(a)

\begin{align} \text{L.H.S } & = \sin (90^\circ + \theta) \phantom{0000000000000000} [\text{Addition formula: } \sin (A + B)] \\ & = \sin 90^\circ \cos \theta + \sin \theta \cos 90^\circ \\ & = (1)(\cos \theta) + (\sin \theta)(0) \\ & = \cos \theta \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \text{L.H.S } & = \cos (90^\circ + \theta) \phantom{0000000000000)} [\text{Addition formula: } \cos (A + B)] \\ & = \cos 90^\circ \cos \theta - \sin 90^\circ \sin \theta \\ & = (0)(\cos \theta) - (1)(\sin \theta) \\ & = -\sin \theta \\ & = \text{R.H.S} \end{align}

(c)

\begin{align} \text{L.H.S } & = \sin \left({3\pi \over 2} - \theta\right) \phantom{00000000000000000} [\text{Addition formula: } \sin (A - B)] \\ & = \sin \left({3\pi \over 2}\right) \cos \theta - \sin \theta \cos \left({3\pi \over 2}\right) \\ & = (-1)(\cos \theta) - (\sin \theta)(0) \\ & = -\cos \theta \\ & = \text{R.H.S} \end{align}

(d)

\begin{align} \text{L.H.S } & = \tan (2\pi - \theta) \phantom{00000000} [\text{Addition formula: } \tan (A - B)] \\ & = {\tan 2\pi - \tan \theta \over 1 + \tan 2\pi \tan\theta} \\ & = {0 - \tan \theta \over 1 + (0)(\tan \theta)} \\ & = {-\tan \theta \over 1} \\ & = -\tan \theta \\ & = \text{R.H.S} \end{align}

(a)

\begin{align} \text{L.H.S } & = \sin \theta \cos 2\theta + \cos \theta \sin 2\theta \\ & = \sin (\theta + 2\theta) \\ & = \sin 3\theta \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \text{L.H.S } & = \sin 2\theta \cos \theta - \cos 2\theta \sin \theta \\ & = \sin (2\theta - \theta) \\ & = \sin \theta \\ & = \text{R.H.S} \end{align}

(c)

\begin{align} \text{L.H.S } & = \cos 3\theta \cos 2\theta + \sin 3\theta \sin 2\theta \\ & = \cos (3\theta - 2\theta) \\ & = \cos \theta \\ & = \text{R.H.S} \end{align}

(d)

\begin{align} \text{L.H.S } & = {\tan 3\theta - \tan 4\theta \over 1 + \tan 3\theta \tan 4\theta} \\ & = {-(\tan 4\theta - \tan 3\theta) \over 1 + \tan 3\theta \tan 4\theta} \\ & = -{\tan 4\theta - \tan 3\theta \over 1 + \tan 3\theta \tan 4\theta} \\ & = -\tan (4\theta - 3\theta) \\ & = -\tan \theta \\ & = \text{R.H.S} \end{align}

(a)

\begin{align} \sin 15^\circ \cos 75^\circ + \cos 15^\circ \sin 75^\circ & = \sin (15^\circ + 75^\circ) \\ & = \sin 90^\circ \\ & = 1 \end{align}

(b)

\begin{align} \sin 50^\circ \cos 20^\circ - \cos 50^\circ \sin 20^\circ & = \sin (50^\circ - 20^\circ) \\ & = \sin 30^\circ \\ & = {1 \over 2} \end{align}

(c)

\begin{align} \cos 70^\circ \cos 40^\circ + \sin 70^\circ \sin 40^\circ & = \cos (70^\circ - 40^\circ) \\ & = \cos 30^\circ \\ & = {\sqrt{3} \over 2} \end{align}

(d)

\begin{align} \sin 15^\circ \sin 30^\circ - \cos 15^\circ \cos 30^\circ & = -(\cos 15^\circ \cos 30^\circ - \sin 15^\circ \sin 30^\circ) \\ & = -[\cos (15^\circ + 30^\circ)] \\ & = -\cos 45^\circ \\ & = -{1 \over \sqrt{2}} \end{align}

(e)

\begin{align} {\tan 35^\circ + \tan 10^\circ \over 1 - \tan 35^\circ \tan 10^\circ} & = \tan (35^\circ + 10^\circ) \\ & = \tan 45^\circ \\ & = 1 \end{align}

(f)

\begin{align} {\tan 30^\circ - \tan 75^\circ \over 1 - \tan 30^\circ \tan 75^\circ} & = {-(\tan 75^\circ - \tan 30^\circ) \over 1 - \tan 30^\circ \tan 75^\circ} \\ & = -{\tan 75^\circ - \tan 30^\circ \over 1 - \tan 30^\circ \tan 75^\circ} \\ & = -\tan (75^\circ - 30^\circ) \\ & = -\tan 45^\circ \\ & = -1 \end{align}

(a)

\begin{align} \cos 75^\circ & = \cos (45^\circ + 30^\circ) \\ & = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ \\ & = \left({1 \over \sqrt{2}}\right)\left({\sqrt{3} \over 2}\right) - \left({1 \over \sqrt{2}}\right)\left({1 \over 2}\right) \\ & = {\sqrt{3} \over 2\sqrt{2}} - {1 \over 2\sqrt{2}} \\ & = {\sqrt{3} - 1 \over 2\sqrt{2}} \end{align}

(b)

\begin{align} \sin 15^\circ & = \sin (45^\circ - 30^\circ) \\ & = \sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ \\ & = \left({1 \over \sqrt{2}}\right)\left({\sqrt{3} \over 2}\right) - \left({1 \over 2}\right)\left({1 \over \sqrt{2}}\right) \\ & = {\sqrt{3} \over 2\sqrt{2}} - {1 \over 2\sqrt{2}} \\ & = {\sqrt{3} - 1 \over 2\sqrt{2}} \end{align}

(a)

\begin{align} \text{L.H.S } & = \sin (A + B) + \sin (A - B) \\ & = (\sin A \cos B + \cos A \sin B) + (\sin A \cos B - \cos A \sin B) \\ & = \sin A \cos B + \cos A \sin B + \sin A \cos B - \cos A \sin B \\ & = 2 \sin A \cos B \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \text{L.H.S } & = \cos (A + B) - \cos (A - B) \\ & = (\cos A \cos B - \sin A \sin B) - (\cos A \cos B + \sin A \sin B) \\ & = \cos A \cos B - \sin A \sin B - \cos A \cos B - \sin A \sin B \\ & = -2 \sin A \sin B \\ & = \text{R.H.S} \end{align}

