# 20. Properties of Triangles’ Solutions Part 3¶

1. The diagram is given below:

Let $BD = DE = EC = x.$ Also let,

$\angle BAD = \alpha, \angle DAE = \beta, EAC = \gamma, CEA = \theta$

Given, $\tan\alpha = t_1, \tan\beta = t_2, \tan\gamma = t_3$

Applying $m:n$ rule in $\triangle ABC,$ we get

$(2x + x)\cot\theta = 2x\cot(\alpha + \beta) - x\cot\gamma$

From $\triangle ADC,$ we get

$2x\cot\theta = x\cot\beta - x\cot\gamma$

$\Rightarrow \frac{3}{2} = \frac{2(\alpha + \beta) - \cot\gamma}{\cot\beta - \cot\gamma}$

$\Rightarrow 3\cot\beta - 3\cot\gamma = 4\cot(\alpha + \beta) - 2\cot\gamma$

$3\cot\beta - \cot\gamma = \frac{4(\cot\alpha\cot\beta - 1)}{\cot\alpha + \cot\beta}$

$3\cot^2\beta - \cot\beta\cot\gamma + 3\cot\alpha\cot\beta - \cot\alpha\cot\gamma = 4\cot\alpha\cot\beta - 4$

$4 + 4\cot^2\beta = \cot^2\beta + \cot\alpha\cot\beta + \cot\beta\cot\gamma + \cot\alpha\cot\gamma$

$4(1 + \cot^2\beta) = (\cot\beta + \cot\alpha)(\cot\beta + \cot\gamma)$

$4\left(1 + \frac{1}{t_2^2}\right) = \left(\frac{1}{t_1} + \frac{1}{t_2}\right)\left(\frac{1}{t_2} + \frac{1}{t_3}\right)$

1. The diagram is given below:

Let the medians be $AD, BE$ and $CF$ meet at $O.$ From question,

$\angle BOC=\alpha, \angle COA = \beta, \angle AOB = \gamma$

Let $AD = p_1, BE=p_2, CF = p_3$

$AO:OD = 2:1 \Rightarrow AO = \frac{2}{3}p_1$

Similalrly, $OB = \frac{2}{3}p_2, OC = \frac{2}{3}p_3$

Applying cosine rule in $\triangle AOC,$

$\cos\beta = \frac{OA^2 + OC^2 - AC^2}{2.OA.OC} = \frac{\frac{4}{9}p_1^2 + \frac{4}{9}p_3^2 - b^2}{2.\frac{2}{3}p_1\frac{2}{3}p_3}$

$\cos\beta = \frac{4p_1^2 + 4p_3^2 - 9b^2}{8p_1p_3}$

$\Delta AOC = \frac{1}{2}.OA.OC.\sin\beta$

$\frac{1}{3}\Delta = \frac{1}{2}\frac{2}{3}p_1\frac{2}{3}p_3\sin\beta$ where $\Delta$ is area of triangle $ABC.$

$\sin\beta = \frac{3\Delta}{2p_1p_3}$

$\Rightarrow \cos\beta = \frac{4p_1^2 + 4p_3^2 - 9b^2}{12\Delta}$

$\because AD$ is mean of $\triangle ABC$

$\therefore AB^2 + AC^2 = 2BD^2 + 2AD^2$

$\Rightarrow b^2 + c^2 = 2\frac{a^2}{4} + 2p_1^2$

$p_1^2 = \frac{2b^2 + 2c^2 - a^2}{4}$

Similarly, $p_2^2 = \frac{2c^2 + 2a^2 - b^2}{4}$

and $p_3^2 = \frac{2a^2 + 2b^2 - c^2}{4}$

$\Rightarrow \cos\beta = \frac{(2b^2 + 2c^2 - a^2) + (2a^2 + 2b^2 - c^2) - 9b^2}{12\Delta}$

$= \frac{a^2 + c^2 - 5b^2}{12\Delta}$

Similarly, $\cos\alpha = \frac{b^2 + c^2 - 5a^2}{12\Delta}$

Similarly, $\cos\gamma = \frac{a^2 + b^2 - 5c^2}{12\Delta}$

$\cos\alpha + \cos\beta + \cos\gamma = \frac{-3(a^2 + b^2 + c^2)}{12\Delta}$

$= -\frac{a^2 + b^2 + c^2}{4\Delta}$

$\cot A + \cot B + \cot C = \frac{b^2 + c^2 - a^2}{2bc\sin A} + \frac{c^2 + a^2 - b^2}{2ca\sin B} + \frac{a^2 + b^2 - c^2}{2ab\sin C}$

