datavorous/entrance-exam-dataset huggingface dataset:
If $\sin \theta+2 \cos \theta=1$, then what is $2 \sin \theta-\cos \theta$ equal to ?

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
Let $\sin \theta+2 \cos \theta=1 \quad$....
.... Consider $2 \sin \theta-\cos \theta=\alpha$ (let)
squaring and adding $\sin ^{2} \theta+4 \cos ^{2} \theta+4 \sin \theta \cos \theta+4 \sin ^{2} \theta+\cos ^{2} \theta-$
$4 \sin \theta \cos \theta=1+\alpha^{2}$
$\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+4\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+\alpha^{2}$
$\Rightarrow 1+4=1+\alpha^{2} \Rightarrow \alpha^{2}=4 \Rightarrow \alpha=2$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
If $\sin 2 \theta=\cos 3 \theta$, where $0 < \theta < \frac{\pi}{2}$, then what is $\sin \theta$ equal
to ?

Options:
A $\frac{\sqrt{5}+1}{4}$
B $\frac{\sqrt{5}-1}{4}$
C $\frac{\sqrt{5}+1}{16}$
D $\frac{\sqrt{5}-1}{16}$

The correct answer is:
$\frac{\sqrt{5}-1}{4}$

Solution:
$\begin{aligned} \sin 2 \theta=\cos 3 \theta \\ \Rightarrow \sin 2 \theta=\sin \left(90^{\circ}-3 \theta\right) \\ 2 \theta=90^{\circ}-3 \theta \quad \therefore \sin \theta=\sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \\ & \Rightarrow 5 \theta=90^{\circ} \\ & \Rightarrow \theta=18^{\circ} \\ & \Rightarrow\left(\frac{1}{q}\right)(p)=1 \end{aligned}$
$\Rightarrow \mathrm{p}=\mathrm{q}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\sin ^{2} x+\sin ^{2} y=1$, then what is the value of $\cot (x+y)$ ?

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

The correct answer is:
0

Solution:
Given $\sin ^{2} x+\sin ^{2} y=1$
$\Rightarrow \sin ^{2} x=1-\sin ^{2} y$
$\Rightarrow \sin ^{2} x=\cos ^{2} y$
$\Rightarrow \sin x=\cos y$
Similarly By considering $\sin ^{2} y=1-\sin ^{2} x$, we have
$\cos x=\sin y$
Now, Consider $\cot (x+y)=\frac{\cos (x+y)}{\sin (x+y)}$
$=\frac{\cos x \cos y-\sin x \sin y}{\sin (x+y)}$
$=\frac{\cos x \cos y-\cos x \cos y}{\sin (x+y)}=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
If $\sin 3 A=1$, then how many distinct values can $\sin A$ assume?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Let $\sin 3 A=1$
$\Rightarrow 3 \sin A-4 \sin ^{3} A=1$
$\Rightarrow 4 \sin ^{3} A-3 \sin A+1=0$
$\Rightarrow(\sin A+1)\left(4 \sin ^{2} A-4 \sin A+1\right)=0$
$\Rightarrow(\sin A+1)(2 \sin A-1)^{2}=0$
$\Rightarrow \sin A=-1$ or $\frac{1}{2}$
Hence, $\sin A$ can take two distinct values

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
If $\sin ^{3} \theta+\cos ^{3} \theta=0$, then what is the value of $\theta$ ?

Options:
A $\frac{-\pi}{4}$
B 0
C $\frac{\pi}{4}$
D $\frac{\pi}{3}$

The correct answer is:
$\frac{-\pi}{4}$

Solution:
Since, $\sin ^{3} \theta+\cos ^{3} \theta=0$
$\Rightarrow(\sin \theta+\cos \theta)\left(\sin ^{2} \theta-\sin \theta \cos \theta+\cos ^{2} \theta\right)=0$
$\left(\because a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right)$
$\Rightarrow(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta)=0$
$\Rightarrow(\sin \theta+\cos \theta)\left(1-\frac{\sin 2 \theta}{2}\right)=0$
$\Rightarrow \sin \theta+\cos \theta=0$
or $\sin 2 \theta=2$
(discarded since $\sin ^{2} \theta=2$ is not possible) $\Rightarrow \sin \theta+\cos \theta=0 \Rightarrow \sin \theta=-\cos \theta$
$\Rightarrow \tan \theta=-1 \Rightarrow \theta=-\pi / 4$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2006 (Phase 2)
If $\sin \theta=3 \sin (\theta+2 \alpha)$, then the value of $\tan (\theta+\alpha)+$
$2 \tan \alpha$ is equal to