(a)

\begin{align} \require{cancel} \text{L.H.S } & = {\sin (A + B) - \sin (A - B) \over \cos (A + B) - \cos (A - B)} \\ & = {(\sin A \cos B + \sin B \cos A) - (\sin A \cos B - \sin B \cos A) \over (\cos A \cos B - \sin A \sin B) - (\cos A \cos B + \sin A \sin B) } \\ & = {\sin A \cos B + \sin B \cos A - \sin A \cos B + \sin B \cos A \over \cos A \cos B - \sin A \sin B - \cos A \cos B - \sin A \sin B } \\ & = {\cancel{2} \cancel{\sin B} \cos A \over -\cancel{2}\sin A \cancel{\sin B}} \\ & = -{\cos A \over \sin A} \\ & = -\cot A \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \require{cancel} \text{R.H.S } & = {\tan A + \tan B \over 1 + \tan A \tan B} \\ & = {{\sin A \over \cos A} + {\sin B \over \cos B} \over 1 + \left( {\sin A \over \cos A} \right)\left( {\sin B \over \cos B} \right)} \\ & = {{\sin A \cos B \over \cos A \cos B} + {\sin B \cos A \over \cos A \cos B} \over 1 + {\sin A \sin B \over \cos A \cos B}} \\ & = {{\sin A \cos B \over \cos A \cos B} + {\sin B \cos A \over \cos A \cos B} \over {\cos A \cos B \over \cos A \cos B} + {\sin A \sin B \over \cos A \cos B}} \phantom{000000} [\text{Combine into a single fraction}] \\ & = {{\sin A \cos B + \sin B \cos A \over \cos A \cos B} \over {\cos A \cos B + \sin A \sin B \over \cos A \cos B}} \\ & = {\sin A \cos B + \sin B \cos A \over \cos A \cos B} \div {\cos A \cos B + \sin A \sin B \over \cos A \cos B} \\ & = {\sin A \cos B + \sin B \cos A \over \cancel{\cos A \cos B}} \times {\cancel{\cos A \cos B} \over \cos A \cos B + \sin A \sin B} \\ & = {\sin A \cos B + \sin B \cos A \over \cos A \cos B + \sin A \sin B} \\ & = {\sin A \cos B + \sin B \cos A \over \cos A \cos B + \sin A \sin B} \\ & = {\sin (A + B) \over \cos (A - B)} \phantom{000000} [\text{Apply addition formula}] \\ & = \text{L.H.S} \end{align}

(c)

\begin{align} \text{L.H.S} & = {\sin (A + B) \over \sin (A - B)} \phantom{0000000000000} [\text{Apply addition formula}] \\ & = {\sin A \cos B + \cos A \sin B \over \sin A \cos B - \cos A \sin B} \\ & = {\sin A \cos B + \cos A \sin B \over \sin A \cos B - \cos A \sin B} \times {{1 \over \cos A} \over {1 \over \cos A}} \\ & = {{\sin A \over \cos A}\cos B + \sin B \over {\sin A \over \cos A}\cos B - \sin B} \\ & = {\tan A \cos B + \sin B \over \tan A \cos B - \sin B} \\ & = {\tan A \cos B + \sin B \over \tan A \cos B - \sin B} \times {{1 \over \cos B} \over {1 \over \cos B}} \\ & = {\tan A + {\sin B \over \cos B} \over \tan A - {\sin B \over \cos B}} \\ & = {\tan A + \tan B \over \tan A - \tan B} \\ & = \text{R.H.S} \end{align}

(d)

\begin{align} \require{cancel} \text{R.H.S } & = {\sec \alpha \sec \beta \over 1 + \tan \alpha \tan \beta} \\ & = { \left({1 \over \cos \alpha}\right) \left({1 \over \cos \beta}\right) \over 1 + \left( {\sin \alpha \over \cos \alpha} \right) \left({\sin \beta \over \cos \beta}\right) } \\ & = {{1 \over \cos \alpha \cos \beta} \over 1 + {\sin \alpha \sin \beta \over \cos \alpha \cos \beta}} \\ & = {{1 \over \cos \alpha \cos \beta} \over {\cos \alpha \cos \beta \over \cos \alpha \cos \beta} + {\sin \alpha \sin \beta \over \cos \alpha \cos \beta}} \phantom{00000000000} [\text{Combine into a single fraction}] \\ & = {{1 \over \cos \alpha \cos \beta} \over {\cos \alpha \cos \beta + \sin \alpha \sin \beta \over \cos \alpha \cos \beta}} \\ & = {1 \over \cos \alpha \cos \beta} \div {\cos \alpha \cos \beta + \sin \alpha \sin \beta \over \cos \alpha \cos \beta} \\ & = {1 \over \cancel{\cos \alpha \cos \beta}} \times { \cancel{\cos \alpha \cos \beta} \over \cos \alpha \cos \beta + \sin \alpha \sin \beta} \\ & = {1 \over \cos \alpha \cos \beta + \sin \alpha \sin \beta} \phantom{000000000} [\text{Apply addition formula: } \cos (A - B)] \\ & = {1 \over \cos (\alpha - \beta) } \\ & = \sec (\alpha - \beta) \\ & = \text{L.H.S} \end{align}

