datavorous/entrance-exam-dataset huggingface dataset:
If $\vec{r}_{1}=\lambda \hat{i}+2 \hat{j}+\hat{k}, \vec{r}_{2}=\hat{i}+(2-\lambda) \hat{j}+2 \hat{k}$ are such that
$\left|\vec{r}_{1}\right|>\left|\vec{r}_{2}\right|$, then $\lambda$ satisfies which one of the following?

Options:
A $\lambda=0$ only
B $\quad \lambda=1$
C $\lambda < 1$
D $\lambda \geq 1$

The correct answer is:
$\lambda \geq 1$

Solution:
Given, $\overrightarrow{\mathbf{r}}_{1}=\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\begin{aligned} & \text { and } \overrightarrow{\mathbf{r}}_{1}=\hat{\mathbf{i}}+(2-\lambda) \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \therefore\left|\overrightarrow{\mathbf{r}}_{1}\right|>\left|\overrightarrow{\mathbf{r}}_{2}\right| \\ \Rightarrow & \sqrt{\lambda^{2}+(2)^{2}+(1)^{2}}>\sqrt{(1)^{2}+(2-\lambda)^{2}+(2)^{2}} \\ \Rightarrow & \lambda^{2}+4+1>1+4+\lambda^{2}-4 \lambda+4 \\ \Rightarrow & 5>9-4 \lambda \\ \Rightarrow & 4 \lambda>4 \\ \Rightarrow & \lambda>1 \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
If $\overrightarrow{\mathrm{r}}_{1}, \overrightarrow{\mathrm{r}_{2}}, \overrightarrow{\mathrm{r}_{3}}$ are the position vectors of three collinear points
and scalars $\mathrm{m}$ and $\mathrm{n}$ exist such that $\overrightarrow{\mathrm{r}}_{3}=\mathrm{m} \overrightarrow{\mathrm{r}}_{1}+\mathrm{n} \overrightarrow{\mathrm{r}}_{2}$, then
what is the value of $(\mathrm{m}+\mathrm{n})$ ?

Options:
A 0
B 1
C -1
D 2

The correct answer is:
1

Solution:
Since $\overrightarrow{\mathrm{r}}_{1}, \overrightarrow{\mathrm{r}}_{2}$ and $\overrightarrow{\mathrm{r}}_{3}$ are the position vector of three collinear points. Thus $\vec{r}_{3}$ is the position vector of the point which divides the joining of points whose
position vectors are $\overrightarrow{\mathrm{r}}_{1}$ and $\overrightarrow{\mathrm{r}}_{2}$ in the ratio $\mathrm{m}: \mathrm{n}$.
So, $\vec{r}_{3}=\frac{\mathrm{mr}_{1}+\mathrm{n} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}+\mathrm{n}}$
But as given, $\overrightarrow{\mathrm{r}}_{3}=\mathrm{mr}_{1}+\overrightarrow{\mathrm{n}} \overrightarrow{\mathrm{r}}_{2}$
So, $\frac{\overline{\mathrm{mr}_{1}}+\overline{\mathrm{nr}_{2}}}{\mathrm{~m}+\mathrm{n}}=\mathrm{mr}_{1}+\mathrm{nr}_{2}$
$\Rightarrow \mathrm{m}+\mathrm{n}=1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 2)
If $\mathrm{r}^{1 / 3}+\frac{1}{r^{1 / 3}}=3$ for a real number $\mathrm{r} \neq 0$, then what is
$\mathrm{r}+1$ equal to?