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

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

Solution:
Given, $\sin \theta=3 \sin (\theta+2 \alpha)$
$\Rightarrow \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{1}{3}$
Apply componendo and divide do rule $\Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{1+3}{1-3}$
$\Rightarrow \frac{2 \sin (\theta+\alpha) \cos \alpha}{2 \cos (\theta+\alpha) \sin \alpha}=\frac{4}{-2}=-2$
$\Rightarrow \frac{\tan (\theta+\alpha)}{\tan \alpha}=-2$
$\Rightarrow \tan (\theta+\alpha)=-2 \tan \alpha \Rightarrow \tan (\theta+\alpha)+2 \tan \alpha=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 1)
If $\sin ^{4} x-\cos ^{4} x=p$, then which one of the following is correct?

Options:
A $\mathrm{p}=1$
B $p=0$
C $|\mathrm{p}|>1$
D $|\mathrm{p}| \leq 1$

The correct answer is:
$|\mathrm{p}| \leq 1$

Solution:
Consider $\sin ^{4} x-\cos ^{4} x=p$
$\Rightarrow\left(\sin ^{2} x\right)^{2}-\left(\cos ^{2} x\right)^{2}=p$
$\Rightarrow\left(\sin ^{2} x-\cos ^{2} x\right)\left(\sin ^{2} x+\cos ^{2} x\right)=p$
$\Rightarrow \sin ^{2} x-\cos ^{2} x=p\left(\because \sin ^{2} x+\cos ^{2} x=1\right)$
$\Rightarrow-\cos 2 x=p \quad\left(\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right)$
$\Rightarrow \cos 2 x=-p$
$\therefore \quad|p| \leq 1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
If sin $A=1 / \sqrt{5}, \cos B=3 / \sqrt{10} ; A, B$ being positive acute angles, then what is $(A+B)$ equal to?

Options:
A $\pi / 6$
B $\pi / 4$
C $\pi / 3$
D $\pi / 2$

The correct answer is:
$\pi / 4$

Solution:
Given, $\sin A=\frac{1}{\sqrt{5}}$ and $\cos B=\frac{3}{\sqrt{10}}$
$\therefore \sin (A+B)=\sin A \cos B+\cos A \sin B$
$=\frac{1}{\sqrt{5}} \times \frac{3}{\sqrt{10}}+\sqrt{1-\frac{1}{5}} \times \sqrt{1-\frac{9}{10}}$
$=\frac{3}{\sqrt{50}}+\frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{10}}=\frac{3+2}{\sqrt{50}}=\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4}$
$\Rightarrow A+B=\frac{\pi}{4}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2009 (Phase 1)
If $\sin \mathrm{A}=\frac{2}{\sqrt{5}}$ and $\cos \mathrm{B}=\frac{1}{\sqrt{10}}$ where $\mathrm{A}$ and $\mathrm{B}$ are acute
$\begin{array}{ll}\text { angles, then what is } \mathrm{A}+\mathrm{B} \text { equal to } ? & {[2012-\mathrm{II}]}\end{array}$

Options:
A $135^{\circ}$
B $90^{\circ}$
C $75^{\circ}$
D $\overline{60^{\circ}}$

The correct answer is:
$135^{\circ}$

Solution:
Given, $\sin \mathrm{A}=\frac{2}{\sqrt{5}}$ and $\cos \mathrm{B}=\frac{2}{\sqrt{10}}$
$\therefore \quad \cos \mathrm{A}=\frac{1}{\sqrt{5}} \quad \sin \mathrm{B}=\frac{3}{\sqrt{10}}$
$\therefore \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$
$=\left(\frac{2}{\sqrt{15}}\right)\left(\frac{1}{\sqrt{10}}\right)+\left(\frac{1}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)$
$=\frac{2}{\sqrt{50}}+\frac{3}{\sqrt{50}}=\frac{5}{\sqrt{50}}=\frac{5}{\sqrt{25 \times 2}}=\frac{5}{5 \sqrt{2}}=\frac{1}{\sqrt{2}}$
$\sin (\mathrm{A}+\mathrm{B})=\frac{1}{\sqrt{2}} \Rightarrow \sin (90+45)=\frac{1}{\sqrt{2}}$
$\left(\because \sin (90+45)=\cos 45^{\circ}\right)$
$\therefore \quad A=90^{\circ}$ and $B=45^{\circ}$
$\therefore \quad A+B=90^{\circ}+45^{\circ}=135^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 2)
If $\sin A=\frac{3}{5}$, where $450^{\circ} < \mathrm{A} < 540^{\circ}$, then $\cos \frac{\mathrm{A}}{2}$ is equal to

