NIMCET 2027 Inverse Trigonometrical Function Previous Year Questions (PYQs) – Page 1 of 2
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If n1 and n2 are the number of real valued solutions x = | sin–1 x | & x = sin (x) respectively, then the value of n2– n1 is
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Solution
If
$ \theta = \tan^{-1}\dfrac{1}{1+2} + \tan^{-1}\dfrac{1}{1+2\cdot3} + \tan^{-1}\dfrac{1}{1+3\cdot4} + \ldots + \tan^{-1}\dfrac{1}{1+n(n+1)} $,
then $\tan\theta$ is equal to:
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Solution
We use identity
$\tan^{-1}a - \tan^{-1}b = \tan^{-1}\dfrac{a-b}{1+ab}$
Here each term satisfies
$\tan^{-1}\dfrac{1}{1+k(k+1)} = \tan^{-1}(k+1) - \tan^{-1}(k)$
So the series telescopes:
$\theta = \tan^{-1}(n+1) - \tan^{-1}(1)$
Thus
$\tan\theta = \dfrac{(n+1)-1}{1+(n+1)(1)} = \dfrac{n}{n+2}$
If
$\sin^{-1}x + \cos^{-1}(1-x) = \sin^{-1}(1-x)$
then $x$ satisfies the equation:
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Solution
Simplifying the given inverse–trigonometric equation gives:
$2x^2 - 3x = 0$
The value of
$\cot^{-1}(21) + \cot^{-1}(13) + \cot^{-1}(-8)$
is:
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Solution
Use identity for inverse cotangent:
$\cot^{-1}a + \cot^{-1}b = \cot^{-1}\left(\dfrac{ab - 1}{a + b}\right)$
Apply to the first two terms:
$\cot^{-1}(21) + \cot^{-1}(13)$
$= \cot^{-1}\left(\dfrac{21 \cdot 13 - 1}{21 + 13}\right)$
$= \cot^{-1}\left(\dfrac{273 - 1}{34}\right)$
$= \cot^{-1}\left(\dfrac{272}{34}\right)$
$= \cot^{-1}(8)$
Now the expression becomes:
$\cot^{-1}(8) + \cot^{-1}(-8)$
But we know:
$\cot^{-1}(x) + \cot^{-1}(-x) = \pi$
So,
$\cot^{-1}(8) + \cot^{-1}(-8) = \pi$
$\therefore$ The value is $\pi$.
If $\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}$, then $x$ is:
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Solution
Use formula:
$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\dfrac{a+b}{1-ab}\right)$
So:
$\tan^{-1}\left(\dfrac{2x+3x}{1 - 6x^2}\right) = \dfrac{\pi}{4}$
Thus:
$\dfrac{5x}{1 - 6x^2} = 1$
Solve:
$5x = 1 - 6x^2$
$6x^2 + 5x - 1 = 0$
Quadratic gives roots:
$x = \dfrac{1}{3}$ or $x = -\dfrac{1}{2}$
Number of solutions for
$\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \dfrac{\pi}{2}$ is:
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Solution
Both square-root expressions must be real.
Domain 1:
$x(x+1) \ge 0 \Rightarrow x \le -1 \text{ or } x \ge 0$
Domain 2:
$x^2 + x + 1 > 0$ for all $x$ (always positive)
Equation transforms into identity impossible to satisfy over real domain.
Therefore no real solution.
If $f(x)=tan^{-1}\left [ \frac{sinx}{1+cosx} \right ]$ , then what is the first derivative of $f(x)$?
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Solution
The correct expression for $cos^{-1} (-x)$ is
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Solution
You should learn it as an important formula.
The value of $\cot \Bigg{(}{cosec}^{-1}\frac{5}{3}+{\tan }^{-1}\frac{2}{3}\Bigg{)}$ is
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Solution
NIMCET 2027
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NIMCET 2027
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and More.
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