15. Trigonometrical Identities Solutions#

$\because A + B + C = \pi \therefore A + B = \pi - C$

$\Rightarrow \cos(A + B) = \cos(\pi - C) = \cos C \Rightarrow \sin A\sin B - \cos C = \cos A\cos B$

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

$\Rightarrow \sin^2A\sin^2B + \cos^2C - 2\sin A\sin B\cos C = (1 - \sin^2A)(1 - \sin^2B)$

$\Rightarrow \sin^A + \sin^2B + \cos^2C - 1 = 2\sin A\sin B\cos C$

$\Rightarrow \sin^2A + \sin^2B - \cos^2C = 2\sin A\sin B\cos C$
$A + B + C = 180^\circ \Rightarrow \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ \Rightarrow \frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2}$

$\Rightarrow \cos\left(\frac{A}{2} + \frac{B}{2}\right) = \cos\left(90^\circ - \frac{C}{2}\right)$

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

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

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

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

$\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2} = 1 - 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
Let $A + B = C \Rightarrow \cos(A + B) = \cos C$

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

$\Rightarrow \sin^2A\sin^2B + \cos^2C + 2\sin A\sin B\cos C = (1 - \sin^2A)(1 - \sin^2B)$

$\Rightarrow \sin^2A + \sin^2B + 2\sin A\sin B\cos C = \sin^2C$

$\Rightarrow \sin^2A + \sin^2B + 2\sin A\sin B\cos(A + B) = \sin^2(A + B)$
Given, $A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \Rightarrow \cos(A + B) = -\cos C$

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

$\Rightarrow \cos^2A\cos^2B + \cos^2C + 2\cos A\cos B\cos C = (1 - \cos^2A)(1 - \cos^2B)$

$\Rightarrow \cos^2A + \cos^2B + \cos^2C + 2\cos A\cos B\cos C = 1$
We have just proved that $\cos^2A + \cos^2B + \cos^2C + 2\cos A\cos B\cos C = 1$

$\Rightarrow 3 - \sin^2A - \sin^2B - \sin^2C + 2\cos A\cos B\cos C = 1$

$\Rightarrow \sin^2A + \sin^2B + \sin^2C = 2(1 + \cos A\cos B\cos C)$
Given, $A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \Rightarrow \cos(A + B) = -\cos C$

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

$\Rightarrow \cos^2A\cos^2B = (1 - \cos^2A)(1 - \cos^2B) + \cos^2C - 2\sin A\sin B\cos C$

$\Rightarrow \cos^2A + \cos^2B - \cos^2C = 1 - 2\sin A\sin B\cos C$
Given, $A + B + C = 180^\circ \Rightarrow \frac{A + B + C}{2} = 90^\circ$

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

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

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

$\Rightarrow \cos^2\frac{A}{2} + \cos^2\frac{B}{2} - \cos^2\frac{C}{2} = 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}$
Given, $A + B + C = 180^\circ \Rightarrow \frac{A + B + C}{2} = 90^\circ$

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

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

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

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

$\cos^2\frac{A}{2} + \cos^2\frac{B}{2} + \cos^2\frac{C}{2} = 2 + 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
Given, $A + B + C = \frac{\pi}{2} \Rightarrow A + B = \frac{\pi}{2} - C \Rightarrow \cos(A + B) = \sin C$

$\Rightarrow \cos A\cos B = \sin A\sin B + \sin C$

$\Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B + \sin^2C + 2\sin A\sin B\sin C$

$\Rightarrow (1 - \sin^2A)(1 - \sin^2B) = \sin^2A\sin^2B + \sin^2C + 2\sin A\sin B\sin C$

$\Rightarrow \sin^2A + \sin^2B + \sin^2C = 1 - 2\sin A\sin B\sin C$
We have just proven that $\sin^2A + \sin^2B + \sin^2C = 1 - 2\sin A\sin B\sin C$ in previous problem.