Options:
A 27
B 36
C 9
D 18

The correct answer is:
18

Solution:
Given equation is
$\mathrm{r}^{1 / 3}+\frac{1}{\mathrm{r}^{1 / 3}}=3$
Cubing both sides, we get
$\left(r^{1 / 3}+\frac{1}{r^{1 / 3}}\right)^{3}=3^{3}\left[(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]$
$\Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+3\left(\mathrm{r}^{1 / 3}+\frac{1}{\mathrm{r}^{1 / 3}}\right)=27$
$\Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+3.3=27 \Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+27-9=18$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 2)
If $r$ and $s$ are roots of $x^{2}+p x+q=0$, then what is the value $\operatorname{of}\left(1 / r^{2}\right)+\left(1 / s^{2}\right) ?$

Options:
A $p^{2}-4 q$
B $\frac{p^{2}-4 q}{2}$
C $\frac{p^{2}-4 q}{q^{2}}$
D $\frac{p^{2}-2 q}{q^{2}}$

The correct answer is:
$\frac{p^{2}-2 q}{q^{2}}$

Solution:
Since $r$ and $s$ are the roots of $x^{2}+p x+q=0$.
Then, $r+s=-p$ and $r s=q$ Now, $\frac{1}{r^{2}}+\frac{1}{s^{2}}=\frac{r^{2}+s^{2}}{(r s)^{2}}=\frac{(r+s)^{2}-2 r s}{(r s)^{2}}$
$=\frac{(-p)^{2}-2 q}{q^{2}}=\frac{p^{2}-2 q}{q^{2}}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 2)
If $\mathrm{R}$ be the set of points inside a rectangle of sides a and $b$ $(\mathrm{a}, \mathrm{b}>1)$ with two sides along the positive direction of $\mathrm{X}$-axis and $\mathrm{Y}$-axis. Then,

Options:
A
$\mathrm{R}=\{(x, y): 0 \leq x \leq a, 0 \leq y \leq b\}$

B
$\mathrm{R}=\{(x, y): 0 \leq x < a, 0 \leq y \leq b\}$
C
$\mathrm{R}=\{(x, y): 0 \leq x \leq a, 0 < y < b\}$
D
$\mathrm{R}=\{(x, y): 0 < x < a, 0 < y < b\}$

The correct answer is:
$\mathrm{R}=\{(x, y): 0 < x < a, 0 < y < b\}$

Solution:
Here, $R$ be the set of points inside a rectangle of sides a and b. So, $\mathrm{R}=\{(x, y): 0 < x < a$ and $0 < y < b\}$

Topic: Mathematics, Sets and Relations
If $\overrightarrow{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$, then what is $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ equal to?

Options:
A $\mathrm{x}$
B $\mathrm{x}+\mathrm{y}$
C $-(x+y+z)$
D $(x+y+z)$

The correct answer is:
$(x+y+z)$

Solution:
$\quad(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=x+y+z$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)
If $\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0$, where $z=x+$ iy is a complex number, then which one of the following is correct?

Options:
A $z=1+i$
B $|z|=2$
C $z=1-i$
D $|z|=1$

The correct answer is:
$|z|=1$

Solution:
$\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}$
$\frac{z-1}{z+1}=\frac{x^{2}+y^{2}-1+2 i y}{x^{2}+y^{2}+2 x+1}$
$\Rightarrow \operatorname{Re}\left(\frac{z-1}{z+1}\right)=\frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+2 x+1}=0$
$\Rightarrow x^{2}+y^{2}-1=0$
$\Rightarrow x^{2}+y^{2}=1$
Also, $z \bar{z}=x^{2}+y^{2}=1$
and $z \bar{z}=|z|^{2}$
$\Rightarrow|z|^{2}=1$
$\Rightarrow|z|=1$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 2)
If roots of an equation $a x^{2}+b x+c=0$ are positive, then which one of the following is correct?

Options:
A Signs of a and $c$ should be like
B Signs of $\mathrm{b}$ and $\mathrm{c}$ should be like
C Signs of a and b should be like
D None of the above

The correct answer is:
Signs of a and $c$ should be like

Solution:
For roots of an equation $a x^{2}+b x+c=0$ to be positive sign of a and c should be like.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 2)
If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$, then what is $\frac{\tan x}{\tan y}$ equal to?

Options:
A $\frac{\mathrm{a}}{\mathrm{b}}$
B $\frac{\mathrm{b}}{\mathrm{a}}$
C $\frac{a+b}{a-b}$
D $\frac{a-b}{a+b}$

The correct answer is:
$\frac{\mathrm{a}}{\mathrm{b}}$

Solution:
\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}
\end{array}\\
\text { Applying componendo and dividendo, we get }\\
\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{a+b+a-b}{a+b-a+b}
\end{array}
$$
$\Rightarrow \frac{2 \sin \left(\frac{x+y+x-y}{2}\right) \cos \left(\frac{x+y-x+y}{2}\right)}{2 \cos \left(\frac{x+y+x-y}{2}\right) \sin \left(\frac{x+y-x+y}{2}\right)}=\frac{2 a}{2 b}$
$\Rightarrow \frac{\sin x \cdot \cos y}{\cos x \cdot \sin y}=\frac{\mathrm{a}}{\mathrm{b}}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 1)
If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$, then what is $\frac{\tan x}{\tan y}$ equal to?

Options:
A $\frac{b}{a}$
B $\frac{a}{b}$
C $a \bar{b}$
D 1

The correct answer is:
$\frac{a}{b}$

Solution:
$\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
Applying componendo and dividendo, we get
$\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2 \sin x \cdot \cos y}{2 \cos x \cdot \sin y}=\frac{2 a}{2 b} \Rightarrow \tan x \cdot \cot y=\frac{a}{b}$
$\therefore \frac{\tan x}{\tan y}=\frac{a}{b}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 2)
If $\mathrm{S}_{\mathrm{n}}=\mathrm{nP}+\frac{\mathrm{n}(\mathrm{n}-1) \mathrm{Q}}{2}$, where $\mathrm{S}_{\mathrm{n}}$ denotes the sum of the
first $\mathrm{n}$ terms of an AP, then the common difference is $[2017-I I]$

Options:
A $\mathrm{P}+\mathrm{Q}$
B $2 \mathrm{P}+3 \mathrm{Q}$
C $2 Q$
D $Q$

The correct answer is:
$Q$

Solution:
$\mathrm{S}_{\mathrm{n}}=\mathrm{np}+\frac{\mathrm{n}(\mathrm{n}-1) \mathrm{Q}}{2}$
We know, $\mathrm{T}_{1}=\mathrm{S}_{1}$ and $\mathrm{T}_{2}=\mathrm{S}_{2}-\mathrm{S}_{1}$
Common difference $(\mathrm{d})=\mathrm{T}_{2}-\mathrm{T}_{1}$
$\therefore \mathrm{S}_{1}=(1) \mathrm{P}+\frac{1(1-1) \mathrm{Q}}{2}=\mathrm{P}+0=\mathrm{P}$
$\mathrm{S}_{2}=(2) \mathrm{P}+\frac{2(2-1) \mathrm{Q}}{2}=2 \mathrm{P}+\frac{2 \mathrm{Q}}{2}=2 \mathrm{P}+\mathrm{Q}$
$\therefore \mathrm{T}_{1}=\mathrm{P} ; \mathrm{T}_{2}=2 \mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{P}+\mathrm{Q}$
$\therefore$ Common difference $(\mathrm{d})=\mathrm{T}_{2}-\mathrm{T}_{1}$
$=\mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{Q}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 2)
If $\mathrm{s}=\sqrt{\mathrm{t}^{2}+1}$, then $\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}$ is equal to

Options:
A $\frac{1}{\mathrm{~s}}$
B $\frac{1}{s^{2}}$
C $\frac{1}{s^{3}}$
D $\frac{1}{s^{4}}$

