k\sin A\cos A+k\sin B\cos B+k\sin C\cos C
\displaystyle =\frac{k}{2}\left(\sin2A+\sin2B+2\sin C\cos C\right)
\displaystyle =\frac{k}{2}\left[2\sin(A+B)\cos(A-B)+2\sin C\cos C\right]
\displaystyle \text{Since }A+B=180^\circ-C,
\displaystyle \sin(A+B)=\sin(180^\circ-C)=\sin C.
\displaystyle \therefore \text{LHS}=k\sin C\left[\cos(A-B)+\cos C\right]
\displaystyle =2k\sin C\cos\frac{A-B+C}{2}\cos\frac{A-B-C}{2}
\displaystyle \text{Since }A+B+C=180^\circ,
\displaystyle \cos\frac{A-B+C}{2}=\cos(90^\circ-B)=\sin B
\displaystyle \text{and }\cos\frac{A-B-C}{2}=\cos(A-90^\circ)=\sin A.
\displaystyle \therefore \text{LHS}=2k\sin A\sin B\sin C
\displaystyle =2a\sin B\sin C
\displaystyle =2b\sin A\sin C
\displaystyle =2c\sin A\sin B=\text{RHS}
\displaystyle \therefore a\cos A+b\cos B+c\cos C=2a\sin B\sin C
\displaystyle =2b\sin A\sin C=2c\sin A\sin B.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }a(\cos B\cos C+\cos A)=b(\cos C\cos A+\cos B)
\displaystyle =c(\cos A\cos B+\cos C)
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle a(\cos B\cos C+\cos A)
\displaystyle =a\left[\cos B\cos C+\cos\left(180^\circ-(B+C)\right)\right]
\displaystyle =a\left[\cos B\cos C-\cos(B+C)\right]
\displaystyle =a\left[\cos B\cos C-\cos B\cos C+\sin B\sin C\right]
\displaystyle =a\sin B\sin C
\displaystyle =k\sin A\sin B\sin C
\displaystyle \text{Similarly,}
\displaystyle b(\cos C\cos A+\cos B)=k\sin A\sin B\sin C
\displaystyle \text{and}
\displaystyle c(\cos A\cos B+\cos C)=k\sin A\sin B\sin C
\displaystyle \therefore a(\cos B\cos C+\cos A)=b(\cos C\cos A+\cos B)
\displaystyle =c(\cos A\cos B+\cos C).\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 24: }a(\cos C-\cos B)=2(b-c)\cos^2\frac{A}{2}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{LHS}=a(\cos C-\cos B)
\displaystyle =2a\sin\frac{B+C}{2}\sin\frac{B-C}{2}
\displaystyle =2k\sin A\sin\frac{180^\circ-A}{2}\sin\frac{B-C}{2}
\displaystyle =2k\sin A\cos\frac{A}{2}\sin\frac{B-C}{2}
\displaystyle =2k\left(2\sin\frac{A}{2}\cos\frac{A}{2}\right)\cos\frac{A}{2}\sin\frac{B-C}{2}
\displaystyle =2k\cos^2\frac{A}{2}\left(2\sin\frac{A}{2}\sin\frac{B-C}{2}\right)
\displaystyle \text{Since }A=180^\circ-(B+C),
\displaystyle \sin\frac{A}{2}=\cos\frac{B+C}{2}.
\displaystyle \therefore \text{LHS}=2k\cos^2\frac{A}{2}\left(2\cos\frac{B+C}{2}\sin\frac{B-C}{2}\right)
\displaystyle =2k\cos^2\frac{A}{2}(\sin B-\sin C)
\displaystyle =2\cos^2\frac{A}{2}(k\sin B-k\sin C)
\displaystyle =2(b-c)\cos^2\frac{A}{2}=\text{RHS}
\displaystyle \therefore a(\cos C-\cos B)=2(b-c)\cos^2\frac{A}{2}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{In }\triangle ABC,\text{ prove that, if }\theta\text{ is any angle, then}
\displaystyle b\cos\theta=c\cos(A-\theta)+a\cos(C+\theta).
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B\text{ and }c=k\sin C.
