Correct answer: 2
Explanation:
Given: \[ \tan \theta + \cot \theta = 2 \] We need to find \( \tan^2 \theta + \cot^2 \theta \). Using the identity: \[ (\tan \theta + \cot \theta)^2 = \tan^2 \theta + \cot^2 \theta + 2 \] Substituting \( \tan \theta + \cot \theta = 2 \): \[ 2^2 = \tan^2 \theta + \cot^2 \theta + 2 \] \[ 4 = \tan^2 \theta + \cot^2 \theta + 2 \] \[ \tan^2 \theta + \cot^2 \theta = 4 - 2 = 2 \]
Hence, \( \tan^2 \theta + \cot^2 \theta = {2} \).
The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.