The correct answer is:
$\frac{1}{s^{3}}$

Solution:
$\mathrm{s}=\sqrt{\mathrm{t}^{2}+1}$
$\Rightarrow \frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\mathrm{t}}{\sqrt{\mathrm{t}^{2}+1}}$
$\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=\frac{1}{\sqrt{\left(\mathrm{t}^{2}+1\right)^{3}}}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=\frac{1}{\mathrm{~s}^{3}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2015 (Phase 2)
If S $=\left(\mathrm{x}: \mathrm{x}^{2}+\mathrm{l}=0, \mathrm{x}\right.$ is real $)$, then $\mathrm{S}$ is

Options:
A $\{-1\}$
B $\{0$
C $\{1\}$
D an empty set

The correct answer is:
an empty set

Solution:
$\mathrm{S}=\left\{\mathrm{x}: \mathrm{x}^{2}+1=0, \mathrm{x}: 5\right.$ real\}
$\mathrm{x}^{2}+1=0 \Rightarrow \mathrm{x}^{2}=-1 \Rightarrow \mathrm{x}=\sqrt{-1} \rightarrow$ complex number
No real numbers. So, $\mathrm{S}$ is empty set

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 1)
If $\sec \alpha=\frac{13}{5}$ where $270^{\circ} < \alpha < 360^{\circ}$ then what is $\sin \alpha$ equal
to ?

Options:
A $\frac{5}{13}$
B $\frac{12}{13}$
C $-\frac{12}{13}$
D $-\frac{13}{12}$

The correct answer is:
$-\frac{12}{13}$

Solution:
We have sec $\alpha=\frac{13}{5}$
Since $\frac{3 \pi}{2} < \alpha < 2 \pi$
$\therefore \sin \alpha < 0$
Now, $\sin \alpha=-\sqrt{1-\frac{1}{\sec ^{2} \alpha}}$
$=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{144}{169}}=-\frac{12}{13}$
133. (b) $\tan \left(-585^{\circ}\right)=-\tan 585^{\circ}$
$=-\tan \left[540^{\circ}+45^{\circ}\right]=-\tan \left[3 \pi+\frac{\pi}{4}\right]$
$=-\left[\frac{\tan 3 \pi+\tan \frac{\pi}{4}}{1-\tan 3 \pi \tan \frac{\pi}{4}}\right]=-\left[\frac{0+1}{1-0 \times 1}\right]=-1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 2)
If $\sec A+\tan A=p$, then what is the value of $\sin A$ ? $[2008-\mathrm{II}]$

Options:
A $\frac{p^{2}-1}{p^{2}+1}$
B $\frac{p^{2}+1}{p^{2}-1}$
C 1
D None of these

The correct answer is:
$\frac{p^{2}-1}{p^{2}+1}$

Solution:
$\sec A+\tan A=P$
$\Rightarrow \frac{1}{\cos A}+\frac{\sin A}{\cos A}=p$
$\Rightarrow \frac{1+\sin A}{\cos A}=p$
$\Rightarrow \frac{(1+\sin A)^{2}}{\cos ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{1-\sin ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{(1+\sin A)(1-\sin A)}=p^{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 2)
If $\sec \theta-\operatorname{cosec} \theta=\frac{4}{3}$, then what is $(\sin \theta-\cos \theta)$ equal
to?

Options:
A $-2$ only
B $\frac{1}{2}$ only
C Both $-2$ and $\frac{1}{2}$
D Neither $\frac{1}{2}$ nor $-2$