\displaystyle \text{RHS}=c\cos(A-\theta)+a\cos(C+\theta)
\displaystyle =c(\cos A\cos\theta+\sin A\sin\theta)
\displaystyle \qquad+a(\cos C\cos\theta-\sin C\sin\theta)
\displaystyle =k\sin C\cos A\cos\theta+k\sin C\sin A\sin\theta
\displaystyle \qquad+k\sin A\cos C\cos\theta-k\sin A\sin C\sin\theta
\displaystyle =k\cos\theta(\sin C\cos A+\sin A\cos C)
\displaystyle =k\cos\theta\sin(A+C)
\displaystyle \text{Since }A+C=180^\circ-B,
\displaystyle \sin(A+C)=\sin(180^\circ-B)=\sin B.
\displaystyle \therefore \text{RHS}=k\sin B\cos\theta
\displaystyle =b\cos\theta=\text{LHS}
\displaystyle \therefore b\cos\theta=c\cos(A-\theta)+a\cos(C+\theta).\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{In }\triangle ABC,\text{ if }\sin^2A+\sin^2B=\sin^2C,\text{ show that the}
\displaystyle \text{triangle is right-angled.}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }\sin A=ka,\qquad\sin B=kb,\qquad\sin C=kc.
\displaystyle \text{Given, }\sin^2A+\sin^2B=\sin^2C
\displaystyle \therefore k^2a^2+k^2b^2=k^2c^2
\displaystyle \therefore a^2+b^2=c^2
\displaystyle \text{By the converse of the Pythagoras Theorem, }\angle C=90^\circ.
\displaystyle \therefore \triangle ABC\text{ is right-angled at }C.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In any }\triangle ABC,\text{ if }a^2,\ b^2,\ c^2\text{ are in A.P., prove that}
\displaystyle \cot A,\ \cot B\text{ and }\cot C\text{ are also in A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a^2,\ b^2,\ c^2\text{ are in A.P.}
\displaystyle \therefore a^2+c^2=2b^2
\displaystyle \text{Let the area of }\triangle ABC=\Delta.
\displaystyle \cot A=\frac{\cos A}{\sin A}
\displaystyle =\frac{\frac{b^2+c^2-a^2}{2bc}}{\frac{2\Delta}{bc}}
\displaystyle =\frac{b^2+c^2-a^2}{4\Delta}
\displaystyle \text{Similarly,}
\displaystyle \cot B=\frac{c^2+a^2-b^2}{4\Delta}
\displaystyle \text{and }\cot C=\frac{a^2+b^2-c^2}{4\Delta}
\displaystyle \cot A+\cot C
\displaystyle =\frac{b^2+c^2-a^2+a^2+b^2-c^2}{4\Delta}
\displaystyle =\frac{2b^2}{4\Delta}
\displaystyle \text{Since }a^2+c^2=2b^2,
\displaystyle \cot A+\cot C=\frac{a^2+c^2}{4\Delta}
\displaystyle =\frac{2(a^2+c^2-b^2)}{4\Delta}
\displaystyle =2\cot B
\displaystyle \therefore \cot A,\ \cot B\text{ and }\cot C\text{ are in A.P. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The upper part of a tree broken by the wind makes an angle of }30^\circ
\displaystyle \text{with the ground. The distance from the root to the point where the top touches}
\displaystyle \text{the ground is }15\text{ m. Using the Sine Rule, find the height of the tree.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be the root, }B\text{ the breaking point and }C\text{ the point where the top touches the ground.}
\displaystyle \text{Then }AC=15\text{ m},\ \angle A=90^\circ,\ \angle C=30^\circ,\ \angle B=60^\circ.
\displaystyle \text{By the Sine Rule,}
\displaystyle \frac{BC}{\sin90^\circ}=\frac{AC}{\sin60^\circ}
\displaystyle \therefore BC=\frac{15}{\sin60^\circ}=\frac{15}{\frac{\sqrt{3}}{2}}=10\sqrt{3}\text{ m}
\displaystyle \text{Again,}
\displaystyle \frac{AB}{\sin30^\circ}=\frac{AC}{\sin60^\circ}
\displaystyle \therefore AB=\frac{15\times\sin30^\circ}{\sin60^\circ}
\displaystyle =\frac{15\times\frac12}{\frac{\sqrt3}{2}}=5\sqrt3\text{ m}
\displaystyle \therefore \text{Height of the tree}=AB+BC
\displaystyle =5\sqrt3+10\sqrt3=15\sqrt3\text{ m}
\displaystyle \therefore \text{The height of the tree is }15\sqrt3\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{At the foot of a mountain, the angle of elevation of its peak is }45^\circ.