Options:
A $\frac{1}{\sqrt{10}}$
B $-\sqrt{\frac{3}{10}}$
C $\frac{\sqrt{3}}{\sqrt{10}}$
D None of the above

The correct answer is:
None of the above

Solution:
$\operatorname{Sin} A=\frac{3}{5} ; 450^{\circ} < \mathrm{A} < 540^{\circ}$
$\Rightarrow 225^{\circ} < \frac{\mathrm{A}}{2} < 270^{\circ}$
$\cos \mathrm{A}=\frac{-4}{5} \quad(\because \mathrm{A}$ lies in Q2 $)$
$\therefore \cos ^{2} \frac{\mathrm{A}}{2}=\frac{1+\cos \mathrm{A}}{2}=\frac{1}{10}$
$\Rightarrow \cos \frac{\mathrm{A}}{2}=\frac{-1}{\sqrt{10}}\left(\because \frac{\mathrm{A}}{2}\right.$ lies in $\left.\mathrm{Q} 3\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
If $\sin \mathrm{A} \sin \left(60^{\circ}-\mathrm{A}\right) \sin \left(60^{\circ}+\mathrm{A}\right)=\mathrm{k} \sin 3 \mathrm{~A}$, then what isk
equal to?

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

The correct answer is:
$1 / 4$

Solution:
$\sin A \cdot \sin \left(60^{\circ}-\mathrm{A}\right) \sin \left(60^{\circ}+\mathrm{A}\right)=\mathrm{k} \sin 3 \mathrm{~A}$
$\begin{aligned} \Rightarrow & \sin \mathrm{A} \cdot \frac{\sin 3 \mathrm{~A}}{4 \sin \mathrm{A}}=\mathrm{k} \cdot \sin 3 \mathrm{~A} \\ & \quad\left[\because \sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)=\frac{\sin 3 \mathrm{~A}}{4 \sin \mathrm{A}}\right] \\ \Rightarrow & \frac{\sin 3 \mathrm{~A}}{4}=\mathrm{k} \cdot \sin 3 \mathrm{~A} \\ \therefore \mathrm{k} &=\frac{1}{4} \end{aligned}$
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 2)
If $\sin \mathrm{A}=\sin \mathrm{B}$ and $\cos \mathrm{A}=\cos \mathrm{B}$, then which one of the
following is correct?

Options:
A $\quad B=n \pi+A$
B $\mathrm{A}=2 \mathrm{n} \pi-\mathrm{B}$
C $\mathrm{A}=2 \mathrm{n} \pi+\mathrm{B}$
D $\quad \mathrm{B}=\mathrm{n} \pi-\mathrm{A} \quad$ (n is an integer)

The correct answer is:
$\quad B=n \pi+A$

Solution:
As given:
$\sin A=\sin B$
and $\cos A=\cos B$
$\Rightarrow \frac{\sin A}{\cos A}=\frac{\sin B}{\cos B}$
$\Rightarrow \tan A=\tan B$ or $\tan B=\tan A=\tan (n \pi+\mathrm{A})$
$\Rightarrow B=n \pi+A$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 2)
If $\sin A+\sin B+\sin C=3$ then what is $\cos A+\cos B+\cos$
C equal to?

Options:
A $-1$
B $\underline{0}$
C 1
D 3

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

Solution:
Let \sin A+\sin B+& \sin C=3 \end{aligned}$
$\Rightarrow \sin A=\sin B=\sin C=1 \quad(\because$ max value of sin is 1$)$
$\therefore \cos A=\sqrt{1-\sin ^{2} A}=\sqrt{1-1}=0$
Similarly, $\cos B=0=\cos C$
Hence, $\cos A+\cos B+\cos C=0+0+0=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
If $\sin \theta$ and $\cos \theta$ are the roots of $a x^{2}+b x+c=0$, then constants $\mathrm{a}, \mathrm{b}, \mathrm{c}$ will satisfy which one of the following conditions?