$\Rightarrow 1 - \cos^2A + 1 - \cos^2B + 1 - \cos^2C = 1 - 2\sin A\sin B\sin C$

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

$\Rightarrow \cos A\cos B - \cos C = \sin A\sin B$

$\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = \sin^2A\sin^2B = (1 - \cos^2A)(1 - \cos^2B)$

$\Rightarrow \cos^2A + \cos^2B + \cos^2C - 2\cos A\cos B\cos C = 1$
Given $A + B = C \Rightarrow \cos(A + B) = \cos C$

$\Rightarrow \cos A\cos B - \cos C = \sin A\sin B$

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

$\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = (1 - \cos^2A)(1 - \cos^2B)$

$\Rightarrow \cos^2A + \cos^2B + \cos^2C - 2\cos A\cos B\cos C = 1$
Given $A + B = \frac{\pi}{3} \Rightarrow \cos(A + B) = \cos\frac{\pi}{3} = \frac{1}{2}$

$\Rightarrow \cos A\cos B - \frac{1}{2} = \sin A\sin B$

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

$\Rightarrow \cos^2A + \cos^2B - \cos A\cos B = \frac{3}{4}$
From problem 12 we have $A + B = C$ and $\cos^2A + \cos^2B + \cos^2C - 2\cos A\cos B\cos C = 1$

Substituting $C = A + B$ we get $\cos^2A + \cos^2B + \cos^2(A + B) - 2\cos A\cos B\cos(A + B) = 1$

$\Rightarrow \cos^2B + \cos^2(A + B) - 2\cos A\cos B\cos(A + B) = 1 - \cos^2A = \sin^2A$ which is independent of $B$
Given $A + B + C = \pi$ and $A + B = 2C \Rightarrow C = \frac{\pi}{3} \Rightarrow A + B = \pi - \frac{pi}{3}$

$\cos(A + B) = -\cos\frac{\pi}{3}\Rightarrow \cos A\cos B = \sin A\sin B - \frac{1}{2}$

$\Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B - \sin A\sin B + \frac{1}{4}$

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

$\Rightarrow 4(\sin^2A + \sin^2B - \sin A\sin B) = 3$
Given $A + B + C = 2\pi \Rightarrow \cos(B + C) = \cos(2\pi - A) = \cos A$

$\Rightarrow \cos B\cos C - \cos A = \sin B\sin C$

$\Rightarrow \cos^B\cos^2C + \cos^2A - 2\cos A\cos B\cos C = \sin^2B\sin^2C = (1 - \cos^2B)(1 - \cos^2C)$

$\Rightarrow \cos^2B + \cos^2C - \sin^2A - 2\cos A\cos B\cos C = 0$
Given $A + B + C = 0 \Rightarrow \cos(A + B) = \cos C$

$\Rightarrow \cos A\cos B - \cos C = \sin A\sin B$

$\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = \sin^2A\sin^2B = (1 - \cos^2A)(1 - \cos^2B)$

$\Rightarrow \cos^2A + \cos^2B + \cos^2C = 1 + 2\cos A\cos B\cos C$
Putting $A = B - C, B = C - A$ and $C = A - B$ in 17 we can obtain the desired result.
Given $A + B + C = \pi,$ we have to prove that $\sin A\cos B\cos C + \sin B\cos C\cos A + \sin C\cos A\cos B= \sin A\sin B\sin C$

Dividing both sides by $\sin A\sin B\sin C,$ we get

$\cot B\cot C + \cot C\cot A + \cot A\cot B = 1$

$A + B = \pi - C\Rightarrow \cot(A + B) = -\cot C$

$\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C$

$\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1$
Given, $A + B + C = \pi \Rightarrow A + B = \pi - C$

$\Rightarrow \tan(A + B) = \tan(\pi - C) = -\tan C$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C$

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

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

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

$\Rightarrow \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1$
Let $B + C - A = \alpha, C + A - B = \beta, A + B - C = \gamma$

$\alpha + \beta + \gamma = A + B + C = \pi$

We have just proven that if $A + B + C = \pi$ then $\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$

