# 18. Properties of Triangles’ Solutions Part 1¶

1. Let $a = 8$ cm, $b = 10$ cm and $c = 12$ cm. So the smallest angle will be $A$ and greatest angle will be $C.$ So from cosine rule,

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{100 + 144 - 64}{2.10.12} = \frac{3}{4}$

$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{64 + 100 - 144}{2.8.10} = \frac{1}{8}$

$\cos 2A = 2\cos^2A - 1 = 2.\frac{9}{16} - 1 = \frac{1}{8} = \cos C$

Thus, we see that greatest angle is double to that of smallest angle.

2. Let $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}, = k$

$\therefore b + c = 11k, c + a = 12k, a + b = 13k \Rightarrow a + b + c = 18k$

$\therefore a = 7k, b = 6k, c = 5k$

$\frac{\cos A}{7} = \frac{b^2 + c^2 - a^2}{2bc} = \frac{36k^2 + 25k^2 - 49k^2}{2.6.7.7k^2} = \frac{1}{35}$

$\frac{\cos B}{19} = \frac{c^2 + a^2 - b^2}{2ca} = \frac{25k^2 + 49k^2 - 36k^2}{2.5.7.19.k^2} = \frac{1}{35}$

$\frac{\cos C}{25} = \frac{a^2 + b^2 - c^2}{2ab} = \frac{49k^2 + 36k^2 - 25k^2}{2.7.6.25.k^2} = \frac{1}{35}$

3. Given, $\Delta = a^2 - (b - c)^2 \Rightarrow \Delta = a^2 - b^2 - c^2 + 2bc$

$\Rightarrow b^2 + c ^2 - a^2 = 2bc - \Delta \Rightarrow 2bc\cos A = 2bc - \frac{1}{2}bc\sin A$

$\Rightarrow 4\cos A + \sin A = 4 \Rightarrow 4\left(1 - 2\sin^2\frac{A}{2}\right) + 2\sin\frac{A}{2}\cos\frac{A}{2} = 4$

$2\sin\frac{A}{2}\left(\cos\frac{A}{2} - 4\sin\frac{A}{2}\right) = 0$

$\Rightarrow$ either $\sin\frac{A}{2} = 0$ or $\cos\frac{A}{2} - 4\sin\frac{A}{2} = 0$

$A = 0$ is not possible. $\therefore \tan\frac{A}{2} = \frac{1}{4}$

$\tan A = \frac{2\tan\frac{A}{2}}{1 - \tan^2\frac{A}{2}} = \frac{8}{15}.$

4. Since $A, B, C$ are in A.P. $\therefore 2B = A + C$

$\because A + B + C = 180^\circ \Rightarrow B = 60^\circ$

$\cos B = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \frac{1}{2} = \frac{c^2 + a^2 - b^2}{2ac} \Rightarrow a^2 - ac + c^2 = b^2$

$\Rightarrow \frac{a + c}{\sqrt{a^2 - ac + a^2}} = \frac{a + c}{b}$

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

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

5. $\Delta = \frac{1}{2}.a.p_1 \therefore \frac{1}{p_1} = \frac{a}{\Delta}$

Similarly, $\frac{1}{p_2} = \frac{b}{\Delta}, \frac{1}{p_3} = \frac{c}{\Delta}$

L.H.S. $= \frac{a + b - c}{2\Delta} = \frac{(a + b)^2 - c^2}{2\Delta(a + b + c)}$

$= \frac{2ab + a^2 + b^2 - c^2}{2\Delta(a + b + c)} = \frac{2ab + 2ab\cos C}{2\Delta(a + b + c)}$

$= \frac{ab(1 + \cos C)}{\Delta(a + b + c)} = \frac{2ab\cos^2\frac{C}{2}}{\Delta(a + b + c)}$

6. $\because :\tan\theta = \frac{2\sqrt{ab}}{a - b}\sin\frac{C}{2}$

$\Rightarrow (a - b)^2\tan^2\theta = 4ab\sin^2\frac{C}{2} \Rightarrow (a - b)^2(\sec^2\theta - 1) = 4ab\sin^2\frac{C}{2}$

