\(sin^{-1}\bigg(\frac{\sqrt3}{2}\bigg)+cos^{-1}\bigg(\frac{-\sqrt3}{2}\bigg)+tan^{-1}(-1)\)
= \(\frac{π}{3}+\frac{5π}{6}-\frac{π}{4}\)
= \(\frac{4π+10π-3π}{12}\)
=\(\frac{11π}{12}\)
Hence, the correct option is (A): \(\frac{11π}{12}\)
Find the area of the region \[ R = \{(x, y) : xy \le 27,\; 1 \le y \le x^2 \}. \]
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 ___________.
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
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
Find the area of the region \[ R = \{(x, y) : xy \le 27,\; 1 \le y \le x^2 \}. \]
An object of uniform density rolls up the curved path with the initial velocity $v_o$ as shown in the figure. If the maximum height attained by an object is $\frac{7v_o^2}{10 g}$ ($g=$ acceleration due to gravity), the object is a _______

A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth ($R_e$). The increase in potential energy will be ____ ($g$ is acceleration due to gravity at the surface of earth)
Various trigonometric identities are as follows:
Cosecant and Secant are even functions, all the others are odd.
T-Ratios of (2x)
sin2x = 2sin x cos x
cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x)
sin 3x = 3sinx – 4sin3x
cos 3x = 4cos3x – 3cosx