Thus, substituting we get, $\Rightarrow \tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma$

$\Rightarrow \tan(B + C - A) + \tan(C + A - B) + \tan(A + B - C) = \tan(B + C - A)\tan(C + A - B)\tan(A + B - C)$
Given $A + B + C = \pi\Rightarrow A + B = \pi - C \Rightarrow \cot(A + B) = \cot(\pi - C)$

$\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C$

$\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1$
From previosu problem if $A + B + C = \pi$ then $\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1$

Given $\cot A + \cot B + \cot C = \sqrt{3}$

$\Rightarrow \cot^2A + \cot^2B + \cot^2C + 2(\cot A\cot B + \cot B\cot C + \cot C\cot A) = 3$

$\cot^2A + \cot^2B + \cot^2C = 1$

$2\cot^2A + 2\cot^2B + 2\cot^2C - 2 = 0$

$2\cot^2A + 2\cot^2B + 2\cot^2C - 2(\cot A\cot B + \cot B\cot C + \cot C\cot A) = 0$

$(\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2 = 0$

This is possible only if $\cot A - \cot B = 0$ i.e. $\cot A = \cot B,$ $\cot B - \cot C = 0$ i.e. $\cot B = \cot C$ and $\cot C - \cot A = 0$ i.e. $\cot C = \cot A$

$\therefore \cot A = \cot B = \cot C \Rightarrow A = B = C$
$\because A + B + C + D = 2\pi \Rightarrow A + B = 2\pi - C - D$

$\Rightarrow \tan(A + B) = -\tan(C + D)$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\frac{\tan C + \tan D}{1 - \tan C\tan D}$

$\Rightarrow (\tan A + \tan B)(1 - \tan C\tan D) = -(1 - \tan A\tan B)(\tan C + \tan D)$

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

Dividing both sides by $\tan A\tan B\tan C\tan D,$ we get

$\frac{\tan A + \tan B + \tan C + \tan D}{\tan A\tan B\tan C\tan D} = \frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C} + \frac{1}{\tan D}$

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

$\Rightarrow \cot(A + B) = \cot\left(\frac{\pi}{2} - C\right)$

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

$\Rightarrow \cot A + \cot B + \cot C = \cot A\cot B\cot C$
We have just proven in 26 that $\Rightarrow \cot A + \cot B + \cot C = \cot A\cot B\cot C$

Dividing both sides by $\cot A\cot B\cot C,$ we get

$\tan A\tan B + \tan B\tan C + \tan C\tan A = 1$
Given $A + B + C = \pi \Rightarrow 3(A + B + C) = 3\pi \Rightarrow 3A + 3B = 3\pi - 3C$

$\Rightarrow \tan(3A + 3B) = \tan(3\pi - 3C) = -\tan3C$

$\Rightarrow \frac{\tan 3A + \tan 3B}{1 - \tan3A\tan3B} = -\tan3C$

$\Rightarrow \tan 3A + \tan 3B + \tan 3C = \tan 3A\tan 3B\tan 3C$
Given $A + B + C = \pi \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}$

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

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

$\Rightarrow \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
We have to prove that $\frac{\cot A + \cot B}{\tan A + \tan B} + \frac{\cot B + \cot C}{\tan B + \tan C} + \frac{\cot C + \cot A}{\tan C + \tan A} = 1$

Putting $\tan A = \frac{1}{\cot A}, \tan B = \frac{1}{\cot B}, \tan C = \frac{1}{\cot C},$ we get

$\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

We have already proven above in problem 19.
Let $A - B = \alpha, B - C = \beta, C - A = \gamma,$ then

$\alpha + \beta + \gamma = 0$

$\Rightarrow \tan(\alpha + \beta) = -\tan\gamma$

$\Rightarrow \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = -\tan\gamma$