$(a - b)^2\sec^2\theta = a^2b^2 -2ab\left(1 - 2\sin^2\frac{C}{2}\right)$

$(a - b)^2\sec^2\theta = a^2 + b^2 - 2ab\cos C = c^2[\because \cos C = \frac{a^2 + b^2 - c^2}{2ab}]$

$\Rightarrow c = (a - b)\sec\theta.$

7. $\Delta ABC = \frac{1}{2}bc\sin A = \frac{1}{2}6.3.\sin C = 9\sin C$

$\tan^2\frac{A - B}{2} = \frac{1 - \cos(A - B)}{1 + \cos(A - B)} = \frac{1}{9}$

$\tan\frac{A - B}{2} = \pm\frac{1}{3}$

$\because 0 < \frac{A - B}{2} < 90^\circ \therefore \tan\frac{A - B}{2} = \frac{1}{3}$

$\tan\frac{A - B}{2} = \frac{a - b}{a + b}\cot\frac{C}{2} \Rightarrow \cot\frac{C}{2} = 1$

$\Rightarrow C = 90^\circ$

Thus, required area $= 8\sin90^\circ = 9$

8. The diagram is given below:

From question, $\angle A = 75^\circ, \angle C = 60^\circ \Rightarrow \angle B = 45^\circ$

Also given, $\frac{\Delta BAD}{\Delta BCD} = \sqrt{3} = \frac{c.x.\sin\theta}{a.x.\sin(45^\circ - \theta)}$ where $BD = x$

$\Rightarrow \frac{c\sin\theta}{a\sin(45^\circ - \theta)} = \sqrt{3}$

$\Rightarrow \frac{\sin C\sin\theta}{\sin A\sin(45^\circ - \theta)} = \sqrt{3}$

$\Rightarrow \frac{\frac{\sqrt{3}}{2}\sin\theta}{\frac{\sqrt{3} + 1}{2\sqrt{2}}\sin(45^\circ - \theta)} = \sqrt{3}$

$\Rightarrow \sqrt{2}\sin\theta = (\sqrt{3} + 1)\sin(45^\circ - \theta)$

$\Rightarrow \sqrt{2}\sin\theta = (\sqrt{3} + 1)\left(\frac{1}{\sqrt{2}}\cos\theta - \frac{1}{\sqrt{2}}\sin\theta\right)$

$\Rightarrow 2\sin\theta = (\sqrt{3} + 1)(\cos\theta - \sin\theta)$

$\tan\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ$

9. We find the largest angle which is opposite to side $7,$ if any angle can be obtuse angle then this one can.

$\cos \theta = \frac{3^2 + 5^2 - 7^2}{2.3.5} = -\frac{15}{30} = -\frac{1}{2}$

$\Rightarrow \theta = 120^\circ$

10. Given $\angle A = 45^\circ, \angle B = 75^\circ \Rightarrow \angle = 60^\circ$

We know that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

$a = \frac{1}{\sqrt{2}}k, b = \frac{\sqrt{3} + 1}{2\sqrt{2}}k, c = \frac{\sqrt{3}k}{2}$

$a + c\sqrt{2} = \frac{1}{\sqrt{\sqrt{2}}}k + \frac{\sqrt{3}}{\sqrt{2}}k = \frac{1 + \sqrt{3}}{\sqrt{2}}k$

$2b = \frac{\sqrt{3} + 1}{\sqrt{2}}$

$\Rightarrow a + \sqrt{2}c = 2b$

11. The diagram is given below:

$\Delta BCD + \Delta ACD = \Delta ABC$

$\Rightarrow \frac{1}{2}3CD\sin30^\circ + \frac{1}{2}4CD\sin60^\circ = \frac{1}{2}3.4$

$\frac{3}{4}CD + \sqrt{3}CD = 6 \Rightarrow CD = \frac{24}{3 + 4\sqrt{3}}$

12. The smallest angle will be opposite to smallest side i.e. $4$ cm. Similarly, greatest angle will be opposite to greatest side i.e. $6$ cm.