(a)

\begin{align} \require{cancel} \text{L.H.S } & = \tan (A + 45^\circ) \tan (A - 45^\circ) \\ & = \left({\tan A + \tan 45^\circ \over 1 - \tan A \tan 45^\circ}\right) \left({\tan A - \tan 45^\circ \over 1 + \tan A \tan 45^\circ}\right) \phantom{000000} [\text{Apply addition formula}] \\ & = \left({\tan A + 1 \over 1 - \tan A}\right) \left({\tan A - 1 \over 1 + \tan A}\right) \phantom{00000000000000000.} [\tan 45^\circ = 1] \\ & = {\tan^2 A - 1 \over 1 - \tan^2 A} \\ & = {-\cancel{(1 - \tan^2 A)} \over \cancel{1 - \tan^2 A}} \\ & = {-1 \over 1} \\ & = -1 \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \text{R.H.S } & = {\sin (A + B) \over \cos A \cos B} \\ & = {\sin A \cos B + \sin B \cos A \over \cos A \cos B} \\ & = {\sin A \cos B \over \cos A \cos B} + {\sin B \cos A \over \cos A \cos B} \\ & = {\sin A \over \cos A} + {\sin B \over \cos B} \\ & = \tan A + \tan B \\ & = \text{L.H.S} \end{align}

(c)

\begin{align} \require{cancel} \text{R.H.S } & = {\cot x \cot y - 1 \over \cot x + \cot y} \\ & = { \left({1 \over \tan x}\right) \left({1 \over \tan y}\right) - 1 \over {1 \over \tan x} + {1 \over \tan y}} \\ & = { {1 \over \tan x \tan y} - 1 \over {1 \over \tan x} + {1 \over \tan y}} \\ & = {{1 \over \tan x \tan y} - {\tan x \tan y \over \tan x \tan y} \over {\tan y \over \tan x \tan y} + {\tan x \over \tan x \tan y}} \phantom{000000} [\text{Combine into a single fraction}] \\ & = {{1 - \tan x \tan y \over \tan x \tan y} \over {\tan x + \tan y \over \tan x \tan y}} \\ & = {1 - \tan x \tan y \over \tan x \tan y} \div {\tan x + \tan y \over \tan x \tan y} \\ & = {1 - \tan x \tan y \over \cancel{\tan x \tan y}} \times {\cancel{\tan x \tan y} \over \tan x + \tan y} \\ & = {1 - \tan x \tan y \over \tan x + \tan y} \\ & = \left( {\tan x + \tan y \over 1 - \tan x \tan y} \right)^{-1} \\ & = [\tan (x + y)]^{-1} \phantom{000000000000} [\text{Apply addition formula: } \tan (A + B)] \\ & = {1 \over \tan (x + y)} \\ & = \cot (x + y) \\ & = \text{L.H.S} \end{align}

(d)

\begin{align} \text{R.H.S } & = {\sin (y - x) \over \sin (y + x)} \\ & = {\sin y \cos x - \sin x \cos y \over \sin y \cos x + \sin x \cos y} \\ & = {\sin y \cos x - \sin x \cos y \over \sin y \cos x + \sin x \cos y} \times {{1 \over \cos y} \over {1 \over \cos y}} \\ & = {{\sin y \over \cos y}\cos x - \sin x \over {\sin y \over \cos y}\cos x + \sin x} \\ & = {\tan y \cos x - \sin x \over \tan y \cos x + \sin x} \\ & = {\tan y \cos x - \sin x \over \tan y \cos x + \sin x} \times {{1 \over \cos x} \over {1 \over \cos x}} \\ & = {\tan y - {\sin x \over \cos x} \over \tan y + {\sin x \over \cos x}} \\ & = {\tan y - \tan x \over \tan y + \tan x} \\ & = \text{L.H.S} \end{align}

(e)

\begin{align} \text{L.H.S } & = 2 \sin (A + 45^\circ) \cos (A + 45^\circ) \phantom{00000000000000000000000000000} [\text{Apply addition formula}] \\ & = 2(\sin A \cos 45^\circ + \sin 45^\circ \cos A)(\cos A \cos 45^\circ - \sin A \sin 45^\circ) \\ & = 2\left({1 \over \sqrt{2}}\sin A + {1 \over \sqrt{2}}\cos A\right) \left({1 \over \sqrt{2}}\cos A - {1 \over \sqrt{2}}\sin A \right) \phantom{0000000} \left[ \sin 45^\circ = \cos 45^\circ = {1 \over \sqrt{2}} \right]\\ & = 2\left[ \left({1 \over \sqrt{2}}\cos A\right)^2 - \left({1 \over \sqrt{2}}\sin A\right)^2 \right] \\ & = 2\left({1 \over 2}\cos^2 A - {1 \over 2}\sin^2 A\right) \\ & = \cos^2 A - \sin^2 A \\ & = \cos^2 A - (1 - \cos^2 A) \phantom{0000000000000} [\text{Since } \sin^2 A + \cos^2 A = 1 \rightarrow \phantom{.} \sin^2 A = 1 - \cos^2 A] \\ & = \cos^2 A - 1 + \cos^2 A \\ & = 2\cos^2 A - 1 \\ & = \text{R.H.S} \end{align}

(i)

\begin{align} \text{L.H.S } & = \cos (A + B + C) \\ & = \cos [A + (B + C)] \phantom{00000000000000000000000000} [\text{Apply addition formula; treat } B + C \text{ as a single term}] \\ & = \cos A \cos (B + C) - \sin A \sin (B + C) \\ & = \cos A(\cos B \cos C - \sin B \sin C) - \sin A(\sin B \cos C + \sin C \cos B) \phantom{000000} [\text{Apply addition formula again}] \\ & = \cos A \cos B \cos C - \cos A \sin B \sin C - \sin A \sin B \cos C - \sin A \cos B \sin C \\ & = \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C \\ & = \text{R.H.S} \end{align}

(ii) The word 'Hence' suggests that we have to apply the results from (i)

\begin{align} \text{L.H.S } & = \cos 3A \\ & = \cos (A + A + A) \\ & = \cos A \cos A \cos A - \sin A \sin A \cos A - \sin A \cos A \sin A - \cos A \sin A \sin A \phantom{000000} [\text{Use result from part i}] \\ & = \cos^3 A - \cos A \sin^2 A - \cos A \sin^2 A - \cos A \sin^ A \\ & = \cos^3 A - 3\cos A \sin^2 A \\ & = \cos^3 A - 3\cos A (1 - \cos^2 A) \phantom{000000000000000000} [\text{Since } \sin^2 A + \cos^2 A = 1 \rightarrow \phantom{.} \sin^2 A = 1 - \cos^2 A] \\ & = \cos^3 A - 3\cos A + 3\cos^3 A \\ & = 4\cos^3 A - 3\cos A \\ & = \text{R.H.S} \end{align}

(i)

\begin{align} \sin (A - B) & = \sin A \cos B - \sin B \cos A \\ {1 \over 2} & = {7 \over 10} - \sin B \cos A \\ \sin B \cos A & = {7 \over 10} - {1 \over 2} \\ \sin B \cos A & = {1 \over 5} \end{align}

(ii)

\begin{align} \sin (A + B) & = \sin A \cos B + \sin B \cos A \\ & = {7 \over 10} + \sin B \cos A \\ & = {7 \over 10} + {1 \over 5} \phantom{00000000} [\text{Use result from part i}] \\ & = {9 \over 10} \end{align}

(iii)

\begin{align} {\tan A \over \tan B} & = \tan A \div \tan B \\ & = {\sin A \over \cos A} \div {\sin B \over \cos B} \\ & = {\sin A \over \cos A} \times {\cos B \over \sin B} \\ & = {\sin A \cos B \over \cos A \sin B} \\ & = {{7 \over 10} \over {1 \over 5}} \\ & = {7 \over 2} \end{align}

\begin{align} {\sin (A + B) \over \sin (A - B)} & = {\sin A \cos B + \sin B \cos A \over \sin A \cos B - \sin B \cos A} \\ {2 \over 3} & = {\sin A \cos B + \sin B \cos A \over \sin A \cos B - \sin B \cos A} \\ 2(\sin A \cos B - \sin B \cos A) & = 3(\sin A \cos B + \sin B \cos A) \\ 2\sin A \cos B - 2\sin B \cos A & = 3\sin A \cos B + 3\sin B \cos A \\ -2 \sin B \cos A - 3 \sin B \cos A & = 3\sin A \cos B - 2\sin A \cos B \\ -5 \sin B \cos A & = \sin A \cos B \\ -5 & = {\sin A \cos B \over \sin B \cos A} \\ -5 & = \left( {\sin A \over \cos A} \right) \left( {\cos B \over \sin B}\right) \\ -5 & = \tan A \cot B \\ \\ \therefore\tan A \cot B & = -5 \end{align}

\begin{align} \require{cancel} \tan (A - B) & = {\tan A - \tan B \over 1 + \tan A \tan B} \\ {25 \over 24} & = {{3 \over 4} - \tan B \over 1 + {3 \over 4}\tan B} \\ 25\left(1 + {3 \over 4}\tan B\right) & = 24\left({3 \over 4} - \tan B\right) \\ 25 + {75 \over 4}\tan B & = 18 - 24 \tan B \\ 100 + 75\tan B & = 72 - 96\tan B \phantom{000000} [\text{Multiply every term by 4}] \\ 75\tan B + 96\tan B & = 72 - 100 \\ 171\tan B & = -28 \\ \tan B & = -{28 \over 171} \end{align}

(i)

\begin{align} \tan A & > 0 \phantom{00000} [\text{1st or 3rd quadrant}] \\ \\ \sin B & < 0 \phantom{00000} [\text{3rd or 4th quadrant}] \\ \\ \therefore A \text{ and } & B \text{ are in the third quadrant} \\ \\ \\ \tan A & = {Opp \over Adj} \\ & = {4 \over 3} \\ & = {-4 \over -3} \end{align}

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ x^2 & = (-3)^2 + (-4)^2 \\ & = 25 \\ x & = \pm\sqrt{25}\\ & = 5 \text{ or } -5 \text{ (Reject)} \\ \\ \\ \tan (A + 45^\circ) & = {\tan A + \tan 45^\circ \over 1 - \tan A \tan 45^\circ} \phantom{000000} [\text{Addition formula: } \tan (A + B)] \\ & = {\tan A + 1 \over 1 - (\tan A) (1)} \phantom{000000000.} [\tan 45^\circ = 1]\\ & = {{4 \over 3} + 1 \over 1 - \left({4 \over 3}\right)(1)} \\ & = {{7 \over 3} \over -{1 \over 3}} \\ & = -7 \end{align}

(ii)

\begin{align} \sin B & = {Opp \over Hyp} \\ & = -{3 \over 5} \\ & = {-3 \over 5} \end{align}

\begin{align} 5^2 & = 3^2 + x^2 \\ 25 - 9 & = x^2 \\ x^2 & = 16 \\ x & = \pm\sqrt{16} \\ & = 4 \text{ (Reject) } \phantom{0} \text{ or } -4 \\ \\ \\ \cos (A - B) & = \cos A \cos B + \sin A \sin B \phantom{00000000} [\text{Addition formula}] \\ & = \left(-{3 \over 5}\right)\left(-{4 \over 5}\right) + \left(-{4 \over 5}\right)\left(-{3 \over 5}\right) \\ & = {12 \over 25} + {12 \over 25} \\ & = {24 \over 25} \end{align}

(iii)

\begin{align} \sin (B - A) & = \sin B \cos A - \sin A \cos B \phantom{00000000} [\text{Addition formula}] \\ & = \left(-{3 \over 5}\right)\left(-{3 \over 5}\right) - \left(-{4 \over 5}\right)\left(-{4 \over 5}\right) \\ & = {9 \over 25} - {16 \over 25} \\ & = -{7 \over 25} \end{align}

(i)

\begin{align} \cos A & = {Adj \over Hyp} \\ & = -{4 \over 5} \\ & = {-4 \over 5} \end{align}

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 5^2 & = (-4)^2 + x^2 \\ 25 & = 16 + x^2 \\ x^2 & = 9 \\ x & = \pm\sqrt{9} \\ & = 3 \text{ or } -3 \text{ (Reject)} \\ \\ \\ \sin B & = {Opp \over Hyp} \\ & = {12 \over 13} \end{align}

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 13^2 & = 12^2 + x^2 \\ 169 & = 144 + x^2 \\ x^2 & = 25 \\ x & = \pm\sqrt{25} \\ & = 5 \text{ or } -5 \text{ (Reject)} \\ \\ \\ \sin (A + B) & = \sin A \cos B + \sin B \cos A \phantom{000000} [\text{Addition formula}] \\ & = \left({3 \over 5}\right)\left({5 \over 13}\right) + \left({12 \over 13}\right)\left(-{4 \over 5}\right) \\ & = {3 \over 13} - {48 \over 65} \\ & = - {33 \over 65} \end{align}

