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

Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

NIMCET
Previous Year Questions

NIMCET
Previous Year Questions
NIMCET

NIMCET Previous Year Questions (PYQs)

NIMCET Inverse Trigonometrical Function PYQ

Qus : 1 NIMCET PYQ

The value of

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2020 PYQ

Solution

Qus : 2 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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Qus : 3 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:

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ 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 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

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

Qus : 5 NIMCET PYQ

The value of $\cot^{-1}(21) + \cot^{-1}(13) + \cot^{-1}(-8)$ is:

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

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 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ 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 NIMCET PYQ

Number of solutions for $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \dfrac{\pi}{2}$ is:

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ 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 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution

Qus : 9 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

You should learn it as an important formula.

Qus : 10 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 11 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 12 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 13 NIMCET PYQ

, then

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

Qus : 14 NIMCET PYQ

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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ 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 NIMCET PYQ

Find the principal value of

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

Qus : 16 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:

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ 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 NIMCET PYQ

The value of

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 18 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

1
2
3
4
Go to Discussion

NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

NIMCET

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

NIMCET

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

Please provide more information - at least 10 characters

Feedback

Info.

Information

Aspire's Library

NIMCET Previous Year Questions (PYQs)

NIMCET Inverse Trigonometrical Function PYQ

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

NIMCET

NIMCET

Info.

Confirm Delete

Edit