Mathematics : Inverse Trigonometric Functions

Prove that  ${\tan }^{- 1} \left( \frac{ \cos \mathrm{x}}{ 1 + \sin \mathrm{x}} \right) = \frac{\mathrm{\pi }}{4} - \frac{\mathrm{\pi }}{2}, \mathrm{x} \in \left( - \frac{\mathrm{\pi }}{2}, \frac{\mathrm{\pi }}{2} \right)$

Prove that   ${\sin }^{- 1} \left( \frac{8}{17} \right) + {\sin }^{- 1} \left( \frac{3}{5} \right) = {\cos }^{- 1} \left( \frac{36}{85} \right).$

$\mathrm{In} ∆\mathrm{ABC} \mathrm{if} \mathrm{\theta } \mathrm{is} \mathrm{any} \mathrm{angle}, \mathrm{then} \mathrm{bcos}\left(\mathrm{C} + \mathrm{\theta }\right) + \mathrm{ccos}\left(\mathrm{B} - \mathrm{\theta }\right) = ?$

• $\mathrm{acot}\left(\mathrm{\theta }\right)$

• $\mathrm{acos}\left(\mathrm{\theta }\right)$

• $\mathrm{atan}\left(\mathrm{\theta }\right)$

• $\mathrm{asin}\left(\mathrm{\theta }\right)$

$$\begin{gathered} \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{solutions} \mathrm{of} \mathrm{the} \mathrm{trigonometric} \mathrm{equation} \\ 1 + \cos \left(\mathrm{x}\right) . \cos \left(5\mathrm{x}\right) = {\sin }^2\left(\mathrm{x}\right) \mathrm{in} [0, 2\mathrm{\pi }] \mathrm{is} \end{gathered}$$

• 8

• 12

• 10

• 6

A value of θ for which  is purely imaginary is

• π/3

• π/6

• 
• 

Consider f(x) = tan^-1 . A normal to y = f (x) at x = π/6 also passes through the point

• (0,0)

• (0, 2π/3)

• (π/6 ,0)

• (π/6 ,0)

If 0≤x<2π, then the number of real values of x, which satisfy the equation cosx+cos2x+cos3x+cos4x=0, is :

• 3

• 5

• 7

• 7

• 4

• 3

• 2

• 2

Let  where .Then, a value of y is

If y  = sec (tan^-1 x), then dy/dx at x = 1 is equal to 

• 
• 

• 1

• 1