(ii)

\begin{align} \tan (A + B) & = {\tan A + \tan B \over 1 - \tan A \tan B} \phantom{000000} [\text{Addition formula}] \\ & = {-{3 \over 4} + {12 \over 5} \over 1 - \left(-{3 \over 4}\right)\left({12 \over 5}\right)} \\ & = {{33 \over 20} \over {56 \over 20}} \\ & = {33 \over 56} \end{align}

(iii)

\begin{align} \sec (A - B) & = {1 \over \cos (A - B)} \\ & = {1 \over \cos A \cos B + \sin A \sin B} \phantom{000000} [\text{Addition formula}] \\ & = {1 \over \left(-{4 \over 5}\right)\left({5 \over 13}\right) + \left({3 \over 5}\right)\left({12 \over 13}\right)} \\ & = {1 \over -{4 \over 13} + {36 \over 65}} \\ & = {65 \over 16} \end{align}

(a)

\begin{align} [\text{Apply addition formula}] \phantom{000000} \sin (x + 30^\circ) & = 2\cos x \\ \sin x \cos 30^\circ + \sin 30^\circ \cos x & = 2\cos x \\ (\sin x) \left({\sqrt{3} \over 2}\right) + \left({1 \over 2}\right) \cos x & = 2\cos x \phantom{00000000} [ \text{Special values: } \sin 30^\circ, \cos 30^\circ ] \\ {\sqrt{3} \over 2} \sin x + {1 \over 2}\cos x & = 2\cos x \\ {\sqrt{3} \over 2} \sin x & = 2\cos x - {1 \over 2} \cos x \\\ {\sqrt{3} \over 2}\sin x & = {3 \over 2}\cos x \\ \sqrt{3}\sin x & = 3\cos x \\ \sqrt{3}\left({\sin x \over \cos x}\right) & = 3 \\ \sqrt{3}\tan x & = 3 \\ \tan x & = {3 \over \sqrt{3}} \\ \tan x & = \sqrt{3} \phantom{0000000000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (\sqrt{3}) \\ & = 60^\circ \end{align}

\begin{align} x & = 60^\circ, 180^\circ + 60^\circ \\ & = 60^\circ, 240^\circ \end{align}

(b)

\begin{align} 2\cos (x + 45^\circ) & = 3 \sin x \\ 2(\cos x \cos 45^\circ - \sin x \sin 45^\circ) & = 3 \sin x \\ 2\left[ (\cos x) \left({1 \over \sqrt{2}}\right) - (\sin x) \left({1 \over \sqrt{2}}\right)\right] & = 3\sin x \phantom{00000000} [\text{Special values: } \cos 45^\circ, \sin 45^\circ ] \\ 2\left({1 \over \sqrt{2}}\cos x - {1 \over \sqrt{2}}\sin x\right) & = 3\sin x \\ {2 \over \sqrt{2}}\cos x - {2 \over \sqrt{2}}\sin x & = 3\sin x \\ \sqrt{2}\cos x - \sqrt{2}\sin x & = 3\sin x \\ \sqrt{2}\cos x & = 3\sin x + \sqrt{2}\sin x \\ \sqrt{2}\cos x & = (3 + \sqrt{2})\sin x \\ \sqrt{2} & = {(3 + \sqrt{2})\sin x \over \cos x} \\ {\sqrt{2} \over 3 + \sqrt{2}} & = {\sin x \over \cos x} \\ \\ \therefore \tan x & = {\sqrt{2} \over 3 + \sqrt{2}} \phantom{000000} [\text{1st & 3rd quadrants since } \tan x > 0 ] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left( {\sqrt{2} \over 3 + \sqrt{2}} \right) \\ & = 17.76^\circ \end{align}

\begin{align} x & = 17.76^\circ, 180^\circ + 17.76^\circ \\ & = 17.76^\circ, 197.76^\circ \\ & \approx 17.8^\circ, 197.8^\circ \end{align}

(c)

\begin{align} \cos (x + 10^\circ) & = \cos (x - 50^\circ) \\ \cos x \cos 10^\circ - \sin x \sin 10^\circ & = \cos x \cos 50^\circ + \sin x \sin 50^\circ \\ (\cos x) (0.9848) - (\sin x) (0.1736) & = (\cos x) (0.6428) + (\sin x) (0.766) \\ 0.9848\cos x - 0.1736\sin x & = 0.6428\cos x + 0.766\sin x \\ 0.9848\cos x - 0.6428\cos x & = 0.766\sin x + 0.1736\sin x \\ 0.342\cos x & = 0.9396\sin x \\ 0.342 & = {0.9396\sin x \over \cos x} \\ {0.342 \over 0.9396} & = {\sin x \over \cos x} \\ \\ \therefore \tan x & = 0.36398 \phantom{00000000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (0.36398) \\ & = 20^\circ \end{align}

\begin{align} x & = 20^\circ, 180^\circ + 20^\circ \\ & = 20^\circ, 200^\circ \end{align}

(d)

\begin{align} 2\sec (x + 60^\circ) & = 5\sec (x - 20^\circ) \\ 2[\sec(x + 60^\circ)] & = 5[\sec (x - 20^\circ)] \\ 2\left[ {1 \over \cos (x + 60^\circ)}\right] & = 5\left[ {1 \over \cos (x - 20^\circ)} \right] \\ {2 \over \cos (x + 60^\circ)} & = {5 \over \cos (x - 20^\circ)} \\ 2\cos (x - 20^\circ) & = 5\cos (x + 50^\circ) \\ 2(\cos x \cos 20^\circ + \sin x \sin 20^\circ) & = 5(\cos x \cos 60^\circ - \sin x \sin 60^\circ) \\ 2[(\cos x) (0.9397) + (\sin x) (0.342)] & = 5\left[ (\cos x) \left({1 \over 2}\right) - (\sin x) \left({\sqrt{3} \over 2}\right)\right] \phantom{0000} [\text{Special values: } \cos 60^\circ, \sin 60^\circ] \\ 2(0.9397 \cos x + 0.342 \sin x) & = 5\left[ {1 \over 2} \cos x - {\sqrt{3} \over 2}\sin x \right] \\ 1.8794 \cos x + 0.684 \sin x & = 2.5 \cos x - 4.3301 \sin x \\ 0.684 \sin x + 4.3301 \sin x & = 2.5 \cos - 1.8794 \cos x \\ 5.0141 \sin x & = 0.6206 \cos x \\ {5.0141 \sin x \over \cos x} & = 0.6206 \\ {\sin x \over \cos x} & = {0.6206 \over 5.0141} \\ \tan x & = 0.12377 \phantom{000000} [\text{1st & 3rd quadrants since } \tan x > 0 ] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (0.12377) \\ & = 7.06^\circ \end{align}

\begin{align} x & = 7.06^\circ, 180^\circ + 7.06^\circ \\ & = 7.06^\circ, 187.06^\circ \\ & \approx 7.1^\circ, 187.1^\circ \end{align}