$\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma$

Substituting back the values, we get

$\tan(A - B) + \tan(B - C) + \tan(C - A) = \tan(A - B)\tan(B - C)\tan(C - A)$
We have already proven in problem 19 that if $A + B + C = 0,$ then

$\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

Let $A = x + y - z, B = z + x - y, C = y + z - x,$ then

$A + B + C = x + y + z = 0$

$\Rightarrow \cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

Substituting back the values, we get

$\cot(x + y - z)\cot(z + x - y) + \cot(x + y - z)\cot(y + z - x) + \cot(y + z - x)\cot(z + x - y) = 1$
Given $A + B + C= n\pi \Rightarrow \tan(A + B) = \tan(n\pi - C) = -\tan C$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C$

$\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$
L.H.S $= (\sin2A + \sin2B) + \sin2C = 2\sin(A + B)\cos(A - B) + \sin2C$

$= 2\sin(\pi - C)\cos(A - B) + \sin2C = 2\sin C\cos(A - B) + 2\sin C\cos C$

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

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

$= 2\cos\left(\frac{\pi}{2} - \frac{C}{2}\right)\cos\frac{A - B}{2} + \cos C - 1$

$= 2\sin\frac{C}{2}\cos\frac{A - B}{2} + 1 - 2\sin^2\frac{C}{2} - 1$

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

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

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

$= 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
We have proven in 34 and 35 that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$ and $\cos A + \cos B + \cos C - 1 = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ respectively. Thus,

$\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = \frac{4\sin A\sin B\sin C}{4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

$= \frac{4.2\sin\frac{A}{2}\cos\frac{A}{2}.2\sin\frac{B}{2}\cos\frac{B}{2}.2\sin\frac{C}{2}\cos\frac{C}{2}}{4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

$= 8\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
L.H.S. $= \left(\cos \frac{A}{2} + \cos\frac{B}{2}\right) + \cos\frac{C}{2}$

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

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

$= 2\cos\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} + \cos\left(\frac{\pi}{2} - \frac{\pi - C}{4}\right)\right]$

$= 2\cos\frac{\pi - C}{4}2\cos\frac{\pi + A + C - B}{8}\cos\frac{\pi + C - A + B}{8}$

$= 4\cos\frac{\pi - A}{4}\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4}$
L.H.S. $= \left(\sin\frac{A}{2} + \sin \frac{B}{2}\right) + \sin\frac{C}{2}$

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

$= 2\sin\frac{\pi - C}{4}\cos\frac{A - B}{4} + 1 - 2\sin^2\frac{\pi - C}{4}$

$=1 + 2\sin\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} - \sin\frac{\pi - C}{4}\right]$

$= 1 + 2\sin\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} - \cos\frac{\pi + C}{4}\right]$

$= 1 + 2\sin\frac{\pi - C}{4}.2\sin\frac{\pi + A + C - B}{8}\sin\frac{\pi + C - A + B}{8}$

$= 1 + 4\sin \frac{B + C}{4}\sin \frac{C + A}{4}\sin \frac{A + B}{4}$
L.H.S. $= \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} - \frac{1 - \cos C}{2}$

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

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

$= 2\sin\frac{C}{2}\cos\frac{A - B}{2} - 1 + 2\sin^2\frac{C}{2}$

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

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

$= -1 + 2\sin\frac{C}{2}.2\cos\frac{A}{2}\cos\frac{B}{2}$

$= -1 + 4\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}$

Thus, L.H.S. $= 1 - 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}$
L.H.S. $= 1 + \cos56^\circ + (\cos58^\circ - \cos66^\circ)$

$= 2\cos^228^\circ + 2\sin62^\circ\sin4^\circ$

$=2\cos^228^\circ + 2\cos28^\circ\sin4^\circ$

$= 2\cos28^\circ[\sin4^\circ + \cos28^\circ]$

$= 4\cos28^\circ\cos29^\circ\sin33^\circ$
Given $A + B + C = \pi,$ we have to prove that $\cos 2A + \cos 2B - \cos 2C = 1 - 4\sin A\sin B\cos C$