The correct answer is:
$\frac{1}{2}$ only

Solution:
$\begin{aligned} & \sec \theta-\operatorname{cosec} \theta=\frac{4}{3} \Rightarrow \frac{1}{\cos \theta}-\frac{1}{\sin \theta}=\frac{4}{3} \\ & \Rightarrow \frac{\sin \theta-\cos \theta}{\sin \theta \cos \theta}=\frac{4}{3} \\ & \Rightarrow \frac{(\sin \theta-\cos \theta)^{2}}{(\sin \theta \cos \theta)^{2}}=\frac{16}{9} \\ & \Rightarrow \frac{\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \theta \cos \theta}{\sin ^{2} \theta \cos ^{2} \theta}=\frac{16}{9} \\ & \Rightarrow \frac{1-2 \sin \theta \cos \theta}{\sin ^{2} \theta \cos ^{2} \theta}=\frac{16}{9} \\ & \text { Let } \sin \theta \cos \theta=x \Rightarrow \frac{1-2 x}{x^{2}}=\frac{16}{9} \\ & \Rightarrow 16 x^{2}+18 x-9=0 \\ & \Rightarrow(8 x-3)(2 x+3)=0 \\ & \Rightarrow x=\frac{3}{8}, x=\frac{-3}{2} \\ & \therefore \sin \theta \cos \theta=\frac{3}{8} \\ & \text { from(i), } \sin \theta-\cos \theta=\frac{4}{3} \times \frac{3}{8}=\frac{4}{8}=\frac{1}{2} \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 1)
If sec $(\theta-\alpha)$, sec $\theta$ and sec $(\theta+\alpha)$ are in AP, where cos $\alpha \neq 1$, then what is the value of $\sin ^{2} \theta+\cos \alpha$ ? $[2018-\mathrm{II}]$

Options:
A 0
B 1
C $-1$
D $1 / 2$

The correct answer is:
0

Solution:
$\frac{2}{\cos \theta}=\frac{\cos (\theta+\alpha)+\cos (\theta-\alpha)}{\cos (\theta+\alpha) \cos (\theta-\alpha)}=\frac{2 \cos \theta \cdot \cos \alpha}{\cos ^{2} \theta-\sin ^{2} \alpha}$
$\Rightarrow \cos ^{2} \theta \cos \alpha=\cos ^{2} \theta-\sin ^{2} \alpha$
$\Rightarrow \sin ^{2} \alpha=\cos ^{2} \theta(1-\cos \alpha)$
$\Rightarrow \cos ^{2} \theta=\frac{\sin ^{2} \alpha}{1-\cos \alpha}=1+\cos \alpha$
$\Rightarrow \quad 1-\sin ^{2} \theta=1+\cos \alpha$
$\Rightarrow \sin ^{2} \theta+\cos \alpha=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 2)
If $\int \sec x \operatorname{cosec} x d x=\log |g(x)|+c$, then what is $g(x)$ equal to?

Options:
A $\sin x \cos x$
B $\sec ^{2} x$
C $\tan x$
D $\log |\tan x|$

The correct answer is:
$\tan x$

Solution:
Let $\mathrm{I}=\int \sec x \cdot \operatorname{cosec} x \mathrm{~d} x=\int \frac{2}{2 \sin x \cos x} \mathrm{dx}$
$=2 \int \frac{1}{\sin 2 x} d x=2 \int \frac{1}{\frac{2 \tan x}{1+\tan ^{2} x}}\left[\because \sin 2 x=\frac{2 \tan x}{1+\tan ^{2} x}\right]$
$=\int \frac{\sec ^{2} x}{\tan x} d x$
Let $\tan x=t \Rightarrow \sec ^{2} \mathrm{~d} x=\mathrm{dt}$
So, $I=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log |t|+\mathrm{c}=\log |\tan x|+\mathrm{c}$
$g(x)=\tan x$
But $\int \sec x \operatorname{cosec} x d x=\log |g(x)|+c$
$\therefore \quad g(x)=\tan x$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2006 (Phase 2)
If $\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$,
then what is the value of $x ?$

Options:
A $a / b$
B $a b$
C $b / a$
D $\frac{a-b}{1+a b}$

The correct answer is:
$\frac{a-b}{1+a b}$

Solution:
Given,
$\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\therefore \quad 2 \tan ^{-1} a-2 \tan ^{-1} b=2 \tan ^{-1} x$
$\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} x$
$\Rightarrow \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x$
$\Rightarrow x=\frac{a-b}{1+a b}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2009 (Phase 1)
If $\sin ^{-1}\left(\frac{2 \mathrm{a}}{1+\mathrm{a}^{2}}\right)+\sin ^{-1}\left(\frac{2 \mathrm{~b}}{1+\mathrm{b}^{2}}\right)=2 \tan ^{-1} \mathrm{x}$, then $\mathrm{x}$ is equal to