Options:
A $a^{2}+b^{2}+2 a c=0$
B $\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ac}=0^{2}$
C $a^{2}-b^{2}+2 a c=0$
D $-a^{2}+b^{2}+2 a c=0$

The correct answer is:
$a^{2}-b^{2}+2 a c=0$

Solution:
As given $\sin \theta$ and $\cos \theta$ are the roots of the equation
So, sum of roots $=\sin \theta+\cos \theta=-\frac{b}{a}$...(1)
and product of roots $=\sin \theta \cos \theta=\frac{\mathrm{c}}{a}$...(2)
On squaring both sides in Eq. (1), we get $(\sin \theta+\cos \theta)^{2}=\frac{b^{2}}{a^{2}}$
$\Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}$
$\Rightarrow 1+\frac{2 \mathrm{c}}{\mathrm{a}}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}} \quad$ using Eq. $(2)$
$\Rightarrow a^{2}+2 c a=b^{2}$
$\Rightarrow a^{2}-b^{2}+2 a c=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 2)
If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $p x^{2}+q x+1$ $=0$, then which one of the following is correct?

Options:
A $p^{2}+q^{2}-2 p r=0$
B $p^{2}-q^{2}+2 p r=0$
C $(p+r)^{2}=2\left(p^{2}+r^{2}\right)$
D $(p-r)^{2}=q^{2}+r^{2}$

The correct answer is:
$p^{2}-q^{2}+2 p r=0$

Solution:
$\sin \alpha$ and $\cos \alpha$ are the roots of the equation $p x^{2}+q x+r=0$
$\therefore \sin \alpha+\cos \alpha=\frac{-q}{p}\cdots. (i)$
and $\sin \alpha \cos \alpha=\frac{r}{p}$ ...(ii)
From equation (i)
$(\sin \alpha+\cos \alpha)^{2}=\frac{q^{2}}{p^{2}}$
$\Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=\frac{q^{2}}{p^{2}}$
$\Rightarrow 1+2 \cdot \frac{r}{p}=\frac{q^{2}}{p^{2}}$
$\Rightarrow p^{2}+2 p r=q^{2}$
$p^{2}-q^{2}+2 p r=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 2)
If $\sin \theta+\cos \theta=\sqrt{2} \cos \theta$, then what is $(\cos \theta-\sin \theta)$ equal
to ?

Options:
A $-\sqrt{2} \cos \theta$
B $-\sqrt{2} \sin \theta$
C $\sqrt{2} \sin \theta$
D $2 \sin \theta$

The correct answer is:
$\sqrt{2} \sin \theta$

Solution:
$\sin \theta+\cos \theta=\sqrt{2} \cos \theta$
Let $\cos \theta-\sin \theta=\mathrm{P}$
$(1)^{2}+(2)^{2} \Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta$
$-2 \sin \theta \cos \theta=2 \cos ^{2} \theta+\mathrm{p}^{2}$
$\Rightarrow 2=2 \cos ^{2} \theta+\mathrm{p}^{2}$
$\Rightarrow \mathrm{p}^{2}=2\left(1-\cos ^{2} \theta\right)=2 \sin ^{2} \theta$
$\Rightarrow p=\sqrt{2} \sin \theta$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\sin \theta=\cos ^{2} \theta$, then what is $\cos ^{2} \theta\left(1+\cos ^{2} \theta\right)$ equal to?

Options:
A 1
B 0
C $\cos ^{2} \theta$
D $2 \sin \theta$

The correct answer is:
1

Solution:
Let $\sin \theta=\cos ^{2} \theta$
$\Rightarrow \sin ^{2} \theta=\cos ^{4} \theta$
Consider $\cos ^{2} \theta\left(1+\cos ^{2} \theta\right)=\cos ^{2} \theta+\cos ^{4} \theta$
$=\cos ^{2} \theta+\sin ^{2} \theta \quad$ (using 1)
$=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
If $\sin \alpha+\cos \alpha=\mathrm{p}$, then what is $\cos ^{2}(2 \alpha)$ equal to?

Options:
A $\mathrm{P}^{2}$
B $\mathrm{p}^{2}-1$
C $\mathrm{P}^{2}\left(2-\mathrm{p}^{2}\right)$
D $\mathrm{p}^{2}+1$

The correct answer is:
$\mathrm{P}^{2}\left(2-\mathrm{p}^{2}\right)$

Solution:
\sin \alpha+\cos \alpha=p \\ & \Rightarrow(\sin \alpha+\cos \alpha)^{2}=p^{2} \\ & \Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=p^{2} \\ & \Rightarrow 1+\sin 2 \alpha=p^{2} \\ & \Rightarrow \sin 2 \alpha=\mathrm{p}^{2}-1 \\ & \cos ^{2} 2 \alpha=1-\sin ^{2} 2 \alpha=1-\left(\mathrm{p}^{2}-1\right)^{2} \\ &=1-\left(\mathrm{p}^{4}+1-2 \mathrm{p}^{2}\right)=-\mathrm{p}^{4}+2 \mathrm{p}^{2} \\ &=\mathrm{p}^{2}\left(2-\mathrm{p}^{2}\right) \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\sin (\pi \cos x)=\cos (\pi \sin x)$, then what is one of the values
of $\sin 2 \mathrm{x}$ ?

Options:
A $-\frac{1}{4}$
B $-\frac{1}{2}$
C $-\frac{3}{4}$
D $-1$

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

Solution:
Given that : $\sin (\pi \cos x)=\cos (\pi \sin x)$
$3 u, \cos \left(\frac{\pi}{2}-\pi \cos x\right)=\cos (\pi \sin x)$
$\Rightarrow \frac{\pi}{2}-\pi \cos x=\pi \sin x$
$\Rightarrow \sin x+\cos x=\frac{1}{2}$
Squaring both sides, we get
$\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=\frac{1}{4}$
$\Rightarrow \sin 2 x=\frac{1}{4}-1=-\frac{3}{4}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 1)
If $\sin \beta$ is the harmonic mean of $\sin \alpha$ and $\cos \alpha$, and $\sin \theta$ is the arithmetic mean of $\sin \alpha$ and $\cos \alpha$, then which of the following is/are correct? $[2018-I I]$
1) $\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \sin \beta=\sin 2 \alpha$
2) $\sqrt{2} \sin \theta=\cos \left(\alpha-\frac{\pi}{4}\right)$
Select the correct answer using the code given below:

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
From question, $\sin \alpha, \sin \beta$ and $\cos \alpha$ are in $H . P$
then, $\sin \beta=\frac{2 \sin \alpha \cos \alpha}{\sin \alpha+\cos \alpha}$
and $\sin \alpha, \sin \theta$ and $\cos \alpha$ are in A.P.
then, $2 \sin \theta=\sin \alpha+\cos \alpha$
Statement 1: $\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \cdot \sin \beta$
$=(\sin \alpha+\cos \alpha) \cdot \frac{2 \sin \alpha \cos \alpha}{\sin \alpha+\cos \alpha}$
$=2 \sin \alpha \cos \alpha=\sin 2 \alpha$
Statement $2: \cos \left(\alpha-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\sin \alpha+\cos \alpha)$
$=\frac{2 \sin \theta}{\sqrt{2}}=\sqrt{2} \sin \theta$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 2)
If $\sin \alpha+\sin \beta=0=\cos \alpha+\cos \beta$, where $0 < \beta < \alpha < 2 \pi$, then
which one of the following is correct?

Options:
A $\quad \alpha=\pi-\beta$
B $\alpha=\pi+\beta$
C $\quad \alpha=2 \pi-\beta$
D $2 \alpha=\pi+2 \beta$

The correct answer is:
$\alpha=\pi+\beta$

Solution:
$\sin \alpha+\sin \beta=0=\cos \alpha+\cos \beta$
$\sin \alpha+\sin \beta=0$
$\Rightarrow \sin \alpha=-\sin \beta$
$\Rightarrow \sin \alpha=\sin (\pi+\beta)$
$\Rightarrow \alpha=\pi+\beta$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 1)
If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$, then what is $x$ equal to?

Options:
A 0
B 1
C $\frac{4}{5}$
D $\frac{1}{5}$

The correct answer is:
$\frac{1}{5}$

Solution:
Let $\sin \left[\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x\right]=1$
$\Rightarrow \sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x=\sin ^{-1} 1$
$\Rightarrow \sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x=\frac{\pi}{2}$
$\Rightarrow x=\frac{1}{5}$
$\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2011 (Phase 1)
If $\sin x=\frac{1}{\sqrt{5}}, \sin y=\frac{1}{\sqrt{10}}$, where $0 < x < \frac{\pi}{2}, 0 < y < \frac{\pi}{2}$,
then what is $(\mathrm{x}+\mathrm{y})$ equal to?

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

The correct answer is:
$\frac{\pi}{4}$

Solution:
$\sin x=\frac{1}{\sqrt{5}}, \sin y=\frac{1}{\sqrt{10}}, 0 < x < \frac{\pi}{2}, 0 < y < \frac{\pi}{2}$
$\cos x=\sqrt{1-\sin ^{2} x} \quad \cos y=\sqrt{1-\sin ^{2} y}$
$=\sqrt{1-\frac{1}{5}} \quad=\sqrt{1-\frac{1}{10}}$
$=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}} \quad=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}} .$
$\sin (x+y)=\sin x \cos y+\cos x \sin y$
$=\frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}}+\frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}}$
$=\frac{5}{\sqrt{5} \cdot \sqrt{10}}=\frac{\sqrt{5} \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{10}}=\sqrt{\frac{5}{10}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} .$
$\therefore \quad x+y=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4} .$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 1)
If $\sin x \cos y=\frac{1}{2}$, then what is the value of $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ ?

Options:
A $-4$
B $-2$
C $-6$
D 0

The correct answer is:
$-4$

Solution:
$\sin x \cos y=\frac{1}{2}$
Differentiating both the sides
$\sin x(-\sin y) \frac{d y}{d x}+\cos y \cos x=0$
$\frac{d y}{d x}=\frac{\cos x \cos y}{\sin x \sin y}=\cot x \cot y$
$\frac{d^{2} y}{d x^{2}}=\cot x\left(-\operatorname{cosec}^{2} y\right) \frac{d y}{d x}+\cot y\left(-\operatorname{cosec}^{2} x\right)$
when the point is $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$
then $\mathrm{x}=\frac{\pi}{4}$ and $\mathrm{y}=\frac{\pi}{4}$
$\frac{d y}{d x}=\cot \frac{\pi}{4} \cdot \cot \frac{\pi}{4}=1$
So, $\frac{d^{2} y}{d x^{2}}=\cot \frac{\pi}{4}\left(-\cos e c^{2} \frac{\pi}{4}\right) 1+\cot \frac{\pi}{4}\left(-\cos e c^{2} \frac{\pi}{4}\right)$
$=1(-2)+1 .(-2)=-4$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 1)
If $(\sin x+\operatorname{cosec} x)^{2}+(\cos x+\sec x)^{2}=k+\tan ^{2} x+\cot ^{2} x$
then what is the value of $k ?$

Options:
A \&
B $\underline{7}$
C 4
D 3

The correct answer is:
$\underline{7}$

Solution:
Given, $(\sin x+\operatorname{cosec} x)^{2}+(\cos x+\sec x)^{2}$
$=k+\tan ^{2} x+\cot ^{2} x$
$\Rightarrow \sin ^{2} x+\operatorname{cosec}^{2} x+2 \sin x \operatorname{cosec} x+\cos ^{2} x+\sec ^{2} x$
$+2 \sec x \cos x=k+\tan ^{2} x+\cot ^{2} x$
$\Rightarrow \sin ^{2} x+\operatorname{cosec}^{2} x+2+\cos ^{2} x+\sec ^{2} x+2$
$\quad=k+\tan ^{2} x+\cot ^{2} x$
$(\because \sin x \operatorname{cosec} x=1$ and $\sec x \cos x=1)$
$\Rightarrow 1+\operatorname{cosec}^{2} x-\cot ^{2} x+\sec ^{2} x-\tan ^{2} x+4=k$
$\Rightarrow 1+1+1+4=k \Rightarrow k=7$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2009 (Phase 2)
If $\sin x+\operatorname{cosec} x=2$, then what is the value of $\sin ^{4} x+\operatorname{cosec}^{4} x ?$