L.H.S. $=\cos 2A + \cos 2B - \cos 2C = \cos 2A + \cos 2B - \cos[2\pi - 2(A + B)]$

$= 2\cos(A + B)\cos(A - B) - \cos2(A + B) = 2\cos(A + B)\cos(A - B) - 2\cos^2(A + B) + 1$

$= 1 + 2\cos(A + B)[\cos(A - B) - \cos(A + B)]$

$= 1 - 4\sin A\sin B\cos C[\because\cos(A + B) = \cos(\pi - C) = -\cos C]$
Given $A + B + C = \pi,$ we have to prove that $\sin 2A + \sin 2B - \sin 2C = 4\cos A\cos B\sin C$

L.H.S. $= \sin 2A + \sin 2B - \sin 2C = 2\sin(A + B)\cos(A - B) - 2\sin C\cos C$

$[\because \sin(A + B) = \sin(\pi - C) = \sin C, \cos C = \cos[\pi - (A + B)] = -\cos(A + B)]$

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

$= 4\cos A\cos B\sin C$
Given $A + B + C = \pi,$ we have to prove that $\sin A + \sin B + \sin C = 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

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

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

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

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

$= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
L.H.S. $= \cos A + \cos B - \cos C = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} - 1 + 2\sin^2\frac{C}{2}$

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

$= 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\left(\frac{[pi}{2} - \frac{C}{2}\right)\right] - 1$

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

$= 4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2} - 1$
$B + C - A = \pi - A - A = \pi - 2A, C + A - B = \pi - 2B, A + B - C = \pi - 2C$

$\Rightarrow$ L.H.S. $= \sin 2A + \sin 2B + \sin 2C$

We have proven in problem 34 that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$

$\therefore \sin(B + C - A) + \sin(C + A - B) + \sin(A + B - C) = 4\sin A\sin B\sin C$
L.H.S. $= \frac{\cos A}{\sin B\sin C} + \frac{\cos B}{\sin C\sin A} + \frac{\cos C}{\sin A\sin B} = 2$

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

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

We have proven in problem 34 that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$

$\Rightarrow \frac{\sin 2A + \sin 2B + \sin 2C}{2\sin A\sin B\sin C} = 2$
Given $A + B + C = \pi,$ we have to prove that $\frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

We have proven in problem 34 that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$

We have also proven in problem 43 that $\sin A + \sin B + \sin C = 4\cos \frac{A}{2}\cos \frac{B}{2}\cos\frac{C}{2}$

Thus, L.H.S. $= \frac{4\sin A\sin B\sin C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos\frac{C}{2}}$

$= 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
Given $x + y + z = \frac{\pi}{2},$ we have to prove that $\cos(x - y - z) + \cos(y - z - x) + \cos(z - x - y) - 4\cos x\cos y\cos z = 0$

$x - y - z = x - \frac{\pi}{2} + x = 2x - \frac{\pi}{2}$

Similarly $y - z - x = 2y - \frac{\pi}{2}$ and $z - x - y = 2z - \frac{\pi}{2}$

$\therefore$ L.H.S. $= \sin 2x + \sin 2y + \sin 2z - 4\cos x\cos y \cos z$

Now, $\sin 2x + \sin 2y + \sin 2z = 2\sin(x + y)\cos(x - y) + 2\sin z\cos z$

$= 2\cos z\cos(x - y) + 2\sin\left(\frac{\pi}{2} - x - y\right)\cos z$

$= 2\cos z[\cos(x - y) + \cos(x + y)]$

$= 4\cos x\cos y\cos z$

$\therefore \sin 2x + \sin 2y + \sin 2z - 4\cos x\cos y \cos z = 0$
We have to prove that $\sin(x - y) + \sin(y - z) + \sin(z - x) + 4\sin\frac{x - y}{2}\sin\frac{y - z}{2}\sin \frac{z - x}{2} = 0$