$= \frac{a^2 + b^2 + c^2}{4\Delta}$

$\Rightarrow \cot\alpha + \cot\beta + \cot\gamma + \cot A + \cot B + \cot C = 0$

1. The diagram is given below:

Let $AD$ be the perpendicular from $A$ on $BC.$ When $AD$ is extended it meets the circumscrbing circle at $E.$ Given, $DE=\alpha.$

Since angles in the same segment are equal, $\therefore \angle AEB = \angle ACB = \angle C$

and $\angle AEC = \angle ABC = \angle B$

From right angled $\triangle BDE, \tan C = \frac{BD}{DE}$

From right angled $\triangle CDE, \tan B = \frac{CD}{DE}$

$\tan B + \tan C = \frac{a}{\alpha}$

Similarly, $\tan C + \tan A = \frac{b}{\beta}$

and $\tan A + \tan B = \frac{c}{\gamma}$

$\frac{a}{\alpha} + \frac{b}{\beta} + \frac{c}{\gamma} = 2(\tan A + \tan B + \tan C)$

1. The diagram is given below:

Let $H$ be the orthocenter of triangle $ABC.$

From question, $HA = p, HB = q, HC = r.$

From figure, $\angle HBD = \angle EBC = 90^\circ - C$

$\angle HCD = \angle FCB = 90^\circ - B$

$\therefore \angle BHC = 180^\circ - (\angle HBD + \angle HCD)$

$= 180^\circ - [90^\circ - C + 90^\circ - B] = B + C = \pi - A$

Similarly, $\angle AHC = \pi - B$ and $\angle AHB = \pi - C$

Now $\Delta BHC + \Delta CHA + \Delta AHB = \Delta ABC$

$\Rightarrow \frac{1}{2}[qr\sin BHC + rp\sin CHA + pq \sin AHB] = \Delta$

$\Rightarrow \frac{1}{2}[qr\sin A + rp\sin B + pq\sin C] = \Delta$

$\Rightarrow aqr + brp + cpq = abc$

1. The diagram is given below:

Let $O$ be the center of unit circle and $A$ be the center of circle whose arc $BPC$ divides the unit circle in two equal parts.

i.e area of the curve $ABPCA = \frac{1}{2}$ area of the unit circle $= \frac{\pi}{2}$

Let the radius of this new circle be $r.$

Then, $AC = AB = AP = r$

$\because OB = OC = 1 \therefore \angle OCA = \angle OAC = \theta$

Applying sine rule in $\triangle AOC,$

$\frac{r}{\sin(\pi -2\theta)} = \frac{1}{\sin\theta}$

$r = 2\cos\theta$

Now area of $ABPCA = 2[$ Are of sector $ACP +$ Area of sector $OAC -$ Are of $\triangle OAC]$

$= 2\left[\frac{1}{2}r^2\theta + \frac{1}{2}1^2(\pi - 2\theta) - \frac{1}{2}\sin(\pi -2\theta)\right]$

$=\theta. 4\cos^2\theta + \pi - 2\theta - \sin2\theta [\because r = 2\cos\theta]$

$= 2\theta\cos2\theta - \sin2\theta + \pi$

$\Rightarrow \frac{\pi}{2} = 2\theta\cos2\theta - \sin2\theta + \pi$

$\Rightarrow \frac{\pi}{2} = \sin2\theta - 2\theta\cos2\theta$

2. The diagram is given below:

Let $EF$ be the perpendicular bisector of $BC$ and $O$ the center of the square. From question,

Let $BF = FC = a \Rightarrow BC = EF = 2a$ and $OE=OF = a$

Let $OP = x \Rightarrow OQ = x$

$\Rightarrow PF = a - x, QF = a + x$

From right angled $\triangle BPF,$

$\tan B = \frac{PF}{BF} = \frac{a - x}{x}$

From right angled $\triangle QFC,$

$\tan C = \frac{a + x}{a}$

$\Rightarrow (\tan B - \tan C)^2 = \frac{4x^2}{a^2}$

In triangle $ABC,$

$\tan A = \tan[\pi - (B + C)] = -\tan(B + C) = -\frac{2a^2}{x^2}$

$\Rightarrow \tan A(\tan B - \tan C)^2 + 8 = 0$

3. The diagram is given below:

$\because CD$ is internal bisector of $\angle C$

$\therefore \frac{AD}{DB} = \frac{b}{a}$

$\Rightarrow BD = \frac{ac}{a + b}$

Since angles of the same segment are equal.

