datavorous/entrance-exam-dataset huggingface dataset:
What is the value of $\log _{7} \log _{7} \sqrt{7 \sqrt{7} \sqrt{7}}$ equal to?

Options:
A $3 \log _{2} 7$
B $1-3 \log _{2} 7$
C $1-3 \log _{7} 2$
D $\frac{7}{8}$

The correct answer is:
$1-3 \log _{7} 2$

Solution:
Let $\log _{7} \log _{7} \sqrt{7 \sqrt{7 \sqrt{7}}}=x$
then $7^{x}=\log _{7} \sqrt{7 \sqrt{7 \sqrt{7}}}$
$$
\begin{aligned}
7^{x} &=\frac{1}{2} \cdot \log _{7} 7 \sqrt{7 \sqrt{7}} \\
7^{x} &=\frac{1}{2}\left[\log _{7} 7+\frac{1}{2} \log _{7} 7 \sqrt{7}\right] \\
&=\frac{1}{2}\left[1+\frac{1}{2} \log _{7} 7+\frac{1}{2} \log _{7} \sqrt{7}\right] \\
&=\frac{1}{2}\left[1+\frac{1}{2}+\frac{1}{4}\right] \\
7^{x} &=\frac{7}{8} \\
x &=\log _{7}\left(\frac{7}{8}\right) \\
x &=\log _{7} 7-\log _{7} 8 \\
x &=1-3 \log _{7} 2
\end{aligned}
$$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 2)
What is the value of $\log _{9} 27+\log _{8} 32$ ?

Options:
A $\frac{7}{2}$
B $\frac{19}{6}$
C 4
D 7

The correct answer is:
$\frac{19}{6}$

Solution:
$\log _{9} 27+\log _{8} 32$
$\quad=\log _{9} 3^{3}+\log _{8} 2^{5}$
$=3 \log _{9} 3+5 \log _{8} 2$
$\left.3 \log _{(3}^{2}\right)^{ \left.3+5 \log _{(2} 3\right)^{3}}$
$\frac{3}{2} \log _{3} 3+\frac{5}{3} \log _{2} 2$
$\frac{3}{2}+\frac{5}{3}=\frac{19}{6}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 2)
What is the value of
$\log _{10}\left(\frac{9}{8}\right)-\log _{10}\left(\frac{27}{32}\right)+\log _{10}\left(\frac{3}{4}\right)$ ?

Options:
A 3
B 2
C 1
D 0

The correct answer is:
0

Solution:
Consider, $\log \frac{9}{8}-\log \frac{27}{32}+\log \frac{3}{4}$
$=\log \frac{9}{8}+\log \frac{32}{27}+\log \frac{3}{4}$
$=\log \left(\frac{9}{8} \times \frac{32}{27}\right)+\log \frac{3}{4}$
$=\log \left(\frac{4}{3}\right)+\log \frac{3}{4}=\log \left(\frac{4}{3} \times \frac{3}{4}\right)=\log 1=0$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
What is the value of $\log _{y} x^{5} \log _{x} y^{2} \log _{z} z$

Options:
A 10
B 20
C 30
D 60

The correct answer is:
30

Solution:
$\log _{y} x^{5} \times \log _{x} y^{2} \log _{z} z^{3}$
$=\frac{5 \ln x}{\ln y} \times \frac{2 \ln y}{\ln x} \times \frac{3 \ln z}{\ln z}=30$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
What is the value of $\mathrm{m}$ if the vectors
$2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $3 \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ are coplanar?

Options:
A $-2$
B 2
C $-4$
D 4

The correct answer is:
$-4$

Solution:
Three vectors $x_{1} \vec{a}+y_{1} \vec{b}+z_{1} \vec{c}, x_{2} \vec{a}+y_{2} \vec{b}+z_{2} \vec{c}$ and
$\mathrm{x}_{3} \overrightarrow{\mathrm{a}}+\mathrm{y}_{3} \overrightarrow{\mathrm{b}}+\mathrm{z}_{3} \overrightarrow{\mathrm{c}}$ will be coplanar iff
$\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & \mathrm{z}_{1} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{z}_{2} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{z}_{3}\end{array}\right|=0$
Here, $\mathrm{x}_{1}=2, \mathrm{y}_{1}=-1, \mathrm{z}_{1}=1$
$\mathrm{x}_{2}=1, \mathrm{y}_{2}=2, \mathrm{z}_{2}=-3$
$\mathrm{x}_{3}=3, \mathrm{y}_{3}=\mathrm{m}, \mathrm{z}_{3}=5$
$\therefore\left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & \mathrm{~m} & 5\end{array}\right|=0$
$\Rightarrow 2(10+3 \mathrm{~m})+1(5+9)+1(\mathrm{~m}-6)=0$
$\Rightarrow 7 \mathrm{~m}+28=0 \Rightarrow \mathrm{m}=-4$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 1)
What is the value of $n$ for which the numbers $1,2,3, \ldots, n$ have variance $2 ?$