Let $a = 4$ cm, $b = 5$ cm and $c = 6$ cm. Also, let opposite angles are $A, B$ and $C.$

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{45}{60} = \frac{3}{4}$

$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{5}{40} = \frac{1}{8}$

$\cos 2A = 2\cos^2A - 1 = 2.\frac{9}{16} - 1 = \frac{1}{8} = \cos C$

13. The diagram is given below:

$\angle A + \angle B + \angle C = 180^\circ = 10\angle B + \angle B + \angle B \Rightarrow \angle B = 15^\circ$

$\Rightarrow \angle A = 150^\circ$

Let $AB = AC = x$

$\therefore \cos 150^\circ = \frac{x^2 + x^2 - a^2}{2.x.x}$

$\Rightarrow -\sqrt{3}x^2 = 2x^2 - a^2 \Rightarrow x = \sqrt{\frac{1}{2 + \sqrt{3}}}a$

14. Let angles are $A = 2k, B=3k, C=7k \therefore 2k + 3k + 7k = 180^\circ \Rightarrow k = 15^\circ$

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = l$

$\sin A = \frac{1}{2}, \sin B= \frac{1}{\sqrt{2}}, \sin C = \frac{\sqrt{2 + \sqrt{3}}}{2}$

Now the ratios of sides can be proven.

15. Clearly, the sides are in the ratio of $7:6:5$

$\therefore \cos A = \frac{6^2 + 5^2 - 7^2}{2.6.5} = \frac{1}{5}$

$\cos B = \frac{7^2 + 5^2 - 6^2}{2.7.5} = \frac{19}{35}$

$\cos C = \frac{7^2 + 6^2 - 5^2}{2.7.6} = \frac{5}{7}$

$\therefore \cos A:\cos B:\cos C = 7:19;25$

16. $\tan \frac{C}{2} = \tan\frac{\pi - (A + B)}{2} = \cot\frac{A + B}{2} = \frac{1 - \tan\frac{A}{2}\tan\frac{B}{2}}{\tan\frac{A}{2} + \tan\frac{B}{2}}$

$= \frac{1 - \frac{5}{6}.\frac{20}{37}}{\frac{5}{6} + \frac{20}{37}}$

$= \frac{\frac{122}{222}}{\frac{305}{222}} = \frac{122}{305} = \frac{2}{5}$

$\sin A = \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{2.\frac{5}{6}}{1 + \frac{25}{36}}$

$= \frac{60}{61}$

$\sin B = \frac{2\tan\frac{B}{2}}{1 + \tan^2\frac{B}{2}} = \frac{2.\frac{20}{37}}{1 + \frac{400}{1369}}$

$= \frac{1480}{1769}$

$\sin B = \frac{2\tan\frac{C}{2}}{1 + \tan^2\frac{C}{2}} = \frac{2\frac{2}{5}}{1 + \frac{4}{25}}$

$= \frac{20}{29}$

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

$= k\left(\frac{60}{61} + \frac{20}{29}\right) = \frac{1740 + 1220}{1769} = \frac{2960}{1769} = 2b.$

17. It is much easier to prove this problem in reverse.

Given, $\frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c}$

Upon solving $a^2 + b^2 - c^2 = ab$

$\Rightarrow \frac{a^2 + b^2 - c^2}{2ab} = \frac{1}{2} = \cos60^\circ = \cos C$

18. Let the sides be $a, b, c.$ We know that $\Delta = \frac{1}{2}a.\alpha$ because area = $\frac{1}{2}\times$ base $\times$ altitude

$\Delta = \frac{1}{2}\alpha \Rightarrow \alpha = \frac{2\Delta}{a} \Rightarrow \frac{1}{\alpha} = \frac{a}{2\Delta}$

$\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{a^2 + b^2 + c^2}{4\Delta^2}$

$\frac{\cot A + \cot B + \cot C}{\Delta} = \frac{\cos A}{\Delta\sin A} + \frac{\cos B}{\Delta\sin B} + \frac{\cos C}{\Delta\sin C}$

$\Delta = \frac{1}{2}bc\sin A \Rightarrow \sin A = \frac{2\Delta}{bc}$

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

$\therefore \frac{\cos A}{\Delta\sin A} + \frac{\cos B}{\Delta\sin B} + \frac{\cos C}{\Delta\sin C} = \frac{a^2 + b^2 + c^2}{4\Delta^2}$

Hence proven.