\displaystyle \text{After ascending }1000\text{ m towards the mountain along a slope inclined at }30^\circ
\displaystyle \text{to the horizontal, the angle of elevation of the peak is }60^\circ.
\displaystyle \text{Find the height of the mountain.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the height of the mountain and }C\text{ be the initial point of observation.}
\displaystyle \text{Let }D\text{ be the point reached after ascending }1000\text{ m along the slope.}
\displaystyle \text{Draw }DE\perp BC\text{ and }DF\perp AB.
\displaystyle DE=1000\sin30^\circ=1000\times\frac{1}{2}=500\text{ m}
\displaystyle EC=1000\cos30^\circ=1000\times\frac{\sqrt3}{2}=500\sqrt3\text{ m}
\displaystyle \text{Since }BE\text{ is horizontal and }DE\text{ is vertical, }FB=DE=500\text{ m}.
\displaystyle \text{Let }AF=x\text{ m}.
\displaystyle \text{In right-angled }\triangle ADF,
\displaystyle \tan60^\circ=\frac{AF}{DF}
\displaystyle \sqrt3=\frac{x}{DF}
\displaystyle \therefore DF=\frac{x}{\sqrt3}
\displaystyle \therefore BE=DF=\frac{x}{\sqrt3}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan45^\circ=\frac{AB}{BC}
\displaystyle 1=\frac{AF+FB}{BE+EC}
\displaystyle 1=\frac{x+500}{\frac{x}{\sqrt3}+500\sqrt3}
\displaystyle \frac{x}{\sqrt3}+500\sqrt3=x+500
\displaystyle x+1500=x\sqrt3+500\sqrt3
\displaystyle 1500-500\sqrt3=x(\sqrt3-1)
\displaystyle 500\sqrt3(\sqrt3-1)=x(\sqrt3-1)
\displaystyle \therefore x=500\sqrt3\text{ m}
\displaystyle \therefore AB=AF+FB
\displaystyle =500\sqrt3+500
\displaystyle =500(\sqrt3+1)\text{ m}
\displaystyle \therefore \text{The height of the mountain is }500(\sqrt3+1)\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{A person observes the angle of elevation of the peak of a hill from a}
\displaystyle \text{station to be }\alpha.\text{ He walks }c\text{ metres along a slope inclined at an angle }\beta
\displaystyle \text{and finds the angle of elevation of the peak to be }\gamma.\text{ Show that the height}
\displaystyle \text{of the peak above the ground is }\frac{c\sin\alpha\sin(\gamma-\beta)}{\sin(\gamma-\alpha)}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }Q\text{ be the first point of observation, }T\text{ the second point and }P\text{ the peak.}
\displaystyle \text{Then }QT=c,\text{ and }QT\text{ is inclined at an angle }\beta\text{ to the horizontal.}
\displaystyle \text{Also, the angles of elevation of }P\text{ from }Q\text{ and }T\text{ are }\alpha\text{ and }\gamma.
\displaystyle \text{In }\triangle QTP,
\displaystyle \angle PQT=\alpha-\beta
\displaystyle \angle QTP=180^\circ-(\gamma-\beta)
\displaystyle \therefore \angle QPT=180^\circ-\left[(\alpha-\beta)+180^\circ-(\gamma-\beta)\right]
\displaystyle =\gamma-\alpha
\displaystyle \text{By the Sine Rule,}
\displaystyle \frac{PT}{\sin(\alpha-\beta)}=\frac{QT}{\sin(\gamma-\alpha)}
\displaystyle \therefore PT=\frac{c\sin(\alpha-\beta)}{\sin(\gamma-\alpha)}
\displaystyle \text{The vertical rise from }Q\text{ to }T=c\sin\beta.
\displaystyle \text{The vertical height of }P\text{ above }T=PT\sin\gamma.