Let $x - y = \alpha, y - z = \beta$ and $z - x = \gamma$ then $\alpha + \beta + \gamma = 0$

The given equation becomes $\sin\alpha + \sin\beta + \sin\gamma + 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2} = 0$

Considering $\sin\alpha + \sin\beta + \sin\gamma$

$= \sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} + 2\sin\frac{\gamma}{2}\cos\frac{\gamma}{2}$

$= -\sin\frac{\gamma}{2}\cos\frac{\alpha - \beta}{2} + 2\sin\frac{\gamma}{2}\cos\frac{\alpha + \beta}{2}$

$= -4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$

Thus, $\sin(x - y) + \sin(y - z) + \sin(z - x) + 4\sin\frac{x - y}{2}\sin\frac{y - z}{2}\sin \frac{z - x}{2} = 0$
$B + 2C = \pi - A + C, C + 2A = \pi - B + A, A + 2B = \pi - C + B$

Thus, L.H.S. $= -[\sin(C - A) + \sin(A - B) + \sin(B - C)]$

Also, note that $A - B + B - C + C - A = 0$ and we have proven in previous problem that $\sin\alpha + \sin\beta + \sin\gamma = 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$ when $\alpha + \beta + \gamma = 0$

Thus, $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4\sin\frac{B - C}{2}\sin\frac{C - A}{2}\sin\frac{A - B}{2}$
L.H.S. $= \sin\frac{\pi - A}{2} + \sin\frac{\pi - B}{2} + \sin\frac{\pi - C}{2}$

Following the result of 43 we can say that

$\sin\frac{\pi - A}{2} + \sin\frac{\pi - B}{2} + \sin\frac{\pi - C}{2} = 4\cos\frac{\pi - A}{4}\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4}$
Let $x = \tan A, y = \tan B, z = \tan C$

Given, $xy + yz + zx = 1$

$\therefore \tan A\tan B + \tan B\tan C + \tan C\tan A = 1$

$\Rightarrow \tan C(\tan A + \tan B) = 1 - \tan A\tan B$

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

$\Rightarrow \tan(A + B) = \tan\left(\frac{\pi}{2} - C\right)$

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

L.H.S. $= \frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2}$

$= \frac{\tan A}{1 - \tan^2A} + \frac{\tan B}{1 - \tan^2B} + \frac{\tan C}{1 - \tan^2C}$

$= \frac{1}{2}(\tan 2A + \tan 2B + \tan 2C)$

We have already proven that if $2A + 2B + 2C = \pi$ then $\tan2A + \tan2B + \tan2C = \tan2A\tan2B\tan2C$

$\therefore \frac{1}{2}(\tan 2A + \tan 2B + \tan 2C) = \frac{1}{2}\tan2A\tan2B\tan2C$

$= \frac{1}{2}\frac{2\tan A}{1 - \tan^2A}.\frac{2\tan B}{1 - \tan^2B}.\frac{2\tan C}{1 - \tan^2C}$

$= \frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}$
Let $x = \tan A, y = \tan B, z = \tan C$

Now, $x + y + z = xyz$

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

$\Rightarrow \tan A + \tan B = \tan C(\tan A\tan B - 1)$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan C} = -\tan C = \tan(\pi - C)$

$\Rightarrow A + B = \pi - C \Rightarrow A + B + C = \pi$

L.H.S. $= \frac{3x - x^3}{1 - 3x^2} + \frac{3y - y^3}{1 - 3y^2} + \frac{3z - z^3}{1 - 3z^2}$

$= \frac{3\tan A - \tan^3A}{1 - 3\tan^2A} + \frac{3\tan B - \tan^3B}{1 - 3\tan^2B} + \frac{3\tan C - \tan^3C}{1 - 3\tan^2C}$

$= \tan 3A + \tan 3B + \tan 3C$

Now following like prebious problem

$\tan3A + \tan3B + \tan3C = \tan3A\tan3B\tan3C$

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

$= \frac{3x - x^3}{1 - 3x^2}.\frac{3y - y^3}{1 - 3y^2}.\frac{3z - z^3}{1 - 3z^2}$
Given $x + y + z = xyz$ , let $x = \tan A, y = \tan B, z = \tan C$

