$\text{If } \tan A = \frac{1}{\sqrt{x(x^2 + x + 1)}}, \; \tan B = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}} \quad \text{and} \quad \tan C = (x^{-3} + x^{-2} + x^{-1})^{1/2}, \; 0 < A, B, C < \frac{\pi}{2}, \text{ then}$ $A + B \text{ is equal to:}$
\( C \)
\( \pi-C \)
\( \dfrac{\pi}{2}-C \)
\( 2\pi-C \)
If \( \sin x = -\frac{3}{5} \), where \( \pi < x < \frac{3\pi}{2} \), then \( 80(\tan^{2} x - \cos x) \) is equal to:
The sum of the solutions \( x \in \mathbb{R} \) of the equation \( \dfrac{3\cos 2x + \cos^{3} 2x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6 \) is:
For $0 < c < b < a$, let $(a+b-2c)x^2 + (b+c-2a)x + (c+a-2b) = 0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements:
(I)$\quad$ If $\alpha \in (-1, 0)$, then $b$ cannot be the geometric mean of $a$ and $c$
(II)$\quad$ If $\alpha \in (0, 1)$, then $b$ may be the geometric mean of $a$ and $c$
The number of solutions of \( \sin^{2}x + (2 + 2x - x^{2})\sin x - 3(x-1)^{2} = 0 \), where \( -\pi \le x \le \pi \), is \( \_\_\_\_\_\_\_ \).
Let \( |\cos \theta\cos(60-\theta)\cos(60+\theta)\le \frac{1}{8},\ \theta\in[0,2\pi] \). Then, the sum of all \( \theta\in[0,2\pi] \), where \( \cos 3\theta \) attains its maximum value, is:
If \( \alpha \), \( -\frac{\pi}{2} < \alpha < \frac{\pi}{2} \) is the solution of \( 4\cos\theta + 5\sin\theta = 1 \), then the value of \( \tan \alpha \) is:
Let a rectangle \(ABCD\) of sides \(2\) and \(4\) be inscribed in another rectangle \(PQRS\) such that the vertices of the rectangle \(ABCD\) lie on the sides of the rectangle \(PQRS\). Let \(a\) and \(b\) be the sides of the rectangle \(PQRS\) when its area is maximum. Then \( (a+b)^2 \) is equal to:
The number of solutions of the equation \( 4\sin^{2}x - 4\cos^{3}x + 9 - 4\cos x = 0 \); \( x \in [-2\pi, 2\pi] \) is:
