\(\dfrac{2\sin 25^\circ \cdot \cos 25^\circ}{-2 + 4\sin^2 25^\circ}\)

\[2^{3\sin^2 x} = 4\cdot 2^{\sin x} = 2^2\cdot 2^{\sin x} = 2^{\sin x + 2}\]\[3\sin^2 x = \sin x + 2\]\[3\sin^2 x - \sin x - 2 = 0\]\[(3\sin x + 2)(\sin x - 1) = 0\]\[\sin x = -\frac{2}{3} \text{ or } \sin x = 1\]

\(\sin x = 1\) is rejected, since \(\sin x \lt 0\)

Reference angle \(= 41{,}81^\circ\); \(\sin\) is negative in quadrants 3 and 4

\(x = 221{,}81^\circ + k\cdot 360^\circ\) or \(x = 318{,}19^\circ + k\cdot 360^\circ\), \(k \in \mathbb{Z}\)

\(\cos\theta \lt 0\) and \(180^\circ \lt \theta \lt 360^\circ\), so \(\theta\) is in the third quadrant.

\(x = -5\), \(r = 13\):

\[y^2 = 13^2 - (-5)^2 = 144\]\[y = -12\]\[\sin\theta = -\frac{12}{13}\]\[\sin^2\theta = \left(-\frac{12}{13}\right)^2\]

\(\sin^2\theta = \frac{144}{169}\)

In \(\triangle ABC\), use the cosine rule with \(BC = x\):

\[19^2 = x^2 + 15^2 - 2(x)(15)\cos 105{,}61^\circ\]\[x^2 + 8{,}07x - 136 = 0\]\[x = \frac{-8{,}07 \pm \sqrt{(8{,}07)^2 - 4(1)(-136)}}{2(1)}\]

\(x = 8{,}30\) m (\(x = -16{,}38\) is not valid)

\(B\hat{E}C = 180^\circ - 122^\circ = 58^\circ\) (\(\angle\)s on a str line)

In right-angled \(\triangle BCE\):

\[\tan 58^\circ = \frac{BC}{EC}\]\[EC = \frac{8{,}30}{\tan 58^\circ} = 5{,}19 \text{ m}\]\[DE = 16 - 5{,}19\]

\(DE = 10{,}81\) m

\[g(-45^\circ) = -\tan(-45^\circ) = 1\]\[f(-45^\circ) = 2\sin(-90^\circ) = -2\]\[1 - (-2) = 3\]

\(3\)

\[\text{Period} = \frac{360^\circ}{2}\]

\(180^\circ\)

\[\text{LHS} = \frac{2\sin x\cos x}{2\cos^2 x} + \frac{\sin x}{\frac{\sin x}{\cos x}(1 + \sin x)}\]\[= \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}\]\[= \frac{\sin x(1 + \sin x) + \cos^2 x}{\cos x(1 + \sin x)}\]\[= \frac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1 + \sin x)}\]\[= \frac{\sin x + 1}{\cos x(1 + \sin x)}\]\[= \frac{1}{\cos x}\]

LHS = RHS