Options:
A 4
B 5
C 6
D 8

The correct answer is:
5

Solution:
Mean of the numbers $=\frac{\frac{n(n+1)}{2}}{n}$
$=\frac{n+1}{2}$
$\therefore$ Variance $=\frac{\left(1-\frac{n+1}{2}\right)^{2}+\left(2-\frac{n+1}{2}\right)^{2}+\left(3-\frac{n+1}{2}\right)^{2}+\ldots}{n}$
$2=\frac{\left(1^{2}+2^{2}+3^{2}+\ldots\right)+n\left(\frac{n+1}{2}\right)^{2}-2\left(\frac{n+1}{2}\right)[1+2+3+\ldots .]}{n}$
$2 n=\frac{1}{6} n(n+1)(2 n+1)+\frac{n(n+1)^{2}}{4}-2\left(\frac{n+1}{2}\right)\left\{\frac{n(n+1)}{2}\right\}$
$2 n=n(n+1)\left[\frac{2 n+1}{6}+\frac{n+1}{4}-\frac{n+1}{2}\right]$
$2=(n+1)\left[\frac{4 n+2-3 n-3}{12}\right]$
$\begin{array}{ll}\Rightarrow 24=(n+1)(n-1) & \Rightarrow n^{2}-1=24 \\ \Rightarrow n^{2}=25 & \Rightarrow n=\pm 5\end{array}$

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 2)
What is the value of $n$, if $P(15, n-1): P(16, n-2)$ $=3: 4 ?

Options:
A 10
B 12
C 14
D 15

The correct answer is:
14

Solution:
Let $\mathrm{P}(15, n-1): \mathrm{P}(16, n-2)=3: 4$
$\Rightarrow \frac{15}{16}{ }^{15} \mathrm{P}_{n-1}=\frac{3}{4}$
$\Rightarrow \quad \frac{15 !}{(15-n+1) !} \times \frac{(16-n+2) !}{16 !}=\frac{3}{4}$
$\Rightarrow \quad \frac{15 !}{(16-n) !} \times \frac{(18-n) !}{16 !}=\frac{3}{4}$
$\Rightarrow \quad \frac{(18-n) !}{16(16-n) !}=\frac{3}{4}$
$\Rightarrow \quad \frac{(18-n)(17-n)(16-n) !}{16(16-n) !}=\frac{3}{4}$
$\Rightarrow \quad(18-n)(17-n)=12$
$\Rightarrow \quad 306-17 n-18 n+n^{2}=12$
$\Rightarrow n^{2}-35 n+294=0$
$\Rightarrow \quad(n-14)(n-21)=0$
$\Rightarrow \quad n=14 \quad(\because n \neq 21)$

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 1)
What is the value of $n$ so that the angle between the lines having direction ratios $(1,1,1)$ and $(1,-1, n)$ is $60^{\circ}$ ?

Options:
A $\sqrt{3}$
B $\sqrt{6}$
C 3
D None of these

The correct answer is:
$\sqrt{6}$

Solution:
If $\left(l_{1}, m_{1}, n_{1}\right)$ and $\left(l_{2}, m_{2}, n_{2}\right)$ are the direction ratios then angle between the lines is
$\cos \theta=\frac{l_{l} l_{2}+m_{1} m_{2}+n_{1} n_{2}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}} \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}$
Here $l_{1}=1, m_{1}=1, n_{1}=1$ and
$l_{2}=1, m_{2}=-1, n_{2}=n$
and $\theta=60^{\circ}$
$\cos 60^{\circ}=\frac{1 \times 1+1 \times(-1)+1 \times n}{\sqrt{1^{2}+1^{2}+1^{2} \times \sqrt{1^{2}+1^{2}+n^{2}}}}$
$\Rightarrow \frac{1}{2}=\frac{n}{\sqrt{3} \sqrt{2+n^{2}}} \Rightarrow 3\left(2+n^{2}\right)=4 n^{2}$
$\Rightarrow n^{2}=6 \Rightarrow n=\pm \sqrt{6}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 2)
What is the value of $p$ for which the function
$$
f(x)=p \sin x+\frac{\sin 3 x}{3}
$$
has an extremum at $x=\frac{\pi}{3} ?$

