Past Exam Questions

\[ \sec^2x+\sec x\tan x=\frac{1}{\cos^2x}+\frac{\sin x}{\cos^2x} =\frac{1+\sin x}{\cos^2x} =\frac{1+\sin x}{1-\sin^2x} =\frac{1}{1-\sin x}. \]

\[ \frac{1-\cos^2x}{\sec^2x-1}=\frac{\sin^2x}{\tan^2x} =\frac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}=\cos^2x=1-\sin^2x. \]

\[ \frac{1+\tan^2x}{\tan x}=\frac{\sec^2x}{\tan x} =\frac{1/\cos^2x}{\sin x/\cos x} =\frac{1}{\sin x\cos x}=\sec x\,\csc x. \]

\[ \frac{\sin x}{1-\cos^2x}=\frac{\sin x}{\sin^2x}=\frac{1}{\sin x}=\csc x. \]

\[ (\tan x+\sec x)^2=\left(\frac{\sin x}{\cos x}+\frac{1}{\cos x}\right)^2 =\frac{(1+\sin x)^2}{\cos^2x} =\frac{(1+\sin x)^2}{1-\sin^2x} =\frac{1+\sin x}{1-\sin x}. \]

\[ \frac{1}{1+\cos x}+\frac{1}{1-\cos x} =\frac{(1-\cos x)+(1+\cos x)}{1-\cos^2x} =\frac{2}{\sin^2x}=2\csc^2x. \]

\[ \frac{\cos x(1-\sin x)+\cos x(1+\sin x)}{1-\sin^2x} =\frac{2\cos x}{\cos^2x}=\frac{2}{\cos x}=2\sec x. \]

Use \(\sin(u\pm v)=\sin u\cos v\pm\cos u\sin v\):
\( \sin(\alpha+\beta)+\sin(\alpha-\beta)\) \(=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)\) \(+(\sin\alpha\cos\beta-\cos\alpha\sin\beta)\) \(=2\sin\alpha\cos\beta. \)

Use \(\cos(u\pm v)=\cos u\cos v\mp\sin u\sin v\):
\( \cos(\alpha+\beta)+\cos(\alpha-\beta)\) \( =(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\) \(+(\cos\alpha\cos\beta+\sin\alpha\sin\beta)\) \(=2\cos\alpha\cos\beta. \)

Recognize the sine–difference pattern \(\sin U\cos V-\cos U\sin V=\sin(U-V)\) with \(U=\alpha-\beta\), \(V=\alpha\):
\( \sin(\alpha-\beta)\cos\alpha-\cos(\alpha-\beta)\sin\alpha\) \(=\sin\big((\alpha-\beta)-\alpha\big)\) \(=\sin(-\beta)=-\sin\beta. \)

Start from the right-hand side and rewrite in sine/cosine:

\[ \frac{\csc\theta\,\csc\varphi}{\cot\theta+\cot\varphi} \;=\; \frac{\dfrac{1}{\sin\theta}\cdot\dfrac{1}{\sin\varphi}} {\dfrac{\cos\theta}{\sin\theta}+\dfrac{\cos\varphi}{\sin\varphi}} \]

Combine the denominator over a common denominator:

\[ =\; \frac{\dfrac{1}{\sin\theta\,\sin\varphi}} {\dfrac{\cos\theta\,\sin\varphi+\cos\varphi\,\sin\theta} {\sin\theta\,\sin\varphi}} \]

Invert and multiply:

\[ =\; \frac{1}{\cos\theta\,\sin\varphi+\cos\varphi\,\sin\theta} \]

Recognize the sine addition formula \(\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi\):

\[ =\; \frac{1}{\sin(\theta+\varphi)} \;=\; \csc(\theta+\varphi). \]

Hence \(\displaystyle \csc(\theta+\varphi)\equiv \dfrac{\csc\theta\,\csc\varphi}{\cot\theta+\cot\varphi}\).

Expand each cosine:

\[ \cos(5x+x) = \cos 5x\cos x - \sin 5x\sin x, \] \[ \cos(5x-x) = \cos 5x\cos x + \sin 5x\sin x. \]

Add the two results:

\[ \cos 6x + \cos 4x = (\cos 5x\cos x - \sin 5x\sin x) + (\cos 5x\cos x + \sin 5x\sin x) = 2\cos 5x\cos x. \]

Hence proved.

Start from the identity \(\cos(x-y)=\cos x\cos y+\sin x\sin y\).

Compute \(p^2+q^2\):

\[ p^2+q^2 =(\sin x+\sin y)^2 + (\cos x+\cos y)^2 \] \[ =(\sin^2x+\sin^2y + 2\sin x\sin y) +(\cos^2x+\cos^2y + 2\cos x\cos y) \] \[ =( \sin^2x+\cos^2x ) + ( \sin^2y+\cos^2y ) + 2(\sin x\sin y+\cos x\cos y) \] \[ =1+1+2\cos(x-y) =2+2\cos(x-y). \]

Solve for \(\cos(x-y)\):

\[ \cos(x-y)=\frac{p^2+q^2-2}{2}. \]

Result: \(\displaystyle \boxed{\cos(x-y)=\tfrac{p^2+q^2-2}{2}}\).