\displaystyle \therefore \text{Height of the peak above the ground}
\displaystyle =c\sin\beta+PT\sin\gamma
\displaystyle =c\sin\beta+\frac{c\sin(\alpha-\beta)\sin\gamma}{\sin(\gamma-\alpha)}
\displaystyle =\frac{c[\sin\beta\sin(\gamma-\alpha)+\sin(\alpha-\beta)\sin\gamma]}{\sin(\gamma-\alpha)}
\displaystyle =\frac{c[\sin\beta(\sin\gamma\cos\alpha-\cos\gamma\sin\alpha)+\sin\gamma(\sin\alpha\cos\beta-\cos\alpha\sin\beta)]}{\sin(\gamma-\alpha)}
\displaystyle =\frac{c\sin\alpha(\sin\gamma\cos\beta-\cos\gamma\sin\beta)}{\sin(\gamma-\alpha)}
\displaystyle =\frac{c\sin\alpha\sin(\gamma-\beta)}{\sin(\gamma-\alpha)}
\displaystyle \therefore \text{The height of the peak above the ground is }\frac{c\sin\alpha\sin(\gamma-\beta)}{\sin(\gamma-\alpha)}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{If the sides }a,\ b,\ c\text{ of }\triangle ABC\text{ are in H.P., prove that}
\displaystyle \sin^2\frac{A}{2},\ \sin^2\frac{B}{2},\ \sin^2\frac{C}{2}\text{ are also in H.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }a,\ b,\ c\text{ are in H.P.,}
\displaystyle \frac{1}{a},\ \frac{1}{b},\ \frac{1}{c}\text{ are in A.P.}
\displaystyle \therefore \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}
\displaystyle \therefore \frac{a-b}{ab}=\frac{b-c}{bc}
\displaystyle \text{By the Sine Rule, }a:b:c=\sin A:\sin B:\sin C.
\displaystyle \therefore \frac{\sin A-\sin B}{\sin A\sin B}
\displaystyle =\frac{\sin B-\sin C}{\sin B\sin C}
\displaystyle \therefore \frac{\sin A-\sin B}{\sin A}=\frac{\sin B-\sin C}{\sin C}
\displaystyle \therefore \frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{\sin A}
\displaystyle =\frac{2\cos\frac{B+C}{2}\sin\frac{B-C}{2}}{\sin C}
\displaystyle \text{Since }A+B+C=180^\circ,
\displaystyle \cos\frac{A+B}{2}=\sin\frac{C}{2}
\displaystyle \text{and }\cos\frac{B+C}{2}=\sin\frac{A}{2}.
\displaystyle \therefore \frac{\sin\frac{C}{2}\sin\frac{A-B}{2}}{\sin A}
\displaystyle =\frac{\sin\frac{A}{2}\sin\frac{B-C}{2}}{\sin C}
\displaystyle \therefore \sin\frac{A-B}{2}\sin\frac{C}{2}\sin C
\displaystyle =\sin\frac{B-C}{2}\sin\frac{A}{2}\sin A
\displaystyle \therefore \sin\frac{A-B}{2}\sin^2\frac{C}{2}\cos\frac{C}{2}
\displaystyle =\sin\frac{B-C}{2}\sin^2\frac{A}{2}\cos\frac{A}{2}
\displaystyle \text{Now, }\cos\frac{C}{2}=\sin\frac{A+B}{2}
\displaystyle \text{and }\cos\frac{A}{2}=\sin\frac{B+C}{2}.
\displaystyle \therefore \sin^2\frac{C}{2}\sin\frac{A-B}{2}\sin\frac{A+B}{2}
\displaystyle =\sin^2\frac{A}{2}\sin\frac{B-C}{2}\sin\frac{B+C}{2}
\displaystyle \therefore \sin^2\frac{C}{2}\left(\sin^2\frac{A}{2}-\sin^2\frac{B}{2}\right)
\displaystyle =\sin^2\frac{A}{2}\left(\sin^2\frac{B}{2}-\sin^2\frac{C}{2}\right)
\displaystyle \text{Dividing by }\sin^2\frac{A}{2}\sin^2\frac{B}{2}\sin^2\frac{C}{2},
\displaystyle \frac{1}{\sin^2\frac{B}{2}}-\frac{1}{\sin^2\frac{A}{2}}
\displaystyle =\frac{1}{\sin^2\frac{C}{2}}-\frac{1}{\sin^2\frac{B}{2}}
\displaystyle \therefore \frac{1}{\sin^2\frac{A}{2}},\ \frac{1}{\sin^2\frac{B}{2}},\ \frac{1}{\sin^2\frac{C}{2}}\text{ are in A.P.}
\displaystyle \therefore \sin^2\frac{A}{2},\ \sin^2\frac{B}{2},\ \sin^2\frac{C}{2}\text{ are in H.P.}
\displaystyle \text{Hence proved.}
\displaystyle \\

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