19. Given, $\frac{a}{b} = 2 + \sqrt{3} = \tan75^\circ = \frac{\sin75^\circ}{\cos75^\circ}$

$\frac{\sin A}{\sin B} = \frac{\sin(90^\circ + 15^\circ)}{\sin(75^\circ - 15^\circ)}$

$\Rightarrow A = 105^\circ, B= 15^\circ$ which satisfied $A + B + C = 180^\circ$

20. $\cos C = \frac{1}{2} = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow (1 + \sqrt{3}^2) + 4 - c^2 = 2(1 + \sqrt{3})$

$\Rightarrow c^2 = 6 \Rightarrow c = \sqrt{6}$

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{4 + 6 - (1 + \sqrt{3})^2}{4\sqrt{6}}$

$= \frac{6 + 2\sqrt{3}}{\sqrt{6}} = \sqrt{6} + \sqrt{2} \Rightarrow A = 75^\circ$

Thus, $\angle B = 45^\circ$

21. Greatest angle will be opposite to greatest side i.e. $\sqrt{x^2 + xy + y^2}$

$\cos\theta = \frac{x^2 + y^2 - x^2 - xy - y^2}{2.x.y} = -\frac{1}{2} \Rightarrow \theta = 120^\circ$

22. Given sides are $2x + 3, x^2 + 3x + 3$ and $x^2 + 2x.$ Since lengths of sides is a positive quantity, therefore $x^2 + 2x > 0 \Rightarrow x > 0$

This leads to the fact that $x^2 + 3x + 3$ will be greatest side.

$\cos\theta = \frac{(2x + 3)^2 + (x^2 + 2x)^2 - (x^2 + 3x + 3)^2}{2(2x + 3)(x^2 + 2x)}$

$= \frac{4x^2 + 12x + 9 + x^4 + 4x^3 + 4x^2 - x^4 - 9x^2 - 9 - 6x^3 - 6x^2 - 18x}{4x^3 + 14x^2 + 12x}$

$= \frac{-2x^3 -7x^2 - 6x}{2(2x^3 + 7x^2 + 6x)} = -\frac{1}{2} = \cos120^\circ$

23. Given, $3a = b + c.$ We know that $s = \frac{a + b + c}{2} \Rightarrow s = 2a$

$\cot\frac{B}{2}\cot\frac{C}{2} = \frac{s(s - b)}{\Delta}.\frac{s(s - c)}{\Delta}$

$=\frac{s^2(s - b)(s - c)}{s(s - a)(s - b)(s - c)} = \frac{s}{s - a} = 2$

24. We have to prove that $a\sin\left(\frac{A}{2} + B\right) = (b + c)\sin\frac{A}{2}$

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

$= \frac{2\sin\frac{B + c}{2}\cos\frac{B - C}{2}}{2\sin\frac{A}{3}\cos\frac{A}{2}}$

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

$= \frac{\cos\left[\frac{B}{2} - \left\{\frac{\pi}{2} - \left(\frac{A + B}{2}\right)\right\}\right]}{}\sin\frac{A}{2}$

$= \frac{\sin\left(\frac{A}{2} + B\right)}{\sin\frac{A}{2}}$

25. Numerator of L.H.S. $= \frac{s(s - a)}{\Delta} + \frac{s(s - b)}{\Delta} + \frac{s(s - c)}{\Delta}$

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

Denominator of R.H.S. $= \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}$

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

[ $\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B = \frac{1}{2}ab\sin C$ ]