(a) Use \(\cos(A+B)=\cos A\cos B-\sin A\sin B\):

\[ \cos 3\theta=\cos(2\theta+\theta) =(\cos2\theta)\cos\theta-(\sin2\theta)\sin\theta . \] With \(\cos2\theta=2\cos^2\theta-1\) and \(\sin2\theta=2\sin\theta\cos\theta\), \[ \cos 3\theta =(2\cos^2\theta-1)\cos\theta-2\sin\theta\cos\theta\sin\theta =4\cos^3\theta-3\cos\theta. \]

(b) Substitute the identity from (a) and \(\cos2\theta=2\cos^2\theta-1\):

\[ (4\cos^3\theta-3\cos\theta)+\cos\theta(2\cos^2\theta-1) =\cos^2\theta . \] \[ \Rightarrow\; 6\cos^3\theta-4\cos\theta=\cos^2\theta \;\Rightarrow\; 6\cos^3\theta-\cos^2\theta-4\cos\theta=0 \] \[ \Rightarrow\; \cos\theta\,(6\cos^2\theta-\cos\theta-4)=0 . \]

Either \(\cos\theta=0\Rightarrow \theta=90^\circ\), or \(6\cos^2\theta-\cos\theta-4=0\). Solving the quadratic in \(c=\cos\theta\): \[ c=\frac{1\pm\sqrt{1+96}}{12} =\frac{1\pm\sqrt{97}}{12} \approx 0.9041\;\text{ or }\;-0.7374. \] Hence \[ \theta=\cos^{-1}(0.9041)\approx 25.3^\circ,\qquad \theta=\cos^{-1}(-0.7374)\approx 137.5^\circ, \] together with \(\theta=90^\circ\) from \(\cos\theta=0\).

Expand each term using \(\sin(u\pm v)=\sin u\cos v\pm\cos u\sin v\) and \(\cos(u\pm v)=\cos u\cos v\mp\sin u\sin v\):

\[ \sin\!\left(x+\tfrac{\pi}{6}\right) -\sin\!\left(x-\tfrac{\pi}{6}\right) =(\sin x\cos\tfrac{\pi}{6}+\cos x\sin\tfrac{\pi}{6}) -(\sin x\cos\tfrac{\pi}{6}-\cos x\sin\tfrac{\pi}{6}) \] \[ =2\cos x\sin\tfrac{\pi}{6} =2\cos x\cdot\tfrac12 =\cos x. \]

\[ \cos\!\left(x+\tfrac{\pi}{3}\right) -\cos\!\left(x-\tfrac{\pi}{3}\right) =(\cos x\cos\tfrac{\pi}{3}-\sin x\sin\tfrac{\pi}{3}) -(\cos x\cos\tfrac{\pi}{3}+\sin x\sin\tfrac{\pi}{3}) \] \[ =-2\sin x\sin\tfrac{\pi}{3} =-2\sin x\cdot\tfrac{\sqrt3}{2} =-\sqrt3\,\sin x. \]

Equate LHS and RHS: \[ \cos x=-\sqrt3\,\sin x \quad\Rightarrow\quad \tan x=\frac{\sin x}{\cos x}=-\frac{1}{\sqrt3}. \]

(b) Since \(\tan x = -\tfrac{1}{\sqrt3}\) (reference angle \(=\tfrac{\pi}{6}\)) and tangent is negative in quadrants II and IV, \[ x = \pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}, \qquad x = 2\pi-\tfrac{\pi}{6}=\tfrac{11\pi}{6}. \]

Final answers

(a) \(\displaystyle \tan x=-\frac{1}{\sqrt3}\).
(b) \(\displaystyle x=\frac{5\pi}{6},\;\frac{11\pi}{6}\) on \(0\le x\le 2\pi\).

Use the tangent addition formula: \( \tan(x+45^\circ)=\frac{\tan x+1}{\,1-\tan x\,}. \) Also \(\cot x=\dfrac{1}{\tan x}\).

Let \(t=\tan x\). Then \[ \frac{t+1}{1-t}=2\cdot\frac{1}{t} \;\Rightarrow\; t(t+1)=2(1-t) \] \[ \Rightarrow\; t^2+t=2-2t \;\Rightarrow\; t^2+3t-2=0. \]

Solve the quadratic: \[ t=\frac{-3\pm\sqrt{9+8}}{2} =\frac{-3\pm\sqrt{17}}{2} \approx 0.5616\;\text{ or }\;-3.5616. \]

\[ \tan x\approx0.5616 \Rightarrow x\approx \arctan(0.5616)=29.3^\circ, \] \[ \tan x\approx-3.5616 \Rightarrow x\text{ in QII } =180^\circ-\arctan(3.5616)=105.7^\circ. \]

Final answer

\(\boxed{x\approx29.3^\circ,\;105.7^\circ}\)

Expand the left side with \(\cos(u-v)=\cos u\cos v+\sin u\sin v\):

\[ \cos(\theta-60^\circ) =\cos\theta\cos60^\circ+\sin\theta\sin60^\circ =\tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta. \]

Set equal to \(3\sin\theta\) and rearrange:

\[ \tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta=3\sin\theta \;\Rightarrow\; \tfrac12\cos\theta=\Big(3-\tfrac{\sqrt3}{2}\Big)\sin\theta. \] \[ \cos\theta=(6-\sqrt3)\,\sin\theta \;\Rightarrow\; \tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{1}{\,6-\sqrt3\,}. \]

Evaluate the principal angle:

\[ \theta=\arctan\!\Big(\frac{1}{6-\sqrt3}\Big) \approx \arctan(0.2343)=13.2^\circ. \]

Tangent is positive in quadrants I and III, so add \(180^\circ\):

\[ \theta=13.2^\circ,\quad 13.2^\circ+180^\circ=193.2^\circ. \]

1. Use the identity \(\cot 2\theta = \frac{1}{\tan 2\theta}\) and \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\) to express \(\cot 2\theta\) in terms of \(\tan \theta\):

\(\cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta}\)

2. Use the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) to express \(\tan(\theta + 45^\circ)\):

\(\tan(\theta + 45^\circ) = \frac{\tan \theta + 1}{1 - \tan \theta}\)

3. Substitute these into the original equation:

\(\tan \theta \cdot \frac{\tan \theta + 1}{1 - \tan \theta} = 2 \cdot \frac{1 - \tan^2 \theta}{2 \tan \theta}\)

4. Simplify and rearrange to form a quadratic equation:

\(2 \tan^2 \theta + \tan \theta - 1 = 0\)

5. Solve the quadratic equation using the quadratic formula \(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 1, c = -1\):

\(\tan \theta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\)

6. This gives \(\tan \theta = \frac{1}{2}\) or \(\tan \theta = -1\).

7. Since \(0^\circ < \theta < 90^\circ\), \(\tan \theta = \frac{1}{2}\) is valid.

8. Therefore, \(\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.6^\circ\).

1. Use the tangent addition and subtraction formulas:

\(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

2. For \(\tan(\theta + 60^\circ)\):

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \cdot \sqrt{3}}\)

3. For \(\tan(60^\circ - \theta)\):

\(\tan(60^\circ - \theta) = \frac{\sqrt{3} - \tan \theta}{1 + \tan \theta \cdot \sqrt{3}}\)

4. Set the equation:

\(\frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \cdot \sqrt{3}} = 2 + \frac{\sqrt{3} - \tan \theta}{1 + \tan \theta \cdot \sqrt{3}}\)

5. Clear the fractions by multiplying through by \((1 - \tan \theta \cdot \sqrt{3})(1 + \tan \theta \cdot \sqrt{3})\).

6. Simplify and rearrange to form a quadratic equation:

\(3\tan^2 \theta + 4\tan \theta - 1 = 0\)

7. Solve the quadratic equation using the quadratic formula:

\(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 3, b = 4, c = -1\).

8. Calculate:

\(\tan \theta = \frac{-4 \pm \sqrt{16 + 12}}{6} = \frac{-4 \pm \sqrt{28}}{6} = \frac{-4 \pm 2\sqrt{7}}{6}\)

9. Simplify:

\(\tan \theta = \frac{-2 \pm \sqrt{7}}{3}\)

10. Find \(\theta\) using \(\tan^{-1}\):

\(\theta \approx 12.1^\circ\) and \(\theta \approx 122.9^\circ\).

(a) Start by expressing the left-hand side as a single fraction:

\(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = \frac{\cos 3x \cdot \cos x + \sin 3x \cdot \sin x}{\sin x \cdot \cos x}\)

Use the angle addition formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\):

\(= \frac{\cos(3x - x)}{\sin x \cos x} = \frac{\cos 2x}{\sin x \cos x}\)

Use the double angle identity \(\sin 2x = 2 \sin x \cos x\):

\(= \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2 \cos 2x}{\sin 2x} = 2 \cot 2x\)

Thus, the identity is proven.

(b) Set \(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = 4\):

\(2 \cot 2x = 4\) \(\cot 2x = 2\)

Then, \(\tan 2x = \frac{1}{2}\).

Find \(2x\):

\(2x = \tan^{-1}\left(\frac{1}{2}\right) \approx 0.464 \text{ or } 0.464 + n\pi\)

For \(0 < x < \pi\), solve for \(x\):

\(x = \frac{0.464}{2} \approx 0.232\)

And for the next solution:

\(x = \frac{0.464 + \pi}{2} \approx 1.80\)

Thus, \(x \approx 0.232, 1.80\).