(a)

\begin{align} [\text{Apply addition formula}] \phantom{000000} {\tan x + \tan 15^\circ \over 1 - \tan x \tan 15^\circ} & = 2 \\ \tan (x + 15^\circ) & = 2 \phantom{000000} [\text{1st & 3rd quadrants since } \tan (x + 15^\circ) > 0 ] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (2) \\ & = 63.43^\circ \end{align}

\begin{align} \text{Since } 0^\circ < x < 360^\circ & \rightarrow \phantom{.} 15^\circ < x + 15^\circ < 375^\circ \\ \\ x + 15^\circ & = 63.43^\circ, 180^\circ + 63.43^\circ \\ & = 63.43^\circ, 243.43^\circ \\ \\ x & = 63.43^\circ - 15^\circ, 243.43^\circ - 15^\circ \\ & = 48.43^\circ, 228.43^\circ \\ & \approx 48.4^\circ, 228.4^\circ \end{align}

(b)

\begin{align} 2\tan x + 3\tan (x - 45^\circ) & = 0 \\ 2\tan x + 3\left( {\tan x - \tan 45^\circ \over 1 + \tan x \tan 45^\circ} \right) & = 0 \\ 2\tan x + 3\left[ {\tan x - 1 \over 1 + (\tan x) (1)} \right] & = 0 \\ 2\tan x + {3(\tan x - 1) \over 1 + \tan x} & = 0 \\ 2\tan x & = -{3(\tan x - 1) \over 1 + \tan x} \\ (2\tan x)(1 + \tan x) & = -3(\tan x - 1) \\ 2\tan x + 2\tan^2 x & = -3\tan x + 3 \\ 2\tan^2 x + 5\tan x - 3 & = 0 \\ (2\tan x - 1)(\tan x + 3) & = 0 \\ \\ 2\tan x - 1 = 0 \phantom{00} & \text{or} \phantom{000} \tan x + 3 = 0 \end{align}

\begin{align} 2\tan x - 1 & = 0 \\ 2\tan x & = 1 \\ \tan x & = {1 \over 2} \\ \tan x & = 0.5 \phantom{000000} [\text{1st & 3rd quadrants since } \tan x > 0 ]\\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (0.5) \\ & = 26.57^\circ \end{align}

\begin{align} x & = 26.57^\circ, 180^\circ + 26.57^\circ \\ & = 26.57^\circ, 206.57^\circ \\ & \approx 26.6^\circ, 206.6^\circ \end{align}

\begin{align} \tan x + 3 & = 0 \\ \tan x & = -3 \phantom{000000} [\text{2nd & 4th quadrants since } \tan x < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (3) \\ & = 71.57^\circ \end{align}

\begin{align} x & = 180^\circ - 71.57^\circ, 360^\circ - 71.57^\circ \\ & = 108.43^\circ, 288.43^\circ \\ & \approx 108.4^\circ, 288.4^\circ \\ \\ \\ \therefore x & = 26.6^\circ, 108.4^\circ, 206.6^\circ, 288.4^\circ \end{align}

(c)

\begin{align} \tan 5x & = \tan 2x \\ \tan (3x + 2x) & = \tan 2x \\ {\tan 3x + \tan 2x \over 1 - \tan 3x \tan 2x} & = \tan 2x \tan 3x + \tan 2x & = \tan 2x(1 - \tan 3x \tan 2x) \\ \tan 3x + \tan 2x & = \tan 2x - \tan 3x \tan^2 2x \\ \tan 3x + \tan 3x \tan^2 2x & = 0 \\ \tan 3x(1 + \tan^2 2x) & = 0 \\ \\ \tan 3x = 0 \phantom{00} & \text{or} \phantom{000} 1 + \tan^2 2x = 0 \end{align}

\begin{align} \tan 3x & = 0 \end{align}

\begin{align} \text{Since } 0^\circ < & x < 360^\circ \rightarrow \phantom{.} 0^\circ < 3x < 1080^\circ \\ \\ 3x & = 0^\circ \text{ (Reject)}, 180^\circ, 360^\circ, 180^\circ + 360^\circ, 360^\circ + 360^\circ \\ & = 180^\circ, 360^\circ, 540^\circ, 720^\circ, 540^\circ + 360^\circ \\ & = 180^\circ, 360^\circ, 540^\circ, 720^\circ, 900^\circ \\ \\ x & = 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ \end{align}

\begin{align} 1 + \tan^2 2x & = 0 \\ \tan^2 2x & = - 1 \\ \tan 2x & = \pm\sqrt{-1} \phantom{0} \text{ (No real roots)} \\ \\ \\ \therefore x & = 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ \end{align}

(a)

\begin{align} [\text{Apply addition formula}] \phantom{000000} \sin x \cos 1 - \cos x \sin 1 & = 0.2 \\ \sin (x - 1) & = 0.2 \phantom{000000} [\text{1st & 2nd quadrants since } \sin (x - 1) > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.2) \\ & = 0.2014 \end{align}

\begin{align} \text{Since } 0 < x < 2\pi \rightarrow & \phantom{.} -1 < x - 1 < 2\pi - 1 \\ & \phantom{.} -1 < x - 1 < 5.283 \\ \\ x - 1 & = 0.2014, \pi - 0.2014 \\ & = 0.2014, 2.940 \\ \\ x & = 0.2014 + 1, 2.940 + 1 \\ & = 1.2014, 3.940 \\ & \approx 1.20, 3.94 \end{align}