$\therefore \angle ABE = \angle ACE = \frac{C}{2}$

and $\angle BEC = \angle BAC = A$

Applying sine rule in $\triangle BEC,$

$\frac{CE}{\sin CBE} = \frac{BC}{\sin BEC} \Rightarrow CE = \frac{a\sin\left(a + \frac{C}{2}\right)}{\sin A}$

Applying sine rule in $\triangle BDE,$

$\frac{DE}{\sin\frac{C}{2}} = \frac{BD}{\sin A}\Rightarrow DE = \frac{ac\sin\frac{C}{2}}{(a + b)\sin A}$

$\Rightarrow \frac{CE}{DE} = \frac{a\sin\left(B + \frac{C}{2}\right)}{ac\sin\frac{C}{2}}(a + b)$

$\Rightarrow \frac{CE}{DE} = \frac{(a + b)\sin\left(B + \frac{C}{2}\right)}{c\sin \frac{C}{2}}$

Now, $\frac{\sin\left(B + \frac{C}{2}\right)}{\sin\frac{C}{2}} = \frac{\sin\left(B + \frac{C}{2}\right).2\cos\frac{C}{2}}{2\sin\frac{C}{2}\cos\frac{C}{2}}$

$= \frac{\sin(B + C)+ \sin B}{\sin C} = \frac{\sin A + \sin B}{\sin C} = \frac{a + b}{c}$

Thus, $\frac{CE}{DE} = \frac{(a + b)^2}{c^2}$

4. The diagram is given below:

$\because AD$ is the interna; bisector of angle $A,$

$\frac{BD}{DC} = \frac{BA}{AC} = \frac{c}{b}$

$\Rightarrow \frac{BD}{c} = \frac{DC}{b} = \frac{BD + DC}{b + c}$

$\Rightarrow \frac{BD}{c} = \frac{a}{b + c}$

Similarly, $\frac{BF}{a} = \frac{c}{a + b}$

Now $\frac{\Delta BDF}{\Delta ABC} = \frac{BD.BF.\sin B}{a.c.\sin B} = \frac{ac}{(a + b)(b + c)}$

Similarly, $\frac{\Delta CDE}{\Delta ABC} = \frac{ab}{(a + c)(b + c)}$

and $\frac{\Delta AEF}{\Delta ABC} = \frac{bc}{(a + b)(a + c)}$

$\therefore \frac{\Delta DEF}{\Delta ABC} = \frac{\Delta ABC - (\Delta BDF + \Delta CDE + \Delta AEF)}{\Delta ABC}$

$= 1 - \frac{ac}{(a + b)(b + c)} - \frac{ab}{(a + c)(b + c)} - \frac{bc}{(a + b)(a + c)}$

$= \frac{2abc}{(a + b)(b + c)(c + a)}$

$\Delta DEF = \frac{2.\Delta .abc}{(a + b)(b + c)(c + a)}$

5. The diagram is given below:

$\because A + B + C = \pi \Rightarrow 3\alpha + 3\beta + 3\gamma = \pi \Rightarrow \alpha + \beta + \gamma = \frac{\pi}{3}$

Clearly, $\angle ADB = 60^\circ$

Applying sine rule in $\triangle ADB,$

$\frac{AR}{\sin\beta} = \frac{c}{\sin[\pi - (\alpha + \beta)]}$

$AR = \frac{c\sin \beta}{\sin(\alpha + \beta)} = \frac{2R\sin C\sin\beta}{\sin(\alpha + \beta)}$

$= \frac{2R\sin3\gamma\sin\beta}{\sin(60^\circ - \gamma)}$

$= \frac{2R(3\sin\gamma - 4\sin^3\gamma)\sin\beta}{\sin(60^\circ - \gamma)}.\frac{\cos(30^\circ - \gamma)}{cos(30^\circ - \gamma}$

$= \frac{4R\sin\beta\sin\gamma.(3 - 4\sin^2\gamma).\cos(30^\circ - \gamma)}{\sin(09^\circ - 2\gamma) + \sin 30^\circ}$

$= \frac{4R\sin\beta\sin\gamma\cos(30^\circ - \gamma)(3 - 4\sin^2\gamma)}{\cos2\gamma + \frac{1}{2}}$

$= \frac{8R\sin\beta\sin\gamma\cos(30^\circ - \gamma)(3 - 4\sin^2\gamma)}{2\cos2\gamma + 1}$

$= \frac{8R\sin\beta\sin\gamma\cos(30^\circ - \gamma)(3 - 4\sin^2\gamma)}{2(1 - 2\sin^2\gamma) + 1}$

$= 8R\sin\beta\sin\gamma\cos(30^\circ - \gamma)$

6. The diagram is given below:

From figure, $\angle AOX = \frac{\pi}{2} - \theta$

Since $OX$ is tangent to the circle, $OB$ will pass through the center $P$ of the circle and hence $OB$ will be the diameter of the given circle.

$\Rightarrow \angle OAB = 90^\circ \Rightarrow \angle OBA = 90^\circ - \theta$

By property of circle, $OAQ = \angle OBA = 90^\circ - \theta$

Also, $AOQ = 90^\circ - theta[\because OQ = OA]$

$\therefore OQA = 2\theta \Rightarrow AQX = \pi - 2\theta$

$\angle BOX = \frac{\pi}{1}$

Applying sine rule in $\triangle ABT, we get$

$\frac{AB}{\sin(\pi - 2\theta)} = \frac{AT}{\sin\theta}$

$\frac{AB}{\sin2\theta} = \frac{t}{\sin\theta} \Rightarrow AB = 2t\cos\theta$

From right angled $\triangle AOB,$

$\tan\theta = \frac{AB}{OA} \Rightarrow AB = c\tan\theta$

$\Rightarrow c\tan\theta = 2t\cos\theta$

$\Rightarrow c\sin\theta - t(1 + \cos2\theta) = 0$

Let $AN\perp OB$

Now, $ON + NB = OB$

$\Rightarrow c\cos\theta + AB\sin\theta = d$

$\Rightarrow c\cos\theta + 2t\sin\theta\cos\theta = d$

$\Rightarrow c\cos\theta + t\sin2\theta = d$

7. Since $AD$ is the median $\therefore BD = DC = \frac{a}{2}$

Also, $\because \angle DAE = \angle CAE = \frac{A}{3}$

$AE$ is common and $\angle AED = angle AEC = 90^\circ$

$\therefore AD = AC = b$

Applying cosine rule in $\triangle ABD,$

$\cos\frac{A}{3} = \frac{AB^2 + AD^2 - BD^2}{2.AB.AD}$

$= \frac{c^2 + b^2 - \frac{a^2}{4}}{2.c.b} = \frac{4b^2 + 4c^2 - a^2}{8bc}$

Applying cosine rule in $\triangle ABC,$

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$

$4\cos^3\frac{A}{3} - 3\cos\frac{A}{3} = \frac{b^2 + c^2 - a^2}{2bc}$

$\Rightarrow 4\cos^3\frac{A}{3} - 4\cos\frac{A}{3} = \frac{b^2 + c^2 - a^2}{2bc} - \frac{4b^2 + 4c^2 - a^2}{8bc}$

$\Rightarrow 4\cos\frac{A}{3}\left(1 - \cos^2\frac{A}{3}\right) = \frac{4b^2 + 4c^2 - a^2}{8bc} - \frac{b^2 + c^2 - a^2}{2bc}$

$\Rightarrow \cos\frac{A}{3}.\sin^2\frac{A}{3} = \frac{3a^2}{32bc}$

8. Given, $\cos A + \cos B + \cos C = \frac{3}{2}$

$\Rightarrow \frac{b^2 + c^2 - a^2}{2bc} + \frac{c^2 + a^2 - b^2}{2ca} + \frac{a^2 + b^2 - c^2}{2ab} = \frac{3}{2}$

$\Rightarrow a(b^2 + c^2 - a^2) + b(c^2 + a^2 - b^2) + c(a^2 + b^2 - c^2) = 3abc$

$\Rightarrow a(b - c)^2 + b(c - a)^2 + c(a - b)^2 = a^3 + b^3 + c^3 - 3abc = \frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2](a + b + c)$

$\Rightarrow \frac{b + c - a}{2}(b - c)^2 + \frac{c + a - b}{2}(c - a)^2 + \frac{a + b - c}{2}(a - b)^2 = 0$