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

The correct answer is:
2

Solution:
Let $\mathrm{f}(\mathrm{x})=\mathrm{p} \sin \mathrm{x}+\frac{\sin 3 \mathrm{x}}{3}$
Differentiate both side w.r.t $(\mathrm{x})$.
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{p} \cos \mathrm{x}+\frac{3 \cos 3 \mathrm{x}}{3}=\mathrm{p} \cos \mathrm{x}+\cos 3 \mathrm{x}$
It is given that $\mathrm{f}(\mathrm{x})$ has extremum value at $\mathrm{x}=\pi / 3$ $\therefore \quad f^{\prime}\left(\frac{\pi}{3}\right)=0$
$\Rightarrow \quad p \cos \frac{\pi}{3}+\cos \pi=0$
$\Rightarrow \frac{\mathrm{p}}{2}-1=0 \quad \Rightarrow \mathrm{p}=2$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
What is the value of $P$ for which the vector $p(2 \hat{i}-\hat{j}+2 \hat{k})$ is of 3 units length ?

Options:
A 1
B 2
C 3
D 6

The correct answer is:
1

Solution:
$|\mathrm{P}(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})|=3$
$\mathrm{P}\left|\sqrt{2^{2}+(-1)^{2}+2^{2}}\right|=3$
$3 \mathrm{P}=3 \Rightarrow \mathrm{P}=1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 2)
What is the value of $r$, if $P(5, r)=P(6, r-1) ?

Options:
A 9
B 5
C 4
D 2

The correct answer is:
4

Solution:
Given $P(5, r)=P(6, r-1)$
$\Rightarrow{ }^{5} P_{r}={ }^{6} P_{r-1}$
$\Rightarrow \frac{5 !}{(5-r) !}=\frac{6 !}{(6-r+1) !}$
$\Rightarrow \frac{5 !}{(5-r) !}=\frac{6 !}{(7-r) !}$
$\Rightarrow \frac{5 !}{(5-r) !}=\frac{6 \times 5 !}{(7-r)(6-r)(5-r) !}$
$\Rightarrow(7-r)(6-r)=6$
$\Rightarrow 42-13 r+r^{2}=6$
$\Rightarrow r^{2}-13 r+36=0$
$\Rightarrow r^{2}-9 r-4 r+36=0$
$\Rightarrow(r-9)(r-4)=0$
$\Rightarrow \mathrm{r}=4 \quad(\because \mathrm{r} \neq 9)$

Topic: Mathematics, Permutation Combination, NDA, NDA 2009 (Phase 1)
What is the value of $\frac{\sin \theta+\cos \theta-\tan \theta}{\sec \theta+\operatorname{cosec}-\cot \theta}$, when $\theta=\frac{3 \pi}{4}$ ?

Options:
A 0
B 1 $[2010-I I]$
C $-1$
D None of these

The correct answer is:
1 $[2010-I I]$

Solution:
Consider \end{array}$
$\frac{\sin \theta+\cos \theta-\tan \theta}{\sec \theta+\cos e c \theta-\cot \theta}$
Now, put value of $\theta=\frac{3 \pi}{4}$, we get
$\frac{\sin \frac{3 \pi}{4}+\cos \frac{3 \pi}{4}-\tan \frac{3 \pi}{4}}{\sec \frac{3 \pi}{4}+\cos e c \frac{3 \pi}{4}-\cot \frac{3 \pi}{4}}$
$=\frac{\sin \frac{\pi}{4}-\cos \frac{\pi}{4}+\tan \frac{\pi}{4}}{-\frac{1}{\cos \frac{\pi}{4}}+\frac{1}{\sin \frac{\pi}{4}}+\frac{1}{\tan \frac{\pi}{4}}}=\frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1}{-\sqrt{2}+\sqrt{2}+1}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
What is the value of $\int \frac{\mathrm{e}^{\mathrm{x}}(1+\mathrm{x})}{\sin ^{2}\left(\mathrm{xe}^{\mathrm{x}}\right)} \mathrm{dx}$?

Options:
A $-e^{x} \cot x+c$
B $\cos ^{2}\left(x e^{x}\right)+c$
C $\log \sin \left(x e^{x}\right)+c$
D $-\cot \left(x e^{x}\right)+c$

The correct answer is:
$-\cot \left(x e^{x}\right)+c$

Solution:
Let the given integral be $\mathrm{I}=\int \frac{\mathrm{e}^{\mathrm{x}}(1+\mathrm{x})}{\sin ^{2}\left(\mathrm{xe}^{\mathrm{x}}\right)} \mathrm{dx}$
Put $x e^{x}=t$ and $e^{x}(1+x) d x=d t$
$\Rightarrow \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\sin ^{2} \mathrm{t}}=\int \operatorname{cosec}^{2} \mathrm{t} \mathrm{dt}$
$=-\cot t+c=-\cot \left(x e^{x}\right)+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2007 (Phase 2)
What is the value of $\sec ^{2} \tan ^{-1}\left(\frac{5}{11}\right) ?$

Options:
A $121 / 96$
B $211 / 921$
C $146 / 121$
D $267 / 121$

The correct answer is:
$146 / 121$

Solution:
Let $\sec ^{2}\left(\tan ^{-1}\left(\frac{5}{11}\right)\right)$
$=1+\tan ^{2}\left(\tan ^{-1}\left(\frac{5}{11}\right)\right) \quad\left(\because \sec ^{2} \theta-\tan ^{2} \theta=1\right)$
$=1+\left[\tan \left(\tan ^{-1}\left(\frac{5}{11}\right)\right)\right]^{2}=1+\left(\frac{5}{11}\right)^{2}$
$=1+\frac{25}{121}=\frac{146}{121}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2012 (Phase 1)
What is the value of $(\sec \theta-\cos \theta)(\operatorname{cosec} \theta-\sin \theta)(\cot \theta+\tan \theta) ?$

Options:
A 1
B 2
C \sin \theta & \text
D \cos \theta

The correct answer is:
1

Solution:
The given expression is:
$(\sec \theta-\cos \theta)(\operatorname{cosec} \theta-\sin \theta)(\cot \theta+\tan \theta)$
$=\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\right)$
$=\left(\frac{1-\cos ^{2} \theta}{\cos \theta}\right)\left(\frac{1-\sin ^{2} \theta}{\sin \theta}\right)\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}\right)$
$=\frac{\sin ^{2} \theta}{\cos \theta} \cdot \frac{\cos ^{2} \theta}{\sin \theta} \times \frac{1}{\sin \theta \cos \theta}=\frac{\sin ^{2} \theta \cdot \cos ^{2} \theta}{\cos ^{2} \theta \cdot \sin ^{2} \theta}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
What is the value of $\left|\begin{array}{rr}\sin 10^{\circ} & -\cos 10^{\circ} \\ \sin 80^{\circ} & \cos 80^{\circ}\end{array}\right| ?$

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

The correct answer is:
1

Solution:
$\left|\begin{array}{lr}\sin 10^{\circ} & -\cos 10^{\circ} \\ \sin 80^{\circ} & \cos 80^{\circ}\end{array}\right|$
$=\sin 10^{\circ} \cos 80^{\circ}+\sin 80^{\circ} \cos 10^{\circ}$
$=\sin 10^{\circ} \sin 10^{\circ}+\cos 10^{\circ} \cos 10^{\circ}$
$=\sin ^{2} 10^{\circ}+\cos ^{2} 10^{\circ}=1$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 2)
What is the value of $\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3} ?$

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{2}$
C $\frac{\pi}{4}$
D $\frac{\pi}{6}$

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

Solution:
Consider $\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}$
$=\tan ^{-1}\left(\frac{4 / 5}{\sqrt{1-16 / 25}}\right)+\tan ^{-1}\left(\frac{2 / 3}{1-1 / 9}\right)$
$\left[\because \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}\right.$ and $\left.2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
$=\tan ^{-1}\left(\frac{4 / 5}{3 / 5}\right)+\tan ^{-1}\left(\frac{2 / 3}{8 / 9}\right)$
$=\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{3}{4}\right)$
$=\tan ^{-1}\left(\frac{4}{3}\right)+\cot ^{-1}\left(\frac{4}{3}\right)=\frac{\pi}{2}$
$\left(\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2010 (Phase 2)
What is the value of $\sin ^{-1} \frac{4}{5}+\sec ^{-1} \frac{5}{4}-\frac{\pi}{2} ?$