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

$\Rightarrow \tan A + \tan B = \tan C(\tan A\tan B - 1)$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan C} = -\tan C = \tan(\pi - C)$

$\Rightarrow A + B = \pi - C \Rightarrow A + B + C = \pi$

L.H.S. $= \frac{2x}{1 - x^2} + \frac{2y}{1 - y^2} + \frac{2z}{1 - z^2}$

$= \frac{2\tan A}{1 - \tan^2A} + \frac{2\tan B}{1 - \tan^2B} + \frac{2\tan C}{1 - \tan^2C}$

$= \tan 2A + \tan 2B + \tan 2C$

Following like problem 52

$\tan 2A + \tan 2B + \tan 2C = \tan2A\tan2B\tan2C = \frac{2x}{1 - x^2}.\frac{2y}{1 - y^2}.\frac{2z}{1 - z^2}$
Given $x + y + z = xyz$ , let $x = \tan A, y = \tan B, z = \tan C$

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

$\Rightarrow \tan A + \tan B = \tan C(\tan A\tan B - 1)$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan C} = -\tan C = \tan(\pi - C)$

$\Rightarrow A + B = \pi - C \Rightarrow A + B + C = \pi$

Given, $x(1 - y^2)(1 - z^2) + y(1 - z^2)(1 - x^2) + z(1 - x^2)(1 - y^2) = 4xyz$

Dividing both sides with $(1 - x^2)(1 - y^2)(1 - z^2),$ we get

$\frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2} = \frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}$

L.H.S. $= \frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2} = \frac{1}{2}[\tan 2A + \tan 2B + \tan 2C]$

$= \frac{1}{2}\tan2A\tan2B\tan2C = \frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}$
L.H.S. $= (\cos A + \cos B) + (\cos C + \cos D)$

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

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

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

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

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

$= 4\cos\frac{A + B}{2}\sin\frac{A + C - (B + C)}{4}\sin\frac{B + C - (A + D)}{4}$

$= 4\cos\frac{A + B}{2}\sin\frac{A + C -(2\pi - A - C)}{4}\sin\frac{B + C - (2\pi - B - C)}{4}$

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

$= 4\cos\frac{A + B}{2}\cos\frac{B + C}{2}\cos\frac{C + A}{2}$
L.H.S. $= \cos^2S + \cos^2(S - A) + \cos^2(S - B) + \cos^2(S - C)$

$= \frac{1 + \cos 2S}{2} + \frac{1 + \cos(2S - 2A)}{2} + \frac{1 + \cos(2S - 2B)}{2} + \frac{1 + \cos(2S - 2C)}{2}$

$= \frac{1}{2}[4 + \{\cos2S + \cos(2S - 2A)\} + \{\cos(2S - 2B) + ]\cos(2S - 2C)\}]$

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

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

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

$= 2 + 2\cos A\cos B\cos C$
If $A + B + C = \pi$ then according to problem 21 $\tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1$

Let $\tan\frac{A}{2} = x, \tan\frac{B}{2} = y, \tan\frac{C}{2} = z$

$\Rightarrow xy + yz + xz = 1$

Now, $(x - y)^2 + (y - z)^2 + (z - x)^2 \geq 0$

$\Rightarrow x^2 + y^2 + z^2 \geq xy + yz + zx$

$\Rightarrow x^2 + y^2 + z^2 \geq 1$

$\tan^2\frac{A}{2} + \tan^2\frac{B}{2} + \tan^2\frac{C}{2}\geq 1$
We have proven that if $A + B + C = \pi$ then $\tan A + \tan B + \tan C = \tan A\tan B\tan C$

