List of top Mathematics Questions on Trigonometry asked in MHT CET

If \( \sin x \cos x = \frac{1}{4} \), then the general solution is:

In \(\triangle ABC\), if \(\angle C = \frac{2\pi}{3}\), then the value of \(\cos^2 A + \cos^2 B - \cos A \cos B\) is:

If $\sin x \cos x = \frac{1}{4}$, then the general solution is:

In a $\triangle ABC$, $a = 1$, $b = \sqrt{3}$ and $\angle C = \dfrac{\pi}{6}$. Then the measure of the third side $c =$}

If the area of triangle $ABC$ is $b^2 - (c-a)^2$, then $\tan B =$

In a triangle $\triangle ABC$, if $a$, $b$, and $c$ are in arithmetic progression, then $\cos A + 2\cos B + \cos C =$

If $\sin x \cos x = \frac{1}{4}$, then the general solution is:

What is the value of \( \sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right) \)?

Find the value of \( \tan(105^\circ) \) using compound angle identities.

The value of (\tan [2 \tan^{-1} \frac{1}{5} - \frac{\pi}{4}]) is

In a triangle (ABC), with usual notations, the sides (a, b, c) are such that they are roots of the equation (x^3 - 11x^2 + 38x - 40 = 0) then (\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = )

The general solutions of the equation (\tan^2 \theta + \sec 2\theta = 1) are

With usual notation, in a triangle ABC $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$, then the value of $\cos B$ is equal to

If $( \sin(\alpha + \beta) = 1, \sin(\alpha - \beta) = \frac{1}{2}, \alpha, \beta \in [0, \pi/2] ), then ( \tan(\alpha + 2\beta) \cdot \tan(2\alpha + \beta) = ) $

In \(\triangle ABC\), with usual notations, if \(a^4 + b^4 + c^4 - 2a^2 c^2 - 2c^2 b^2 = 0\), then \(\angle C = \dots\)

The general solutions of the equation (\tan^2 \theta + \sec 2\theta = 1) are

In a triangle (ABC), with usual notations, the sides (a, b, c) are such that they are roots of the equation (x^3 - 11x^2 + 38x - 40 = 0) then (\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = )

The value of (\tan [2 \tan^{-1} \frac{1}{5} - \frac{\pi}{4}]) is

With usual notation, in a triangle ABC $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$, then the value of $\cos B$ is equal to

If $( \sin(\alpha + \beta) = 1, \sin(\alpha - \beta) = \frac{1}{2}, \alpha, \beta \in [0, \pi/2] ), then ( \tan(\alpha + 2\beta) \cdot \tan(2\alpha + \beta) = ) $

The general solutions of the equation (\tan^2 \theta + \sec 2\theta = 1) are

In a triangle (ABC), with usual notations, the sides (a, b, c) are such that they are roots of the equation (x^3 - 11x^2 + 38x - 40 = 0) then (\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = )

The value of (\tan [2 \tan^{-1} \frac{1}{5} - \frac{\pi}{4}]) is

With usual notation, in a triangle ABC $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$, then the value of $\cos B$ is equal to

If $( \sin(\alpha + \beta) = 1, \sin(\alpha - \beta) = \frac{1}{2}, \alpha, \beta \in [0, \pi/2] ), then ( \tan(\alpha + 2\beta) \cdot \tan(2\alpha + \beta) = ) $