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

L.H.S. $= \frac{(a + b + c)^2}{a^2 + b^2 + c^2}$

26. First term of L.H.S. $= \frac{b^2 - c^2}{a^2}\sin2A = \frac{b^2 - c^2}{a^2}2\sin A\cos A$

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

$= \frac{1}{Kabc}[(b^2 - c^2)(b^2 + c^2 - a^2)] = \frac{1}{Kabc}[b^4 - c^4 -a^2(b^2 - c^2)]$

Similarly, second term $= \frac{1}{Kabc}[c^4 - a^4 - b^2(c^2 - a^2)]$

27. First term of L.H.S $= a^3\cos(B - C) = a^2[a\cos(B - C)]$

$= Ra^2[2\sin A\cos(B - C)] = Ra^2[2\sin(B + C)\cos(B - C)] = Ra^2[\sin2B + \sin2C]$

$=Ra^2[2\sin B\cos B + 2\sin C\cos C] = a^2[b\cos B + c\cos C]$

Similarly, second term $= b^2[a\cos A + c\cos C]$

and third term $= c^2[a\cos A + b\cos B]$

Adding, $ab[a\cos B + b\cos A] + ca[c\cos A + a\cos C] + bc[b\cos C + c\cos B]$

$= 3abc =$ R.H.S.

28. L.H.S. $= \frac{\cos^2\frac{B - C}{2}}{K^2[\sin B + \sin C]^2} + \frac{\sin^2\frac{B - C}{2}}{K^2[\sin B - \sin C]^2}$

$= \frac{1}{K^2}\left(\frac{\cos^2\frac{B - C}{2}}{4\sin^2\frac{B + C}{2}\cos^2\frac{B - C}{2}} + \frac{\sin^\frac{B - C}{2}}{4\cos^2\frac{B + C}{2}\sin^2\frac{B - C}{2}}\right)$

$= \frac{1}{4k^2}\left(\frac{1}{\sin^2\frac{B + C}{2}} + \frac{1}{\cos^2\frac{B + C}{2}}\right)$

$= \frac{1}{4K^2}\left(\frac{1}{\cos^2\frac{A}{2}} + \frac{1}{\sin^2\frac{A}{2}}\right)$

$= \frac{1}{k^2}.\frac{1}{4\sin^2\frac{A}{2}\cos^2\frac{A}{2}} = \frac{1}{a^2}$

29. First term of L.H.S. $= \frac{a}{\cos B\cos C} = \frac{2R\sin A}{\cos B\cos C}$

$= \frac{2R\sin(B + C)}{\cos B\cos C} = 2R(\tan B + \tan C)$

Similarly, second term $= 2R(\tan C + \tan A)$ and third term $= 2R(\tan A + \tan B)$

L.H.S. $= 4R[\tan A + \tan B + \tan C]$

$= 4R.\tan A\tan B\tan C[\because A + B + C = \pi \therefore \tan A + \tan B + \tan C = \tan A\tan B\tan C]$

$= 2.a\tan B\tan C\sec A =$ R.H.S.

30. We have to prove that $(b - c)\cos\frac{A}{2} = a\sin\frac{B - C}{2}$

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

$= \frac{2\sin\frac{A}{2}\sin\frac{B - C}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}} = \frac{\sin\frac{B - C}{2}}{\cos\frac{A}{3}}$

31. We have to prove that $\tan\left(\frac{A}{2} + B\right) = \frac{c + b}{c - b}\tan \frac{A}{2}$

$\frac{c + b}{c - b} = \frac{\sin C + \sin B}{\sin C - \sin B}$

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

$= \frac{\tan\frac{B + C}{2}}{\tan\frac{C - B}{2}} = \frac{\cot\frac{A}{2}}{\tan\frac{\pi - B - A - B}{2}}$

$= \frac{\tan\left(\frac{A}{2} + B\right)}{\tan\frac{A}{3}}$

32. We have to prove that $\tan\frac{A - B}{2} = \frac{a - b}{a + b}\cot\frac{C}{2}$

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

$= \frac{\tan\frac{A - B}{2}}{\tan \frac{A + B}{2}} = \frac{\tan\frac{A - B}{2}}{\cot\frac{C}{2}}$

33. L.H.S. $= (b + c)\cos A + (c + a)\cos B + (a + b)\cos C$

$= (a\cos B + b\cos A) + (b\cos C + c\cos B) + (a\cos C + c\cos A)$

$= c + a + b =$ R.H.S.

