List of top Questions asked in TS EAMCET- 2025

If $2 \sin\theta+3 \cos\theta=2$ and $\theta \neq (2n+1)\frac{\pi}{2}$ then $\sin\theta+\cos\theta=$

$4 \cos\frac{70}{2}\cos\frac{30}{2} - \sin 50 =$

\( \alpha, \beta \) are the roots of the equation \( \sin^2 x + b\sin x + c = 0 \). If \( \alpha + \beta = \frac{\pi}{2} \) then \( b^2 - 1 = \)

$\frac{\cos 15^\circ \cos^2 22\frac{1}{2}^\circ - \sin 75^\circ \sin^2 52\frac{1}{2}^\circ}{\cos^2 15^\circ - \cos^2 75^\circ} =$

\( \frac{\sin 1^\circ + \sin 2^\circ + \dots + \sin 89^\circ}{2(\cos 1^\circ + \cos 2^\circ + \dots + \cos 44^\circ) + 1} = \)

Number of solutions of the equation \( \tan^2 x + 3\cot^2 x = 2\sec^2 x \) lying in the interval \( [0, 2\pi] \) is

The number of values of x satisfying the equation $\text{Tan}^{-1}(x+\frac{\sqrt{2}}{x}) + \text{Tan}^{-1}(x-\frac{\sqrt{2}}{x}) = \text{Tan}^{-1}(x)$ is

If \( 0 \le A, B \le \frac{\pi}{4} \) and \( \cot A + \cot B + \tan A + \tan B = \cot A \cot B - \tan A \tan B \) then \( \sin(A+B) = \)

If the extreme values of the function \( f(x) = (2\sqrt{6}+1)\cos x + (2\sqrt{2}-\sqrt{3})\sin x - 6 \) are m and M, then \( \sqrt{|M^2-m^2|} = \)

In a triangle ABC, \( (r_2 + r_3)\sec^2\left(\frac{A}{2}\right) = \)
(Note: Based on the answer key and standard identities, the function is interpreted as \( \sec^2 \)).

If $\tan(\frac{\pi}{4}+\frac{\alpha}{2}) = \tan^3(\frac{\pi}{4}+\frac{\beta}{2})$, then $\frac{3+\sin^2\beta}{1+3\sin^2\beta}=$

If $x \in (-\pi,\pi)$ then the number of solutions of the equation $2 \sin x \sin 3x \sin 5x + \sin 5x \cos 4x = 0$ is

If $\sin A = -\frac{24}{25}$, $\cos B = \frac{15}{17}$, A does not belong to 4\textsuperscript{th} quadrant and B does not belong to 1\textsuperscript{st} quadrant then $(A + B)$ lies in the quadrant

If \( \sin A = -\frac{60}{61} \), \( \cot B = -\frac{40}{9} \) and neither A nor B is in \( 4^{\text{th}} \) quadrant then \( 6\cot A + 4\sec B = \)

The number of real solutions of $\tan^{-1}x + \tan^{-1}(2x) = \frac{\pi}{4}$ is

If a particle is moving in a straight line so that after \(t\) seconds its distance \(S\) (in cms) from a fixed point on the line is given by \(S = f(t) = t^3 - 5t^2 + 8t\) then the acceleration of the particle at \(t=5\) sec is (in cm/sec\(^2\))

\( \int \frac{x+\cos x}{1-\sin x} dx = \)

If \( \int x^3 \sin(3x) dx = \frac{1}{27} [f(x)\cos(3x) + g(x)\sin(3x)] + c \) then f(1)+g(1)=

If \( \cos x \frac{dy}{dx} = y \sin x - 1, x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} \) is the differential equation corresponding to the curve \( y = f(x) \) and \( f(0)=1 \) then \( f(x)= \)

\(\int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx =\)

If \(\int \frac{dx}{(x^2+9)\sqrt{x^2+16}} = \frac{1}{3\sqrt{7}} \tan^{-1} \left( K \frac{x}{\sqrt{16+x^2}} \right) + c\), then \(K =\)

\( \int_{-2 \pi}^{2 \pi} \sin ^{4} x \cos ^{6} x d x= \)

If the rate of change of the slope of the tangent drawn to the curve \( y = x^3 - 2x^2 + 3x - 2 \) at the point (2,4) is k times the rate of change of its abscissa, then k =

A rod of length 41 m with an end A on the floor and another end B on the wall perpendicular to the floor is sliding away horizontally from the wall at the rate of 3 ft/min. When the end B is at the height of 9 ft from the floor, then the rate at which the area of the triangle formed by the rod with wall and floor changes at that instant is (in ft/min)