$\Rightarrow (a - b)^2 = (b - c)^2 = (c - a)^2 = 0$

$\Rightarrow a = b = c$

9. If the $\triangle ABC$ is equilateral $\Rightarrow A = B = C = 60^\circ$

$\Rightarrow \tan A + \tan B + \tan C = 3\sqrt{3}$

If $\tan A + \tan B + \tan C = 3\sqrt{3}$

then $\tan A\tan B\tan C = 3\sqrt{3}$

Thus, A.M. of $\tan A, \tan B, \tan C =$ G.M. of $\tan A, \tan B, \tan C$

$\Rightarrow \tan A = \tan B = \tan C$

10. L.H.S. $= (a + b + c)\tan\frac{C}{2} = 2R(\sin A + \sin B + \sin C)\frac{\sin\frac{C}{2}}{\cos\frac{C}{2}}$

$= 2R\left(2\sin\frac{A + B}{2}\cos\frac{A - B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2}\right)\frac{\sin\frac{C}{2}}{\cos\frac{C}{2}}$

$= 2R\left(2\cos\frac{C}{2}\cos\frac{A - B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2}\right)\frac{\sin\frac{C}{2}}{\cos\frac{C}{2}}$

$= 2R\left(2\sin\frac{C}{2}\cos\frac{A - B}{2} + 2\sin^2\frac{C}{2}\right)$

$= 2R\left(2\cos\frac{A + B}{2}\cos\frac{A - B}{2} + 2\sin^2\frac{C}{2}\right)$

$= 2R\left(\cos A + \cos B + 2\sin^2\frac{C}{2}\right)$

R.H.S $= a\cot\frac{A}{2} + b\cot\frac{B}{2} - c\cot\frac{C}{2}$

$= 2R\left(\sin A\cot\frac{A}{2} = \sin B\cot\frac{B}{2} - \sin C\cot\frac{C}{2}\right)$

$= 2R\left(2\cos^2\frac{A}{2} + 2\cos^2\frac{B}{2} - 2\cos^2\frac{C}{2}\right)$

$= 2R\left(2\cos^2\frac{A}{2} + 2\cos^2\frac{B}{2} - 2 + 2\sin^2\frac{C}{2}\right)$

$= 2R\left(\cos A + \cos B + 2\sin^2\frac{C}{2}\right)$

Thus, L.H.S. = R.H.S.

11. $\sin^2\theta = \frac{1 - \cos2\theta}{2} \Rightarrow \sin^4\theta = \frac{(1 - \cos2\theta)^2}{4}$

Also, for a triangle $\cos 2A + \cos 2B + \cos 2C = -1 -4\cos A\cos B\cos C$

and $\cos^22A + \cos^2B + \cos^2C = 1 + 2\cos 2A\cos 2B\cos 2C$

L.H.S. $= \frac{(1 - \cos2A)^2}{4} + \frac{(1 - \cos2B)^2}{4} + \frac{(1 - \cos 2C)^2}{4}$

$= \frac{1}{4}[3 - 2(\cos2A +\cos 2B + \cos 2C) + \cos^22A + \cos^22B + \cos^22C]$

$= \frac{1}{4}[3 - 2(-1 - 4\cos A\cos B\cos C) + 1 + 2\cos 2A\cos 2B\cos 2C]$

$= \frac{3}{2} + 2\cos A\cos B\cos C + \frac{1}{2}\cos 2A\cos 2B\cos 2C =$ R.H.S.

12. Observe the relations in previous problem.

L.H.S. $= \frac{(1 + \cos2A)^2}{4} + \frac{(1 + \cos2B)^2}{4} + \frac{(1 + \cos2C)^2}{4}$

$= \frac{1}{4}[3 + 2(\cos2A +\cos 2B + \cos 2C) + \cos^22A + \cos^22B + \cos^22C]$

$= \frac{1}{4}[3 + 2(-1 - 4\cos A\cos B\cos C) + 1 + 2\cos 2A\cos 2B\cos 2C]$

$= \frac{1}{2} - 2\cos A\cos B\cos C + \frac{1}{2}\cos2A\cos2B\cos2C =$ R.H.S.

13. L.H.S. $= \cot B + \frac{\cos C}{\cos A\sin B} = \frac{\cos B\cos A + \cos[\pi - (A + B)]}{\cos A\sin B}$

$= \frac{\cos B\cos A - \cos(A + B)}{\cos A\sin B} = \frac{\sin A\sin B}{\cos A\sin B}$

$= \tan A$

R.H.S. $= \cot C + \frac{\cos B}{\cos A\sin C} = \frac{\cos C\cos A + \cos[\pi - (A + C)]}{\cos A\sin C}$

$= \frac{\sin A\sin C}{\cos A\sin C} = \tan A$

Thus, L.H.S. = R.H.S.

