$$ \begin{eqnarray} &\ & tan^2 x^\circ\ cos^2\ x^\circ \\[9pt] &=& \left( \frac{sin\ x^\circ}{cos\ x^\circ} \right)^2\ cos^2 x^\circ \\[9pt] &=& \left( \frac{sin^2x^\circ}{cos^2x^\circ} \right)\ cos^2 x^\circ \\[9pt] &=& \left( \frac{sin^2x^\circ}{\cancel{cos^2}x^\circ} \right)\ \cancel{cos^2} x^\circ \\[9pt] &=& sin^2\ x^\circ \end{eqnarray} $$

$$ \begin{eqnarray} &\ & sin x^\circ\ cos x^\circ\ tan x^\circ \\[9pt] &=& \left( sin x^\circ\right)\left(cos x^\circ\right)\left(\frac{sin x^\circ}{cos x^\circ}\right) \\[9pt] &=& \left( sin x^\circ\right)\left(\cancel{cos x^\circ}\right)\left(\frac{sin x^\circ}{\cancel{cos x^\circ}}\right) \\[9pt] &=& sin^2 x^\circ \end{eqnarray} $$

$$ \begin{eqnarray} &\ & \left(sin x^\circ+cos x^\circ\right)^2 \\[9pt] &=& \left(sin x^\circ+cos x^\circ\right)\left(sin x^\circ+cos x^\circ\right) \\[9pt] &=& sin^2 x^\circ + 2\small\ \normalsize sin x^\circ\small\ \normalsize cos x^\circ + cos^2x^\circ \\[9pt] &=& \left(sin^2 x^\circ + cos^2x^\circ\right) + 2\small\ \normalsize sin x^\circ\small\ \normalsize cos x^\circ \\[9pt] &=& 1+2\small\ \normalsize sin x^\circ\small\ \normalsize cos x^\circ \end{eqnarray} $$

$$ \begin{eqnarray} cos\ x^\circ\ \left(tan\ x^\circ +1\right) &=& cos\ x^\circ\ tan\ x^\circ + cos\ x^\circ \\[6pt] &=& \cancel{cos\ x^\circ} \left(\frac{sin\ x^\circ}{\cancel{cos\ x^\circ}}\right) + cos\ x^\circ \\[6pt] &=& sin\ x^\circ + cos\ x^\circ \\[6pt] \end{eqnarray} $$

There is only one term in the denominator, so we can split the fraction as follows, and then simplify:

$$ \begin{eqnarray} \frac{sin\ x+2\ cos\ x}{cos\ x} &=& \frac{sin\ x}{cos\ x} + \frac{2\ cos\ x}{cos\ x} \\[6pt] &=& tan\ x + \frac{2\ \cancel{cos\ x}}{\cancel{cos\ x}} \\[6pt] &=& tan\ x + 2 \\[6pt] \end{eqnarray} $$

This surprisingly simple question is just double one of the identities!

$$ \begin{eqnarray} 2\,sin^2\,x+2\,cos^2\,x &=& 2(sin^2\,x+cos^2\,x) \\[6pt] &=& 2\times 1 \\[6pt] &=& 2 \end{eqnarray} $$

The SQA has never combined the trig equations and trig identities topics, but this is within the scope of the course.

First we rearrange:

$$ \begin{eqnarray} 3\tiny\ \normalsize sin\ x &=& 2\tiny\ \normalsize cos\ x \\[6pt] \frac{3sin\ x}{cos\ x} &=& 2 \\[6pt] \frac{sin\ x}{cos\ x} &=& \frac{2}{3} \\[6pt] \ tan\ x &=& \frac{2}{3}\:\:\small \textsf{(using the 2nd identity)} \normalsize \end{eqnarray} $$

Then we solve this trig equation in the usual way:

The related acute angle \( \alpha = tan^{-1} \large \frac{2}{3} \normalsize = 33.7^\circ \) (to 1 d.p.)

Because \(tan\ x\) > 0, there are solutions in the 1^st (A) and 3^rd (T) quadrants.

So \(x=\alpha\)  or  \(x=180+\alpha\)

\(x=33.7^\circ\)  or  \(x=180+33.7=213.7^\circ\) (to 1 d.p.)

This example requires us to prove that the given trigonometric identity is true.

We expand the left hand side (LHS) and then simplify it, to show that it equals the right hand side (RHS).

$$ \begin{eqnarray} LHS &=& cos^2\, x\ \left(1+tan^2\,x\right) \\[6pt] &=& cos^2\, x\ + cos^2\, x\ tan^2\, x \\[6pt] &=& cos^2\, x\ + cos^2\, x\ \left(\frac{sin^2\,x}{cos^2\,x}\right) \\[6pt] &=& cos^2\, x\ + \cancel{cos^2\, x}\ \left(\frac{sin^2\,x}{\cancel{cos^2\,x}}\right) \\[6pt] &=& cos^2\, x\ + sin^2\ x \\[6pt] &=& 1\\[6pt] &=& RHS \end{eqnarray} $$