(b)

\begin{align} [\text{Apply addition formula}] \phantom{000000} {\tan 2x - \tan 1 \over 1 + \tan 2x \tan 1} & = {1 \over 2} \\ \tan (2x - 1) & = 0.5 \phantom{000000} [\text{1st & 3rd quadrants since } \tan (2x - 1) > 0 ] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (0.5) \\ & = 0.4636 \end{align}

\begin{align} \text{Since } 0 < x < 2\pi \rightarrow \phantom{.} 0 < \phantom{.} & 2x < 4\pi \rightarrow \phantom{.} -1 < 2x - 1 < 4\pi - 1 \\ & \phantom{2x < 4\pi \rightarrow . .} -1 < 2x - 1 < 11.57 \\ \\ 2x - 1 & = 0.4636, \pi + 0.4636 \\ & = 0.4646, 3.605, 0.4636 + 2\pi, 3.605 + 2\pi \\ & = 0.4646, 3.605, 6.747, 9.888 \\ \\ 2x & = 1.4636, 4.605, 7.747, 10.888 \\ \\ x & = 0.7318, 2.3025, 3.8735, 5.444 \\ & \approx 0.732, 2.30, 3.87, 5.44 \end{align}

(c)

\begin{align} \cos (x - 0.4) & = 3 \sin x \\ \cos x \cos 0.4 + \sin x \sin 0.4 & = 3 \sin x \\ (\cos x) (0.9211) + (\sin x) (0.3894) & = 3 \sin x \\ 0.9211\cos x + 0.3894\sin x & = 3 \sin x \\ 0.9211\cos x & = 3\sin x - 0.3894\sin x \\ 0.9211\cos x & = 2.6106\sin x \\ 0.9211 & = {2.6106 \sin x \over \cos x} \\ {0.9211 \over 2.6106} & = {\sin x \over \cos x} \\ \\ \therefore \tan x & = 0.3528 \phantom{000000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (0.3528) \\ & = 0.3392 \end{align}

\begin{align} x & = 0.3392, \pi + 0.3392 \\ & \approx 0.339, 3.48 \end{align}

(d)

\begin{align} 3 \cos x & = 4 \sin (x - 2) \\ 3 \cos x & = 4(\sin x \cos 2 - \sin 2 \cos x) \\ 3 \cos x & = 4[\sin x (-0.4161) - (0.9093) \cos x] \\ 3 \cos x & = 4(-0.4161\sin x - 0.9093 \cos x) \\ 3 \cos x & = -1.6644\sin x -3.6372 \cos x \\ 6.6372 \cos x & = -1.6644 \sin x \\ 6.6372 & = {-1.6644 \sin x \over \cos x} \\ {6.6372 \over -1.6644} & = {\sin x \over \cos x} \\ \\ \therefore \tan x & = -3.9877 \phantom{000000} [\text{2nd & 4th quadrants since } \tan x < 0 ] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (3.9877) \\ & = 1.325 \end{align}

\begin{align} x & = \pi - 1.325, 2\pi - 1.325 \\ & \approx 1.82, 4.96 \end{align}

(i)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ (\sqrt{5})^2 & = 2^2 + x^2 \\ 5 & = 4 + x^2 \\ x^2 & = 1 \\ x & = \pm \sqrt{1} \\ x & = 1 \text{ or } - 1 \text{ (Reject)} \\ \\ \tan A & = {Opp \over Adj} \\ & = {1 \over 2} \end{align}

(ii)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ (\sqrt{10})^2 & = 3^2 + x^2 \\ 10 & = 9 + x^2 \\ x^2 & = 1 \\ x & = \pm \sqrt{1} \\ x & = 1 \text{ or } -1 \text{ (Reject)} \\ \\ \tan B & = {Opp \over Adj} \\ & = {3 \over 1} \\ & = 3 \end{align}

(iii)

\begin{align} \tan (B - A) & = {\tan B - \tan A \over 1 + \tan B \tan A} \\ & = {3 - {1 \over 2} \over 1 + (3)\left({1 \over 2}\right)} \\ & = 1 \\ \\ \text{Since } \tan 45^\circ = 1, & \phantom{.} B - A = 45^\circ \end{align}

(i)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 5^2 & = 3^2 + x^2 \\ 25 & = 9 + x^2 \\ x^2 & = 16 \\ x & = \pm \sqrt{16} \\ x & = 4 \text{ or } -4 \text{ (Reject)} \end{align}

(Given that angles A and B are both acute and cos (A + B) < 0, we can conclude that angle (A + B) is in the 2nd quadrant)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 13^2 & = (-5)^2 + x^2 \\ 169 & = 25 + x^2 \\ x^2 & = 144 \\ x & = \pm \sqrt{144} \\ x & = 12 \text{ or } -12 \text{ (Reject)} \\ \\ \sin B & = \sin [(A + B) - A] \\ & = \sin (A + B) \cos A - \sin A \cos (A + B) \\ & = \left({12 \over 13}\right)\left({3 \over 5}\right) - \left({4 \over 5}\right)\left(-{5 \over 13}\right) \\ & = {36 \over 65} + {4 \over 13} \\ & = {56 \over 65} \end{align}

(ii)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 13^2 & = 12^2 + x^2 \\ 169 & = 144 + x^2 \\ x^2 & = 25 \\ x & = \pm \sqrt{25} \\ x & = 5 \text{ or } - 5 \text{ (Reject)} \\ \\ \sin (A + B + C) & = \sin [(A + B) + C] \\ & = \sin (A + B) \cos C + \sin C \cos (A + B) \\ & = \left({12 \over 13}\right)\left({5 \over 13}\right) + \left({12 \over 13}\right)\left(-{5 \over 13}\right) \\ & = 0 \end{align}

(iii)

\begin{align} \sin (A + B + C) & = 0 \phantom{000} [\text{From part ii}] \\ \\ \text{Since } \sin 180^\circ = 0, & \phantom{.} A + B + C = 180^\circ \\ \\ \therefore A, B \text{ and } & C \text{ are the angles of a triangle} \end{align}