14. $\frac{a\sin(B - C)}{b^2 - c^2} = \frac{1}{2R}.\frac{\sin A\sin(B - C)}{\sin^2B - \sin^2C}$

$= \frac{1}{2R}.\frac{\sin[\pi - (B + C)]\sin(B - C)}{\sin(B + C)\sin(B - C)}$

$= \frac{1}{2R}[\because \sin\{\pi - (B + C) = \sin(B + C)\}]$

Similarly, $\frac{b\sin(C - A)}{c^2 - a^2} = \frac{c\sin(A - B)}{a^2 - b^2} = \frac{1}{2R}$

15. R.H.S. $= \frac{b - c}{a}\cos\frac{A}{2} = \frac{\sin B - \sin C}{\sin A}\cos\frac{A}{2}$

$= \frac{2\cos\frac{B + C}{2}\sin\frac{B - C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}\cos\frac{A}{2}$

$= \frac{\sin\frac{A}{2}\sin\frac{B - C}{2}}{\sin\frac{A}{2}}$

$= \sin\frac{B - C}{2} =$ L.H.S.

16. L.H.S. $= \sin^3A\cos(B - C) + \sin^3B\cos(C - A) + \sin^3C\cos(A - B) = 3\sin A\sin B\sin C$

$= \sin^2A\sin(B + C)\cos(B - C) + \sin^2B\sin(C + A)\cos(C - A) + \sin^2C\sin(A + B)\cos(A - B)$

$= \frac{1}{2}[\sin^2A(\sin 2B + \sin 2C) + \sin^2B(\sin 2C + \sin 2A) + \sin^2C(\sin2A + \sin 2B)]$

$= \sin^2A(\sin B\cos B + \sin C\cos C) + \sin^2B(\sin C\cos C + \sin A\cos A) + \sin^2C(\sin A\cos A + \sin B\cos B)$

$= \sin A\sin B(\sin A\cos B + \cos A\sin B) + \sin B\sin C(\sin B\cos C + \cos B\sin C) + \sin A\sin C(\sin A\cos C + \cos A\sin C)$

$= \sin A\sin B\sin(A + B) + \sin B\sin C\sin(B + C) + \sin A\sin C\sin(A + C)$

$= 3\sin A\sin B\sin C =$ R.H.S.

17. L.H.S. $= \sin^3A + \sin^3B + \sin^3C = \frac{3}{4}[\sin A + \sin B + \sin C] - \frac{1}{3}[\sin 3A + \sin 3B + \sin 3C]$

$\sin A + \sin B + \sin C = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2} + 2\sin \frac{C}{2}\cos \frac{C}{2}$

$= 2\cos\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\frac{A - B}{2}\right]$

$= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

Similarly, $\sin3A + \sin3B + \sin3C = 4\cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2}$

18. $\sin3A\sin^3(B - C) = \sin3A\frac{3\sin(B - C) - \sin3(B - C)}{4}$

Now $\sin 3A\sin3(B - C) = \sin3(B + C)\sin3(B - C) = \sin^23B - \sin^23C$

and $\sin 3A\sin(B - C) = (3\sin A - 4\sin^3A)\sin(B - C)$

$= 3\sin(B + C)\sin(B - C) - 4\sin^2A\sin(B + C)\sin(B - C)$

$= 3[\sin^2B - \sin^2C] - 4\sin^2A(\sin^2B - \sin^2C)$

Thus, $\sin3A\sin^3(B - C) + \sin3B\sin^3(C - A) + \sin3C\sin^3(A - B) = 0$

19. $\sin3A\cos^3(B - C) = \sin3A.\frac{3\cos(B - C) + \cos3(B - C}{4}$

Now, $\frac{1}{4}\sin3A \cos3(B - C) = \frac{1}{8}2\sin3(B + C)\cos3(B - C) = \frac{1}{8}(\sin 6B + \sin 6C)$

So $\sum \sin3A \cos3(B - C) = \frac{1}{4}(\sin 6A + \sin 6B + \sin 6C)$

Again, $\frac{3}{4}\sin3A.\cos(B - C) = \frac{3}{4}(3\sin A - 4\sin^3A)\cos(B - C)$