Options:
A $\frac{a-b}{1+a b}$
B $\frac{a-b}{1-a b}$
C $\frac{2 a b}{1+a b}$
D $\frac{a+b}{1-a b}$

The correct answer is:
$\frac{a+b}{1-a b}$

Solution:
Since a $>0, b>0$ and $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$\therefore$ Given expression is $2 \tan ^{-1} \mathrm{a}+2 \tan ^{-1} \mathrm{~b}=2 \tan ^{-1} \mathrm{x}$
$\Rightarrow 2 \tan ^{-1}\left(\frac{a+b}{1-a b}\right)=2 \tan ^{-1} x$
$\Rightarrow x=\frac{a+b}{1-a b}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2013 (Phase 1)
If $\sin ^{-1} \frac{2 p}{1+p^{2}}-\cos ^{-1} \frac{1-q^{2}}{1+q^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$, then what is $\mathrm{x}$
equal to ?

Options:
A $\frac{p+q}{1+p q}$
B $\frac{p-q}{1+p q}$
C $\frac{p q}{1+p q}$
D $\frac{p+q}{1-p q}$

The correct answer is:
$\frac{p-q}{1+p q}$

Solution:
$\sin ^{-1} \frac{2 p}{1+p^{2}}-\cos ^{-1} \frac{1-q^{2}}{1+q^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$
$\Rightarrow 2 \tan ^{-1} \mathrm{p}-2 \tan ^{-1} \mathrm{q}=2 \tan ^{-1} \mathrm{x}$
$\Rightarrow \tan ^{-1} \mathrm{p}-\tan ^{-1} \mathrm{q}=\tan ^{-1} \mathrm{x}$
$\Rightarrow \tan ^{-1} \frac{p-q}{1+p q}=\tan ^{-1} \mathrm{x}$
$\Rightarrow \mathrm{x}=\frac{p-q}{1+p q}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2019 (Phase 1)
If $\sin ^{-1} 1+\sin ^{-1} \frac{4}{5}=\sin ^{-1} x$, then what is $x$ equal to?

Options:
A $3 / 5$
B $4 / 5$
C 1
D 0

The correct answer is:
$3 / 5$

Solution:
Let $\sin ^{-1}(1)+\sin ^{-1}\left(\frac{4}{5}\right)=\sin ^{-1} \mathrm{x}$
$\operatorname{Let} \sin ^{-1}(1)=\theta \Rightarrow \sin \theta=1 \Rightarrow \cos \theta=0$
and $\sin ^{-1}\left(\frac{4}{5}\right)=\phi \Rightarrow \sin \phi=\left(\frac{4}{5}\right) \Rightarrow \cos \phi=\sqrt{1-\frac{16}{25}}$
$\begin{aligned} &=\sqrt{\frac{9}{25}}=\frac{3}{5} \\ \therefore & \sin ^{-1} \mathrm{x}=\theta+\phi \\ \Rightarrow & \mathrm{x}=\sin (\theta+\phi)=\sin \theta \cos \phi+\cos \theta \sin \phi \\ &=1 \times \frac{3}{5}+0 \times \frac{4}{5} \\ \Rightarrow & \mathrm{x}=\frac{3}{5} \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2012 (Phase 1)
If $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$, then what is the value of $\sin 81^{\circ}$ ?