Prove question

\begin{align} \text{L.H.S } & = \sin (A + B) \sin (A - B) \\ & = (\sin A \cos B + \sin B \cos A) (\sin A \cos B - \sin B \cos A) \\ & = \sin^2 A \cos^2 B - \sin^2 B \cos^2 A \\ & = \sin^2 A (1 - \sin^2 B) - \sin^2 B (1 - \sin^2 A) \phantom{000000} [\text{Since } \sin^2 \theta + \cos^2 \theta = 1 \rightarrow \cos^2 \theta = 1 - \sin^2 \theta] \\ & = \sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B \\ & = \sin^2 A - \sin^2 B \\ & = \text{R.H.S} \end{align}

(i)

\begin{align} \sin^2 3x & = \sin^2 x \\ \sin^2 3x - \sin^2 x & = 0 \\ \sin (3x + x) \sin (3x - x) & = 0 \phantom{000000} [\text{Use identity proved earlier} ] \\ \sin 4x \sin 2x & = 0 \\ \\ \sin 4x = 0 \phantom{00} & \text{or} \phantom{000} \sin 2x = 0 \end{align}

\begin{align} \sin 4x & = 0 \end{align}

\begin{align} \text{Since }0^\circ \le \phantom{.} & x \le 180^\circ \rightarrow \phantom{.} 0^\circ \le 4x \le 720^\circ \\ \\ 4x & = 0^\circ, 180^\circ, 360^\circ, 180^\circ + 360^\circ, 360^\circ + 360^\circ \\ & = 0^\circ, 180^\circ, 360^\circ, 540^\circ, 720^\circ \\ \\ x & = 0^\circ, 45^\circ, 90^\circ, 135^\circ, 180^\circ \end{align}

\begin{align} \sin 2x & = 0 \end{align}

\begin{align} \text{Since }0^\circ \le \phantom{.} & x \le 180^\circ \rightarrow \phantom{.} 0^\circ \le 2x \le 360^\circ \\ \\ 2x & = 0^\circ, 180^\circ, 360^\circ \\ \\ x & = 0^\circ, 90^\circ, 180^\circ \\ \\ \\ \therefore x & = 0^\circ, 45^\circ, 90^\circ, 135^\circ, 180^\circ \end{align}

(ii)

\begin{align} \sin (A + B) \sin (A - B) & = \sin 75^\circ \sin 15^\circ \\ \\ A + B & = 75 \phantom{0} \text{--- (1)} \\ \\ A - B & = 15 \\ A & = B + 15 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (B + 15) + B & = 75 \\ 2B & = 75 - 15 \\ 2B & = 60 \\ B & = {60 \over 2} \\ B & = 30 \\ \\ \text{Substitute } & B = 30 \text{ into (2),} \\ A & = 30 + 15 \\ & = 45 \\ \\ \\ \sin (A + B) \sin (A - B) & = \sin^2 A - \sin^2 B \phantom{00000000} [\text{First part}] \\ \\ \therefore \sin 75^\circ \sin 15^\circ & = \sin^2 45^\circ - \sin^2 30^\circ \\ & = (\sin 45^\circ)^2 - (\sin 30^\circ)^2 \\ & = \left({1 \over \sqrt{2}}\right)^2 - \left({1 \over 2}\right)^2 \phantom{000000} [\text{Special values: } \sin 45^\circ, \sin 30^\circ] \\ & = {1 \over 2} - {1 \over 4} \\ & = {1 \over 4} \end{align}

(i)

\begin{align} \tan \alpha & = {5 \over x} \\ \\ \tan (\alpha + \beta) & = {10 \over x} \\ {\tan \alpha + \tan \beta \over 1 - \tan \alpha \tan \beta} & = {10 \over x} \\ { {5 \over x} + \tan \beta \over 1 - {5 \over x} \tan \beta } & = {10 \over x} \\ { {5 \over x} + \tan \beta \over 1 - {5 \over x} \tan \beta } \times {x \over x} & = {10 \over x} \\ { 5 + x \tan \beta \over x - 5 \tan \beta} & = {10 \over x} \\ x(5 + x \tan \beta) & = 10(x - 5 \tan \beta) \\ 5x + x^2 \tan \beta & = 10x - 50 \tan \beta \\ x^2 \tan \beta + 50 \tan \beta & = 10x - 5x \\ \tan \beta (x^2 + 50) & = 5x \\ \tan \beta & = {5x \over x^2 + 50} \end{align}

(ii)

\begin{align} \tan (2\alpha + \beta) & = \tan [\alpha + (\alpha + \beta)] \\ & = { \tan \alpha + \tan (\alpha + \beta) \over 1 - \tan \alpha \tan (\alpha + \beta) } \\ & = { {5 \over x} + {10 \over x} \over 1 - \left(5 \over x\right)\left(10 \over x\right) } \\ & = { {15 \over x} \over 1 - {50 \over x^2} } \times {x^2 \over x^2} \\ & = { 15x \over x^2 - 50 } \\ \\ \\ \text{For } \tan (2 \alpha + \beta) & \text{ to be defined, } x^2 - 50 \ne 0 \\ \\ \text{Consider } x^2 - 50 & = 0 \\ x^2 & = 50 \\ x & = \pm \sqrt{50} \\ \\ \\ \therefore & \phantom{.} x \in \mathbb{R}, x \ne \pm \sqrt{50} \end{align}

\begin{align} A + B + C & = 180^\circ \\ A + B & = 180^\circ - C \\ \tan (A + B) & = \tan (180^\circ - C) \\ {\tan A + \tan B \over 1 - \tan A \tan B} & = { \tan 180^\circ - \tan C \over 1 + \tan 180^\circ \tan C} \\ {\tan A + \tan B \over 1 - \tan A \tan B} & = { 0 - \tan C \over 1 + (0)(\tan C) } \\ {\tan A + \tan B \over 1 - \tan A \tan B} & = { - \tan C \over 1 } \\ \tan A + \tan B & = - \tan C (1 - \tan A \tan B) \\ \tan A + \tan B & = - \tan C + \tan A \tan B \tan C \\ \tan A + \tan B + \tan C & = \tan A \tan B \tan C \phantom{00} \text{ (Shown)} \end{align}