The answer is \(8\pi^2-40\pi+50\).
Prove that:
\( \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1}\left( \frac{1 - x}{1 + x} \right), \quad x \in [0, 1] \)
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
The inverse trigonometric functions are also called arcus functions or anti trigonometric functions. These are the inverse functions of the trigonometric functions with suitably restricted domains. Specifically, they are the inverse functions of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle’s trigonometric ratios. Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.
Considering the domain and range of the inverse functions, following formulas are important to be noted:
Also, the following formulas are defined for inverse trigonometric functions.
cosec−1(cosec y) = y if -π/2 ≤ y ≤ π/2, y ≠ 0