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{2}$
C $\pi$
D 0

The correct answer is:
0

Solution:
$\begin{aligned} & \sin ^{-1} \frac{4}{5}+\sec ^{-1} \frac{5}{4}-\frac{\pi}{2}=\sin ^{-1} \frac{4}{5}+\cos ^{-1} \frac{4}{5}-\frac{\pi}{2} \\=& \frac{\pi}{2}-\frac{\pi}{2}=0 \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2019 (Phase 1)
What is the value of $\sin 15^{\circ} \sin 75^{\circ}$ ? $\quad$

Options:
A $1.4$
B $1 / 8$
C $1 / 16$
D 1

The correct answer is:
$1.4$

Solution:
Consider $\sin 15^{\circ} \sin 75^{\circ}$
$=\sin \left(45^{\circ}-30^{\circ}\right) \sin \left(45^{\circ}+30^{\circ}\right)$
$=\left(\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}\right)\left(\sin 45^{\circ} \cos 30^{\circ}\right.$
$\left.\quad+\cos 45^{\circ} \sin 30^{\circ}\right)$
$($ using $\sin (\mathrm{A}+\mathrm{B})=\sin A \cos B$
$+\cos A \sin B$ and $\sin (A-B) \sin A \cos B-\cos A \sin B)$
$=\left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right)\left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right)$
$=\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)=\frac{3-1}{8}=\frac{2}{8}=\frac{1}{4}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
What is the value of $\sin 15^{\circ} ?$

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

The correct answer is:
$\frac{\sqrt{3}-1}{2 \sqrt{2}}$

Solution:
$\begin{aligned} & \sin 15^{\circ}=\sin \left[45^{\circ}-30^{\circ}\right] \\ &=\sin 45^{\circ} \cdot \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\ &=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ &=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}-1}{2 \sqrt{2}} \\ \text { 130. (c) } & 4 \sin ^{2} \theta=1 \\ & \Rightarrow \sin ^{2} \theta=\frac{1}{4} \Rightarrow \sin \theta=\pm \frac{1}{2} \Rightarrow \theta=\pm \frac{\pi}{6} \end{aligned}$
Hence $\theta$ take 4 values. (one value for each quadrant)

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 2)
What is the value of $\sin 18^{\circ} \cos 36^{\circ}$ equal to?

Options:
A 4
B 2
C 1
D $1 / 4$

The correct answer is:
$1 / 4$

Solution:
We know that $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$
Now, $\cos 36^{\circ}=1-2 \sin ^{2} 18^{\circ}$
$=1-\frac{2}{16}(\sqrt{5}-1)^{2}=1-\frac{(3-\sqrt{5})}{4}$
$\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$
Now, $\sin 18^{\circ} \cos 36^{\circ}=\frac{(\sqrt{5})^{2}-(1)^{2}}{16}=\frac{4}{16}=\frac{1}{4}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 2)
What is the value of $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) ?$

Options:
A $-\frac{\pi}{3}$
B $\frac{2 \pi}{3}$
C $-\frac{2 \pi}{3}$
D $\frac{\pi}{3}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left[\sin \left(\pi-\frac{2 \pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)$
$=\frac{\pi}{3} \quad\left[\right.$ Since, $\left.\sin \frac{2 \pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)\right]$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2006 (Phase 1)
What is the value of $\left(\sin ^{4} \theta-\cos ^{4} \theta+1\right) \operatorname{cosec}^{2} \theta$ ?

Options:
A -2
B 0
C 1
D 2

The correct answer is:
2

Solution:
Consider $\left(\sin ^{4} \theta-\cos ^{4} \theta+1\right) \operatorname{cosec}^{2} \theta .$
$\quad=\left[\left(\sin ^{2} \theta\right)^{2}-\left(\cos ^{2} \theta\right)^{2}+1\right] \operatorname{cosec}^{2} \theta$
$=\left[\left(\sin ^{2} \theta-\cos ^{2} \theta\right)\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+1\right] \operatorname{cosec}^{2} \theta$
$=\left[1-\cos ^{2} \theta+\sin ^{2} \theta\right] \operatorname{cosec}^{2} \theta$
$=\left(\sin ^{2} \theta+\sin ^{2} \theta\right) \operatorname{cosec}^{2} \theta$
$=2 \sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}=2$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
What is the value of $\left(\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}\right) ?$

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

The correct answer is:
0

Solution:
Consider $\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}$
$=2 \cos \frac{70^{\circ}+50^{\circ}}{2} \cdot \sin \frac{50^{\circ}-70^{\circ}}{2}+\sin 10^{\circ}$
$\left[\because \operatorname{Sin} A-\operatorname{Sin} B=2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \operatorname{Sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right]$
$=-2 \cos 60^{\circ} \sin 10^{\circ}+\sin 10^{\circ}=-\sin 10^{\circ}+\sin 10^{\circ}=0$
$\left(\because \cos 60^{\circ}=\frac{1}{2}\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
What is the value of $\sin \left(\frac{5 \pi}{12}\right)$ ?

Options:
A $\frac{\sqrt{3}+1}{2}$
B $\frac{\sqrt{6}+\sqrt{2}}{4}$
C $\frac{\sqrt{3}+\sqrt{2}}{4}$
D $\frac{\sqrt{6}+1}{2}$

The correct answer is:
$\frac{\sqrt{6}+\sqrt{2}}{4}$

Solution:
$\begin{aligned} & \sin \frac{5 \pi}{12}=\sin 75^{\circ} \\ &=\sin \left(45^{\circ}+30^{\circ}\right) \\ &=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ} \\ &=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}=\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}+1}{2}\right) \\ &=\frac{\sqrt{3}+1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4} \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 1)
What is the value of $\left(\sin 22 \frac{1^{\circ}}{2}+\cos 22 \frac{1^{\circ}}{2}\right)^{4}$ ?

Options:
A $\frac{3+2 \sqrt{2}}{2}$
B $\frac{1+2 \sqrt{2}}{2}$
C $\frac{3 \sqrt{2}+2}{2}$
D 1

The correct answer is:
$\frac{3+2 \sqrt{2}}{2}$

Solution:
Let $x=\left(\sin 22 \frac{1^{\circ}}{2}+\cos ^{2} 22 \frac{1^{\circ}}{2}\right)^{4}$
$=\left\{\left(\sin 22 \frac{1^{\circ}}{2}+\cos 22 \frac{1}{2}^{-}\right)^{2}\right\}^{2}$
$=\left(\sin ^{2} 22 \frac{1^{\circ}}{2}+\cos ^{2} 22 \frac{1^{\circ}}{2}+2 \sin 22 \frac{1^{\circ}}{2} \cos 22 \frac{1^{\circ}}{2}\right)^{2}$
$=\left(1+\sin 45^{\circ}\right)^{2}$
$(\because 2 \sin \theta \cos \theta=\sin 2 \theta)$
$=\left(1+\frac{1}{\sqrt{2}}\right)^{2}=\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)^{2}$
$=\frac{2+1+2 \sqrt{2}}{2}=\frac{3+2 \sqrt{2}}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 1)
What is the value of $\sin \left(1920^{\circ}\right) ?$

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

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

Solution:
$\begin{aligned} & \sin \left(1920^{\circ}\right)=\sin \left(360 \times 5+120^{\circ}\right) \\ &=\sin 120^{\circ} \quad\left(\because \sin \left(360^{\circ}+\theta\right)=\sin \theta\right) \\ \text { 122. (b) } & \sin (\mathrm{A}+\mathrm{B})=1 \\ & \Rightarrow \sin (\mathrm{A}+\mathrm{B})=\sin 90^{\circ} \\ & \Rightarrow \mathrm{A}+\mathrm{B}=90^{\circ} \\ & \text { Given } \sin (\mathrm{A}-\mathrm{B})=\frac{1}{2}=\sin 30^{\circ} \\ & \Rightarrow \mathrm{A}-\mathrm{B}=30^{\circ} \end{aligned}$
On solving (1) and (2), we get $\mathrm{A}=60$
$\mathrm{B}=30$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What is the value of $\sin 292 \frac{1}{2}^{\circ}$ ?