34. First term of L.H.S. $= \frac{\cos^2B - \cos^2C}{b + c} = \frac{1}{k}\left[\frac{(\cos B + \cos C)(\cos B - \cos C)}{\sin B + \sin C}\right]$

$= \frac{1}{k}\left[\frac{2\cos\frac{B+C}{2}\cos\frac{B - C}{2}.2\sin\frac{B + C}{2}\sin\frac{C - B}{2}}{2\sin\frac{B + C}{2}\cos\frac{B - C}{3}}\right]$

$= \frac{1}{k}\left[2\cos\frac{B + C}{2}\sin\frac{C - B}{2}\right] = \frac{1}{k}[\sin C - \sin B]$

Similarly, second term $= \frac{1}{k}[\sin A - \sin C]$ and third term $= \frac{1}{k}[\sin B - \sin A]$

Thus, L.H.S. = R.H.S. = 0

35. First term of L.H.S. $= a^3\sin(B - C) = Ra^2.2\sin A\sin(B - C) = Ra^2.2\sin(B + C)\sin(B - C)$

$= Ra^2[\cos 2C - \cos 2B] = Ra^2(1 - \sin^2C - 1 + \sin^2B) = R[(2R\sin B)^2 - (2R\sin C)^2]$

$= R[b^2 - c^2]$

Similarly, second term $= R[c^2 - a^2]$ and third term $= R[a^2 - b^2]$

Thus, L.H.S. $= 0 =$ R.H.S.

36. Consider first term i.e. $(b + c - a)\tan \frac{A}{2}$

$b + c - a = 2s - 2a = 2(s - a)$

$\tan\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$

$\therefore (b + c - a)\tan \frac{A}{2} = 2\sqrt{\frac{(s - a)(s - b)(s - a)}{s}}$

Similarly, $(c + a - b)\tan \frac{B}{2} = 2\sqrt{\frac{(s - a)(s - b)(s - a)}{s}} = (a + b - c)\tan\frac{C}{2}$

37. $1 - \tan\frac{A}{2}\tan\frac{B}{2} = 1 =- \sqrt{\frac{(s - b)(s - c)}{s(s - a)}.\frac{(s - a)(s - c)}{s(s - b)}}$

$= 1 - \frac{s - c}{s} = \frac{c}{s} = \frac{2c}{a + b + c} =$ R.H.S.

38. L.H.S. $= \frac{\cos2A}{a^2} - \frac{\cos2B}{b^2}$

$= \frac{1 - 2\sin^2A}{a^2} - \frac{1 - 2\sin^2B}{b^2}$

$= \frac{1 - 2.\frac{a^2}{4r^2}}{a^2} - \frac{1 - 2.\frac{b^2}{4r^2}}{b^2}$

$= \frac{1}{a^2} - \frac{1}{b^2} =$ R.H.S.

39. We have to prove that $a^2(\cos^2B - \cos^2C) + b^2(\cos^2C - \cos^2A) + c^2(\cos^2A - \cos^2B) = 0$

L.H.S. $= a^2(\sin^2C - \sin^2B) + b^2(\sin^2A - \sin^2C) + c^2(\sin^2B - \sin^2A)$

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

$= 0 =$ R.H.S.

40. First term of L.H.S. $= \frac{a^2\sin(B - C)}{\sin B + \sin C}$

$= \frac{2Ra\sin A\sin(B - C)}{\sin B + \sin C} = \frac{2Ra\sin(B + C)\sin(B - C)}{\sin B + \sin C}$