Q1. Evaluate $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$.
Q2. The value of $\cos^2 30^\circ - \sin^2 30^\circ$ is:
Q3. If $\sin 2A = 2 \sin A$, then the value of A is:
Q4. Assertion (A): The value of $\tan 60^\circ$ is greater than 1. Reason (R): $\tan \theta$ increases as $\theta$ increases from $0^\circ$ to $90^\circ$.
Q5. Evaluate: $2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$.
Solve on white paper.
Q6. If $\sin(A - B) = \frac{1}{2}$ and $\cos(A + B) = \frac{1}{2}$, where $0^\circ < A + B \le 90^\circ$ and $A > B$, find A and B.
Q7. Find the value of $x$ if $\tan 3x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ$.
Q8. Verify that $\cos 60^\circ = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ}$.
Q9. In $\triangle ABC$, right-angled at B, $AB = 5$ cm and $\angle ACB = 30^\circ$. Determine the lengths of the sides BC and AC.
Q10. Evaluate: $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$.
Q11. Case Study:
An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work.
(i) What should be the length of the ladder that she should use which, when inclined at an angle of $60^\circ$ to the horizontal, would enable her to reach the required position?
(ii) How far from the foot of the pole should she place the foot of the ladder?
(iii) If the angle of inclination is changed to $30^\circ$, what will be the length of the ladder required?