NIMCET Trigonometry Previous Year Questions (PYQs) – Page 1 of 8
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The expression $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$ can be written as
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Solution
Angle of elevation of the top of the tower from 3
points (collinear) A, B and C on a road leading to the
foot of the tower are 30°, 45° and 60°, respectively.
The ratio of AB and BC is
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Solution
According to the given information, the figure should be as follows. 
Let the height of tower = h
If $3 sin x + 4 cos x = 5$, then $6tan\frac{x}{2}-9tan^2\frac{x}{2}$
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Solution
Largest value of $cos^2\theta -6sin\theta cos\theta+3sin^2\theta+2 $ is
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Solution
Number of point of which f(x) is not differentiable $f(x)=|cosx|+3$ in $[-\pi, \pi]$
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Solution
Points of Non-Differentiability of \( f(x) = |\cos x| + 3 \)
Step 1: \( \cos x \) is differentiable everywhere, but \( |\cos x| \) is not differentiable where \( \cos x = 0 \).
Step 2: In the interval \( [-\pi, \pi] \), we have:
\[ \cos x = 0 \Rightarrow x = -\frac{\pi}{2},\ \frac{\pi}{2} \]
So \( f(x) = |\cos x| + 3 \) is not differentiable at these two points due to sharp turns.
✅ Final Answer: \( \boxed{2 \text{ points}} \)
A square with side $a$ is revolved about its centre through $45^\circ$.
What is the area common to both the squares?
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Solution
The overlapping area of a square rotated by $45^\circ$ is:
Common area $= 2(\sqrt{2}-1)a^2$
The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for:
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Solution
Simplify:
$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$
$= 1 - \dfrac{1}{2}\sin^2(2x)$
So equation becomes:
$1 - \dfrac{1}{2}\sin^2(2x) + \sin(2x) + \alpha = 0$
Let $t = \sin(2x)$, where $t \in [-1,1]$
Expression becomes:
$1 + \alpha + t - \dfrac{t^2}{2} = 0$
Multiply by 2:
$t^2 - 2t - 2(1+\alpha) = 0$
For real $t$ in $[-1,1]$, discriminant must allow roots inside interval.
Solving gives the range:
${-\dfrac{3}{2} \le \alpha \le \dfrac{1}{2}}$
If $\sin^2 x = 1 - \sin x$, then $\cos^4 x + \cos^2 x$ is equal to:
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Solution
From the given equation:
$\sin^2 x = 1 - \sin x$
So:
$\cos^2 x = 1 - (1 - \sin x)$
$\cos^2 x = \sin x$
$\cos^4x+\cos^2x$
=$\sin^2 x + \sin x$
$ = 1$
If $A = \cos^2\theta + \sin^4\theta$, then for all values of $\theta$:
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Solution
$A = \cos^2\theta + \sin^4\theta$
Let $\sin^2\theta = t$, $0 \le t \le 1$
Then
$A = (1 - t) + t^2 = t^2 - t + 1$
This is a quadratic in $t$.
Minimum value at $t = \dfrac{1}{2}$:
$A_{\min} = \left(\dfrac{1}{2}\right)^2 - \dfrac{1}{2} + 1 = \dfrac{3}{4}$
Maximum value at endpoints $t=0$ or $t=1$:
$A_{\max} = 1$
Thus:
${\dfrac{3}{4} \le A \le 1}$
The value of
$ \sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ $
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Solution
Expression is $\sin(30^\circ + 45^\circ)$
$ = \sin 75^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{\sqrt{3}+1}{2\sqrt{2}} $
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