Thus, L.H.S. $= \tan A\tan B\tan C(\cot A + \cot B + \cot C)$

$= \tan B\tan C + \tan C\tan A + \tan A\tan B$

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

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

$= \frac{\sin C[\cos A\sin B + \sin A\cos B] + \cos C\sin A\sin B}{\cos A\cos B\cos C}$

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

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

$= \frac{1 - \cos^2C + \cos C\sin A\sin B}{\cos A\cos B\cos C}$

$= \frac{1 + \cos C[\sin A\sin B - \cos C]}{]\cos A\cos B\cos C}$

$= \frac{1 + \cos C[\sin A\sin B - \cos(\pi - A - B)]}{\cos A\cos B\cos C}$

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

$= \frac{1 + \cos A\cos B\cos C}{\cos A\cos B\cos C} = 1 + \sec A\sec B\sec C$
$\cot B\ + \cot C = \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}$

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

$= \frac{\sin(\pi - A)}{\sin B\sin C} = \frac{\sin A}{\sin B\sin C}$

Similarly, $\cot C + \cot A = \frac{\sin B}{\sin C\sin A}$ and $\cot A + \cot B = \frac{\sin C}{\sin A + \sin B}$
L.H.S. $= \sin^2A\sin B\cos B + \sin^2A \sin C\cos C + \sin^2B\sin A\cos A + \sin^2B\sin C\cos C + \sin^2C\sin A\cos A + \sin^2C\sin B\cos B$

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

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

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

$= 3\sin A\sin B\sin C$
L.H.S. $= \cos A - \cos B + \cos C - \cos D = 2\sin\frac{A + B}{2}\sin\frac{B - A}{2} + 2\sin\frac{C + D}{2}\sin\frac{D - C}{2}$

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

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

$= 2\sin\frac{A + B}{2}.2\sin\frac{B + D - (A + C)}{4}\cos\frac{B + C - (A + D)}{4}$

$= 4\sin\frac{A + B}{2}\sin\frac{2\pi - 2(A + C)}{4}\cos\frac{2\pi - 2(A + D)}{4}$

$= 4\sin\frac{A + B}{2}\sin\frac{A + D}{2}\cos \frac{A + C}{2}$
Since $A, B, C, D$ are angles of a cyclic quadrilateral $\therefore A + B + C + D = 2\pi, A + C = \pi, B + D = \pi$

We have proven in problem 56 that $\cos A + \cos B + \cos C + \cos D = 4\cos\frac{A + B}{2}\cos\frac{B + C}{2}\cos\frac{C + A}{2}$

$\cos\frac{A + C}{2} = \cos\frac{\pi}{2} = 0$

$\therefore \cos A + \cos B + \cos C + \cos D = 0$
We know that $(\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2 \geq 0$

Also, $\because A + B + C = \pi \therefore \cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

$\therefore 2\cot^2A + 2\cot^2B + 2\cot^2C \geq 2(\cot A\cot B + \cot B\cot C + \cot C\cot A)$

$\cot^2A + \cot^2B + \cot^2C \geq 1$
$\cos\frac{A}{2}\cos\frac{B - C}{2} = \cos\left(\frac{\pi}{2} - \frac{B + C}{2}\right)\cos\frac{B - C}{2}$

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

$= \frac{1}{2}(\sin B + \sin C)$

Similarly $\cos\frac{B}{2}\cos\frac{C - A}{2} = \frac{1}{2}(\sin A + \sin C)$

and $\cos\frac{C}{2}\cos\frac{A - B}{2} = \frac{1}{2}(\sin A + \sin B)$

Adding all these we have desired result.
$\sin 3A\sin(B - C) = (3\sin A - 4\sin^3A)\sin(B - C)$

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

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

$= \frac{3}{2}(\cos2C - \cos2B) - 2\sin^2A(\cos2C - \cos2B)$

Now, $2\sin^2A(\cos2C - \cos2B) = (1 - \cos 2A)(\cos2C - \cos2B)$

$= \cos 2C - \cos 2B - \cos2C \cos 2A + \cos 2A\cos 2B$

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