Options:
A 2
B 4
C 8
D 16

The correct answer is:
2

Solution:
Given $\sin x+\operatorname{cosec} x=2$
$\begin{aligned} \text { Consider } \sin ^{4} x+\operatorname{cosec}^{4} x=\left(\sin ^{2} x+\operatorname{cosec}^{2} x\right)^{2} \\ &-2\left(\sin ^{2} x \cos ^{2} x\right) \\=& \left.\left[(\sin x+\operatorname{cosec} x)^{2}-2\right)\right]^{2}-2 \\=(4-2)^{2}-2=2 & \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
If $\sin x+\sin y=a$ and $\cos x+\cos y=b$, then
$\tan ^{2}\left(\frac{x+y}{2}\right)+\tan ^{2}\left(\frac{x-y}{2}\right)$ is equal to

Options:
A $\frac{a^{4}+b^{4}+4 b^{2}}{a^{2} b^{2}+b^{4}}$
B $\frac{a^{4}-b^{4}+4 b^{2}}{a^{2} b^{2}+b^{4}}$
C $\frac{a^{4}-b^{4}+4 a^{2}}{a^{2} b^{2}+a^{4}}$
D None of the above

The correct answer is:
$\frac{a^{4}-b^{4}+4 b^{2}}{a^{2} b^{2}+b^{4}}$

Solution:
$\sin x+\sin y=a$
$\Rightarrow 2 \sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{a}$
$\cos x+\cos y=b$
$\Rightarrow \quad 2 \cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{b}$
dividing eq $(1) \&(2)$ $\tan \left(\frac{x+y}{2}\right)=\frac{a}{b}$
Squaring eq (1) \& (2) and adding $4 \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{a}^{2}+\mathrm{b}^{2}$
$\sec ^{2}\left(\frac{x-y}{2}\right)=\frac{4}{a^{2}+b^{2}}$
again, $\tan ^{2}\left(\frac{x+y}{2}\right)+\tan ^{2}\left(\frac{x-y}{2}\right)$
$=\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}+\sec ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)-1$
$=\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}+\frac{4}{\mathrm{a}^{2}+\mathrm{b}^{2}}-1=\frac{\mathrm{a}^{4}-\mathrm{b}^{4}+4 \mathrm{~b}^{2}}{\mathrm{a}^{2} \mathrm{~b}^{2}+\mathrm{b}^{4}}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 2)
If $\sin x+\sin y=a, \cos x+\cos y=b$, then what is the value
of $\cos (x-y)$ ?

Options:
A $a^{2}-1$
B $b^{2}-1$
C $\frac{1}{2}\left(a^{2}+b^{2}-2\right)$
D $\frac{1}{2}\left(a^{2}+b^{2}\right)$

The correct answer is:
$\frac{1}{2}\left(a^{2}+b^{2}-2\right)$

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

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 2)
If $\sin \theta=x+\frac{a}{x}$ for all $x \in R-\{0\}$, then which one of the following is correct?

Options:
A $a \geq 4$
B $a \geq \frac{1}{2}$
C $a \leq \frac{1}{4}$
D $a \leq \frac{1}{2}$

The correct answer is:
$a \leq \frac{1}{4}$

Solution:
Given equation is
$\sin \theta=x+\frac{a}{x}, x \in R-\{0\}$
$\Rightarrow x^{2}+a=x \sin \theta$
$\Rightarrow x^{2}-x \sin \theta+a=0$
Now, discriminant $=\sqrt{\sin ^{2} \theta-4 a}$
For $x$ to be real root, discriminant $\geq 0$
$\Rightarrow \sqrt{\sin ^{2} \theta-4 a} \geq 0$
$\Rightarrow \sin ^{2} \theta-4 a \geq 0 \Rightarrow \sin ^{2} \theta \geq 4 a$
$\Rightarrow \frac{1}{\sin ^{2} \theta} \leq \frac{1}{4 a} \Rightarrow a \leq \frac{\sin ^{2} \theta}{4}$
$\Rightarrow a \leq \frac{1}{4}\left(\because \sin ^{2} \theta\right.$ lies between 0 and 1$)$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
If soap has high alkali content it irritates skin. How can the amount of excess alkali be determined? What can be the source of excess alkali?

Options:

None

Solution:
Acid-base titration can be used to determine the excess amount of alkali in soap. A soap solution is titrated with standard hydrochloric acid, phenolphthalein is used as an indicator. The source of this excess alkali is the alkali left unused when the soap is prepared by hydrolysis of fat.

Topic: Chemistry, Chemistry in Everyday Life