Options:
A $\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$
B $\frac{\sqrt{3+\sqrt{5}}+\sqrt{5+\sqrt{5}}}{4}$
C $\frac{\sqrt{3-\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$
D $\frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4}$

The correct answer is:
$\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$

Solution:
$$
\begin{aligned}
\because & \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \\
& x^{2}=4^{2}-(\sqrt{5}-1)^{2} \\
\Rightarrow & x=\sqrt{10+2 \sqrt{5}} \\
\Rightarrow & \cos 18^{\circ}=\frac{\sqrt{10+2 \sqrt{5}}}{4} \\
\Rightarrow & 2 \cos ^{2} 9-1=\frac{\sqrt{10+2 \sqrt{5}}}{4} \\
& \cos ^{2} 9=\frac{\sqrt{10+2 \sqrt{5}}+4}{8} \\
\Rightarrow & \sin ^{2} 81=\frac{4+\sqrt{10+2 \sqrt{5}}}{8}
\end{aligned}
$$
After squaring all the options available, we come to a conclusion that option (a) is correct.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
If $\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}+\sin \theta_{4}=4$, then what is the value
of $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}+\cos \theta_{4}=?

Options:
A $\underline{0}$
B 1
C 2
D 4

The correct answer is:
$\underline{0}$

Solution:
\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}+\sin \theta_{4}=4 & \ldots(1) \end{aligned}$
Since max. value of $\sin \theta=1$ We have four terms in LHS of eq. (1). $\Rightarrow$ Every term should be $=1$ $\Rightarrow \sin \theta_{1}=\sin \theta_{2}=\sin \theta_{3}=\sin \theta_{4}=1$
$\Rightarrow \theta_{1}=\theta_{2}=\theta_{3}=\theta_{4}=90^{\circ}$
Now, $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}+\cos \theta_{4}=\left(\cos 90^{\circ}\right) \times 4=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
If $\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}$, then what is the value of $x$ ?

Options:
A 1
B 7
C 13
D 17

The correct answer is:
13

Solution:
$\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} \frac{5}{x}+\cos ^{-1} \frac{\sqrt{x^{2}-144}}{x}=\frac{\pi}{2}$
$\left(\because \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}\right)$
But we know $\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}$
$\therefore \quad \frac{5}{x}=\frac{\sqrt{x^{2}-144}}{x}$
$\Rightarrow 5=\sqrt{x^{2}-144}$
$\Rightarrow 25=x^{2}-144 \Rightarrow x^{2}=169$
$\Rightarrow \quad x=13$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2010 (Phase 1)
If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then what is the value of $x$ ?

Options:
A $\mathrm{x}=-\frac{1}{2}$
B $x=1$
C $\mathrm{x}=\frac{1}{2}$
D $\mathrm{x}=\frac{\sqrt{3}}{2}$

The correct answer is:
$\mathrm{x}=\frac{\sqrt{3}}{2}$

Solution:
As given:
$\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ ...(1)
and we know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ ...(2)
On adding Eqs. (1) and (2) we get
$2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2 \pi}{3}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{3} \Rightarrow x=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2008 (Phase 1)
If $\sin ^{-1} x+\cot ^{-1}(1 / 2)=\pi / 2$, then what is the value of $x$ ?

Options:
A 0
B $1 / \sqrt{5}$
C $2 / \sqrt{5}$
D $\sqrt{3} / 2$

The correct answer is:
$1 / \sqrt{5}$

Solution:
Let $\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$
As we know $\cot ^{-1} x=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)$
$\therefore \sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x+\sin ^{-1}\left(\frac{1}{\sqrt{1+\frac{1}{4}}}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x+\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\frac{\pi}{2}$
Now, $\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}$
$\therefore \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\cos ^{-1} \sqrt{1-\frac{4}{5}}=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)$
From equation (1), we have
$\sin ^{-1} x+\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{\pi}{2}$
since, $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$x=\frac{1}{\sqrt{5}}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2009 (Phase 2)
If $\sin ^{-1} x+\sin ^{-1} y=\pi / 2$ and $\cos ^{-1} x-\cos ^{-1} y=0$, then
values $x$ and $y$ are respectively

