MCA NIMCET Inverse Trigonometrical Function Previous Year Question Paper and Solution with PDF

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MCA NIMCET Inverse Trigonometrical Function PYQ

Qus : 1 MCA NIMCET PYQ

The value of

2tan^-1[cosec(tan^-1x) - tan(cot^-1x)]

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 2 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 3 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

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

$\tan\theta = \dfrac{(n+1)-1}{1+(n+1)(1)} = \dfrac{n}{n+2}$

Qus : 4 MCA NIMCET PYQ

If $\sin^{-1}x + \cos^{-1}(1-x) = \sin^{-1}(1-x)$ then $x$ satisfies the equation:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Simplifying the given inverse–trigonometric equation gives: $2x^2 - 3x = 0$

Qus : 5 MCA NIMCET PYQ

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 $\pi$.

Qus : 6 MCA NIMCET PYQ

If $\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}$, then $x$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

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 $x = -\dfrac{1}{2}$

Qus : 7 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

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.

Qus : 8 MCA NIMCET PYQ

If $f(x)=tan^{-1}\left [ \frac{sinx}{1+cosx} \right ]$ , then what is the first derivative of $f(x)$?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 9 MCA NIMCET PYQ

The correct expression for $cos^{-1} (-x)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

You should learn it as an important formula.

Qus : 10 MCA NIMCET PYQ

The value of $\cot \Bigg{(}{cosec}^{-1}\frac{5}{3}+{\tan }^{-1}\frac{2}{3}\Bigg{)}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 11 MCA NIMCET PYQ

Solutions of the equation ${\tan }^{-1}\sqrt[]{{x}^2+x}+{\sin }^{-1}\sqrt[]{{x}^2+x+1}=\frac{\pi}{2}$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 12 MCA NIMCET PYQ

If $cos^{-1} \frac{x}{2}+cos^{-1} \frac{y}{3}=\phi$, then $9x^2-12xy cos\phi+4y^2$ is

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Solution

Qus : 13 MCA NIMCET PYQ

, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 14 MCA NIMCET PYQ

The value of A that satisfies the equation asinA + bcosA = c is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Let

$R = \sqrt{a^2 + b^2}$

Choose $\theta$ such that

$\cos \theta = \frac{b}{R}, \quad \sin \theta = \frac{a}{R}$

Then

$a\sin A + b\cos A = R(\sin A \sin \theta + \cos A \cos \theta) = R\cos(A - \theta)$

Given that

$a\sin A + b\cos A = c,$

we get

$R\cos(A - \theta) = c \quad \Rightarrow \quad \cos(A - \theta) = \frac{c}{\sqrt{a^2 + b^2}}$

Hence,

$A - \theta = \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right) + 2\pi n, \quad n \in \mathbb{Z}$

and since $\tan \theta = \dfrac{a}{b}$, we have $\theta = \tan^{-1}\!\left(\dfrac{a}{b}\right)$

Therefore,

$\boxed{A = \tan^{-1}\!\left(\frac{a}{b}\right) \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right)}$

Qus : 15 MCA NIMCET PYQ

Find the principal value of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 16 MCA NIMCET PYQ

If $\sin^{-1}\frac{2a}{1+a^2} - \cos^{-1}\frac{1-b^2}{1+b^2} = \tan^{-1}\frac{2x}{1-x^2}$ then $x$ equals:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

$\sin^{-1}\frac{2a}{1+a^2} = 2\tan^{-1} a$ $\cos^{-1}\frac{1-b^2}{1+b^2} = 2\tan^{-1} b$ So equation becomes: $2\tan^{-1} a - 2\tan^{-1} b = \tan^{-1}\frac{2x}{1-x^2}$ Use identity: $\tan^{-1}u - \tan^{-1}v = \tan^{-1}\frac{u-v}{1+uv}$ Thus $2\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1}\frac{2x}{1-x^2}$ This gives: $x=\frac{a-b}{1+ab}$

Qus : 17 MCA NIMCET PYQ

An engineer standing at a point $P$ wishes to determine the width of a new rectangular pond. She finds the distance to the western-most point $A$ of the pond from $P$ to be $60$ m, while the distance to the northern-most point $B$ of the pond from $P$ is $80$ m. If the angle between the two lines of sight at $P$ is $60^\circ$, then the width $AB$ in metres of the pond is, where $AB$ is not parallel to line of North-South:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

Given,

$PA=60$

$PB=80$

$\angle APB=60^\circ$

Using cosine rule in triangle $APB$,

$AB^2=PA^2+PB^2-2(PA)(PB)\cos 60^\circ$

$AB^2=60^2+80^2-2(60)(80)\cdot \frac{1}{2}$

$AB^2=3600+6400-4800$

$AB^2=5200$

$AB=\sqrt{5200}$

$AB=\sqrt{400\cdot 13}$

$AB=20\sqrt{13}$

Qus : 18 MCA NIMCET PYQ

The number of solutions of the equation $\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

Given,

$\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}$

Taking tangent on both sides,

$\tan\left(\tan^{-1}(3x)+\tan^{-1}(2x)\right)=\tan\frac{\pi}{4}$

Using,

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

we get,

$\frac{3x+2x}{1-(3x)(2x)}=1$

$\frac{5x}{1-6x^2}=1$

So,

$5x=1-6x^2$

$6x^2+5x-1=0$

Factorizing,

$6x^2+6x-x-1=0$

$6x(x+1)-1(x+1)=0$

$(x+1)(6x-1)=0$

So,

$x=-1$ or $x=\frac{1}{6}$

Now check both values.

For $x=\frac{1}{6}$,

$\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{4}$

So, $x=\frac{1}{6}$ is valid.

For $x=-1$,

$\tan^{-1}(-3)+\tan^{-1}(-2)$ is negative, so it cannot be equal to $\frac{\pi}{4}$.

Therefore, only one solution exists.

Qus : 19 MCA NIMCET PYQ

The value of $\cot\!\left(cosec^{-1}\dfrac{5}{3} + \tan^{-1}\dfrac{2}{3}\right)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 20 MCA NIMCET PYQ

The value of $sin^{-1}\frac{1}{\sqrt{2}}+sin^{-1}\frac{\sqrt{2}-\sqrt{1}}{\sqrt{6}}+sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+...$ to infinity , is equal to

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Solution

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