$= \frac{9}{8}[(\sin 2B + \sin 2C) -3\sin^3A\cos(B - C)$

We have just proved that $\sum \sin^3A\cos(B - C) = 3\sin A\sin B\sin C$

$\therefore \frac{9}{8}\sum(\sin2B + \sin 2C) = \frac{9}{4}(\sin 2A + \sin 2B + \sin 2C)$

and $3\sum\sin^3A\cos(B - C) = 9\sin A\sin B\sin C$

Now, $\sin2A + \sin2B + \sin2C = 4\sin A\sin B\sin C$

and $\sin6A + \sin 6B + \sin6C = 4\sin3A\sin3B\sin3C$

Thus, the sum would be $\sin 3A\sin3B\sin3C$

20. L.H.S. $= \left(\frac{s(s - a) + s(s - b)}{\Delta}\right)\left(\frac{a.(s - a)(s - c)}{ac} + \frac{b(s - b)(s - c)}{bc}\right)$

$= \frac{s(2s - a - b)}{\Delta}\left(\frac{(s - c)(2s - a - b)}{c}\right)$

$= c\cot\frac{C}{2} =$ R.H.S.

21. Given $a,b,c$ are in A.P. $\therefore 2b = a + c$

$2\sin B = \sin A + \sin C \Rightarrow 4\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A + C}{2}\cos\frac{A - C}{2}$

$\Rightarrow 2\cos\frac{A + C}{2} = \cos\frac{A - C}{2}$

L.H.S. $= 4(1 - \cos A)(1 - \cos C) = 4.2\sin^2\frac{A}{2}.2\sin^2\frac{C}{2}$

$4\left(2\sin\frac{A}{2}\sin\frac{C}{2}\right)^2 = 4\left(\cos\frac{A - C}{2} - \cos\frac{A + C}{2}\right)^2$

$= 4\left(2\cos\frac{A + C}{2} - \cos\frac{A + C}{2}\right)^2 = 4\cos^2\frac{A + C}{2}$

R.H.S. $= \cos A + \cos C = 2\cos\frac{A + C}{2}\cos\frac{A - C}{2} = 4\cos^2\frac{A + C}{2}$

Thus, L.H.S. = R.H.S.

22. Given, $a, b, c$ are in H.P.

$\Rightarrow \frac{1}{a}. \frac{1}{b}, \frac{1}{c}$ are in A.P.

$\Rightarrow \frac{s}{a}, \frac{s}{b}, \frac{s}{c}$ are in A.P.

$\Rightarrow \frac{s}{a} -1, \frac{s}{b} - 1, \frac{s}{c} - 1$ are in A.P.

$\Rightarrow \frac{bc}{(s - b)(s - c), \frac{ca}{(s - c)(s - a)}}, \frac{ab}{(s - a)(s - c)}$ are in A.P.

$\Rightarrow \frac{1}{\sin^2\frac{A}{2}}, \frac{1}{\sin^2\frac{B}{2}}, \frac{1}{\sin^2\frac{C}{2}}$ are in A.P.

$\Rightarrow \sin^2\frac{A}{2}, \sin^2\frac{B}{2}, \sin^2\frac{C}{2}$ are in H.P.

23. We have to prove that $\cos A\cot\frac{A}{2}, \cos B\cot\frac{B}{2}, \cot C\cot\frac{C}{2}$ are in A.P.

$\Rightarrow \left(1 - 2\sin^2\frac{A}{2}\right)\cot\frac{A}{2} \left(1 - 2\sin^2\frac{B}{2}\right)\cot\frac{B}{2}, \left(1 - 2\sin^2\frac{C}{2}\right)\cot\frac{C}{2}$ are in A.P.

$\Rightarrow \cot\frac{A}{2} - \sin A, \cot\frac{B}{2} - \sin B, \cot\frac{C}{2} - \sin C$ are in A.P.

Thus if we prove that $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ and $\sin A, \sin B, \sin C$ are in A.P. separately then we would have prove the above in A.P.

Now, $\cot \frac{A}{2} + \cot \frac{C}{2} = \frac{s(s - a)}{\Delta} + \frac{s(s - c)}{\Delta} = \frac{s}{\Delta}[2s - a - c]$

$= \frac{s}{\Delta}(2s - 2b)[\because 2b = a + c] = 2\cot \frac{B}{2}$