Options:
A $\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$
B $\frac{1}{2}, \frac{1}{2}$
C $\frac{1}{2},-\frac{1}{2}$
D $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$

The correct answer is:
$\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$

Solution:
Given, $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$ ...(i)
and $\cos ^{-1} x-\cos ^{-1} y=0$
$\Rightarrow\left(\frac{\pi}{2}-\sin ^{-1} x\right)-\left(\frac{\pi}{2}-\sin ^{-1} y\right)=0$
$\Rightarrow \sin ^{-1} y-\sin ^{-1} x=0$
$\Rightarrow \sin ^{-1} y=\sin ^{-1} x$ ...(ii)
From equations (i) and (ii), we get
$2 \sin ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{4}$
$\Rightarrow x=\frac{1}{\sqrt{2}}$
From equation (ii)
$y=\frac{1}{\sqrt{2}}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2008 (Phase 2)
If $\sin ^{-1} \mathrm{x}=\tan ^{-1} \mathrm{y}$, what is the value of $\frac{1}{\mathrm{x}^{2}}-\frac{1}{\mathrm{y}^{2}}$ ?

Options:
A 1
B $-1$
C 0
D 2

The correct answer is:
1

Solution:
Let, $\sin ^{-1} x=\tan ^{-1} y=\theta$
$\Rightarrow \mathrm{x}=\sin \theta$ and $\mathrm{y}=\tan \theta$
$\frac{1}{x^{2}}=\frac{1}{\sin ^{2} \theta}=\operatorname{cosec}^{2} \theta$
and $\frac{1}{\mathrm{y}^{2}}=\frac{1}{\tan ^{2} \theta}=\cot ^{2} \theta$.
$\Rightarrow \frac{1}{\mathrm{x}^{2}}-\frac{1}{\mathrm{y}^{2}}=\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2007 (Phase 2)
If $\sin \theta=-\frac{1}{2}$ and $\tan \theta=\frac{1}{\sqrt{3}}$, then in which quadrant does
$\theta$ lie?

Options:
A First
B Second
C Third
D Fourth

The correct answer is:
Third

Solution:
$\sin \theta=\frac{-1}{2}, \tan \theta=\frac{1}{\sqrt{3}}$
$\sin \theta$ is negative, $\tan \theta$ is positive
$\theta$ lies in third quadrant.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\sin 2 \mathrm{~A}=\frac{4}{5}$, then what is the value of tan $\mathrm{A}\left(0 \leq \mathrm{A} \leq \frac{\pi}{4}\right)$ ?

Options:
A 1
B $-1$
C $\frac{1}{2}$
D 2

The correct answer is:
$\frac{1}{2}$

Solution:
As given : $\sin 2 \mathrm{~A}=\frac{4}{5}$
$\sin 2 \mathrm{~A}=\frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{~A}}$
$\Rightarrow \frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{~A}}=\frac{4}{5}$
$\Rightarrow 10 \tan \mathrm{A}=4+4 \tan ^{2} \mathrm{~A}$
$\Rightarrow 5 \tan \mathrm{A}=2+2 \tan ^{2} \mathrm{~A}$
$\Rightarrow 2 \tan ^{2} \mathrm{~A}-5 \tan \mathrm{A}+2=0$
$\Rightarrow 2 \tan ^{2} \mathrm{~A}-4 \tan \mathrm{A}-\tan \mathrm{A}+2=0$
$\Rightarrow 2 \tan \mathrm{A}(\tan \mathrm{A}-2)-1(\tan \mathrm{A}-2)=0$
$\Rightarrow(2 \tan \mathrm{A}-1)(\tan \mathrm{A}-2)=0$
$\Rightarrow \tan \mathrm{A}=\frac{1}{2}$ (since $\left.\mathrm{A} \leq \frac{\pi}{4} \Rightarrow \tan \mathrm{A} \neq 2\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 1)