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

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

Solution:
$\quad \sin \left(292 \frac{1}{2}\right)^{\circ}=\sin \frac{585^{\circ}}{2}$
$=\sqrt{\frac{1-\cos 585^{\circ}}{2}}$
$=\left(\because \cos 2 \theta=1-2 \sin ^{2} \theta \Rightarrow \sin \theta=\sqrt{\frac{1-\cos 2 \theta}{2}}\right)$
$=\sqrt{\frac{1-\cos \left(360^{\circ}+225^{\circ}\right)}{2}}=\sqrt{\frac{1-\cos 225^{\circ}}{2}}$
$=\sqrt{\frac{1-\cos \left(180^{\circ}+45^{\circ}\right)}{2}}$
96. (b) We know that, $1^{\circ} < 1$ radian $\quad\left(\because 1^{\circ}=\frac{\pi}{180}\right.$ radian $)$
$\Rightarrow \sin 1^{\circ} < \sin 1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
What is the value of $\sin 420^{\circ} \cdot \cos 390^{\circ}+\cos \left(-300^{\circ}\right) \cdot \sin \left(-330^{\circ}\right) ?

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

The correct answer is:
1

Solution:
Given expression $=\sin \left(360^{\circ}+60^{\circ}\right) \cdot \cos \left(360^{\circ}+30^{\circ}\right)+\cos \left(360^{\circ}-60^{\circ}\right)$
$\left(-\sin \left(360^{\circ}-30^{\circ}\right)\right.$
$=\sin 60^{\circ} \cdot \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}=\sin \left(60^{\circ}+30^{\circ}\right)$
$=\sin 90^{\circ}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What is the value of $\sin (A+B) \sin (A-B)+\sin (B+C) \sin (B-C)+\sin (C+A)$
$\sin (\mathrm{C}-\mathrm{A}) ?

Options:
A 0
B $\sin A+\sin B+\sin C$
C $\cos A+\cos B+\cos C$
D 1

The correct answer is:
0

Solution:
$\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})$
$=\frac{1}{2}\{2 \sin (\mathrm{A}+\mathrm{B}) \cdot \sin (\mathrm{A}+\mathrm{B})\}$
$=\frac{1}{2}\{\cos (\mathrm{A}-\mathrm{B}-\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}-\mathrm{B}+\mathrm{A}+\mathrm{B})\}$
$[$ Since $2 \sin \mathrm{X} \sin \mathrm{Y}=\cos (\mathrm{X}-\mathrm{Y})-\cos (\mathrm{X}+\mathrm{Y})]$
$=\frac{1}{2}\{\cos 2 \mathrm{~B}-\cos 2 \mathrm{~A}\}$
Also, $\sin (\mathrm{B}+\mathrm{C}) \sin (\mathrm{B}-\mathrm{C})=\frac{1}{2}\{\cos 2 \mathrm{C}-\cos 2 \mathrm{~B}\}$
and $\sin (C+A) \cdot \sin (C-A)$
$\begin{aligned}=\frac{1}{2}\{\cos 2 \mathrm{~A}-\cos 2 \mathrm{C}\} & \\ \therefore \quad \sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})+\sin (\mathrm{B}+\mathrm{C}) \sin (\mathrm{B}-\mathrm{c}) \\ &+\sin (\mathrm{C}+\mathrm{A}) \cdot \sin (\mathrm{C}-\mathrm{A}) \\=\frac{1}{2}\{\cos 2 \mathrm{C}-\cos 2 \mathrm{~B}+\cos 2 \mathrm{~A}-\cos 2 \mathrm{C}\\+&\cos 2 \mathrm{~B}-\cos 2 \mathrm{~A}\}=0 \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
What is the value of $\sum_{r=1}^{n} \frac{P(n, r)}{r !} ?

Options:
A $2^{n}-1$
B $2^{n}$
C $2^{n-1}$
D $2^{n}+1$

The correct answer is:
$2^{n}-1$

Solution:
Consider $\sum_{r=1}^{n} \frac{P(n, r)}{r !} \equiv \sum_{r=1}^{n} C(n, r)$
Now, consider $(1+1)^{n}={ }^{n} \mathrm{C}_{0}+{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{2}+\ldots \ldots . .+{ }^{n} \mathrm{C}_{n}$
$\Rightarrow 2^{n}=1+{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{2}+\ldots \ldots . .+{ }^{n} \mathrm{C}_{n}$
$\Rightarrow 2^{n}-1={ }^{n} C_{1}+{ }^{n} C_{2}+\ldots \ldots . .+{ }^{n} C_{n}$
$\Rightarrow 2^{n}-1=\sum_{r=1}^{n}{ }^{n} C_{r} \equiv \sum_{r=1}^{n} \frac{P(n, r)}{r !}$

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 2)