List of top Mathematics Questions on Trigonometry asked in BITSAT

If \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \), where \( A + B + C = \pi \), then what is the value of \( \tan A \tan B + \tan B \tan C + \tan C \tan A \)?

Given that \( f(x) = \sin x + \cos x \) and \( g(x) = x^2 - 1 \), find the conditions under which \( g(f(x)) \) is invertible.

If \( \cos \cot^{-1} \left( \frac{1}{2} \right) = \cot (\cos^{-1} x) \), then the value of \( x \) is:

If \( \cot(\cos^{-1} x) = \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right) \), then:

ABC is a triangular park with \( AB = AC = 100 \) m. A TV tower stands at the midpoint of \( BC \). The angles of elevation of the top of the tower at \( A, B, C \) are \( 45^\circ, 60^\circ, 60^\circ \) respectively. The height of the tower is:

From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be \(30^\circ\) and \(45^\circ\). The height of the tower is:

Number of solutions of equations \(\sin(9\theta) = \sin(\theta)\) in the interval \([0,2\pi]\) is:

The sum of all values of \(x\) in \([0, 2\pi]\), for which \(x + \sin(2x) + \sin(3x) + \sin(4x) = 0\) is equal to:

Let \(A\), \(B\) and \(C\) are the angles of a triangle and \(\tan \frac{A}{2} = 1/3\), \(\tan \frac{B}{2} = \frac{2}{3}\). Then, \(\tan \frac{C}{2}\) is equal to:

If \[ y = \tan^{-1} \left( \frac{1}{x^2 + x + 1} \right) + \tan^{-1} \left( \frac{1}{x^2 + 3x + 3} \right) + \tan^{-1} \left( \frac{1}{x^2 + 5x + 7} \right) + \cdots { (to n terms)} \], then \(\frac{dy}{dx}\) is:

If \( \tan^{-1}\left(\frac{1}{1+1\cdot2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+n(n+1)}\right) = \tan^{-1}(x) \), then \( x \) is equal to:

In a △ ABC, if (cos A)/(a)=(cos B)/(b)=(cos C)/(c), and the side a=2, then area of the triangle is:

If sin⁻1((2a)/(1+a²)) -cos⁻1((1-b²)/(1+b²)) =tan⁻1((2x)/(1-x²)), then what is the value of x?

If A and B are positive acute angles satisfying 3cos²A+2cos²B=4 and (3sin A)/(sin B)=(2cos B)/(cos A), then the value of A+2B is equal to:

The general solution of the equation sin 2x + 2sin x + 2cos x +1 =0 is:

If \[ \cos^{-1} x - \frac{\cos^{-1} y}{2} = \alpha, \] then \[ 4x^2 - 4xy \cos \alpha + y^2 \] is equal to: