datavorous/entrance-exam-dataset huggingface dataset:
What is $\int \sec ^{n} x \tan x d x$ equal to?

Options:
A $\frac{\sec ^{n} x}{n}+c$
B $\frac{\sec ^{n-1} x}{n-1}+c$
C $\frac{\tan ^{n} x}{n}+c$
D $\frac{\tan ^{n-1} x}{n-1}+c$
Where $^{\prime} \mathrm{c}$ ' is a constant of integration.

The correct answer is:
$\frac{\sec ^{n} x}{n}+c$

Solution:
Let $\mathrm{I}=\int \sec ^{n} x \tan x d x$
Put, sec $x=t \Rightarrow \sec x \tan x d x=d t \Rightarrow d x=\frac{d t}{t}$
$\therefore \quad I=\int t^{n} \cdot \frac{d t}{t}=\int t^{n-1} d t=\frac{t^{n}}{n}+c=\frac{\sec ^{n} x}{n}+c$
where ' $\mathrm{c}$ ' is a constant of integration.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2010 (Phase 2)
What is secondary growth in plants?

Options:
A Increase in cell number
B Increase in length
C Increase in size of the cell
D Increase in girth of the organs

The correct answer is:
Increase in girth of the organs

Solution:
Secondary growth in plants is caused by lateral meristems, vascular cambium, and cork-cambium. It results in an increase in girth of the organs in which they are active.

Topic: Biology, Plant - Growth and Development, NEET
What is secondary metabolism?

Options:

None

Solution:
Secondary metabolism is also known as specialised metabolism. It is a term for pathways and small molecule metabolites produced by metabolism that are not absolutely essential for the organism's survical. Metabolites aid in the growth and development in case of plants. They also facilitie the primary metabolism in plants.

Topic: Biology, Human Health and Diseases
What is secondary productivity in an ecosystem?

Options:
A The rate of production of organic matter during photosynthesis
B The rate of formation of new organic matter by consumers
C The rate of biomass production
D Neither A) nor B)

The correct answer is:
The rate of formation of new organic matter by consumers

Solution:
Secondary productivity is defined as the rate of formation of new organic matter by consumers.

Topic: Biology, Ecosystem, NEET
What is $\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{4}{5}$ equal to ?

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

The correct answer is:
$\pi / 2$

Solution:
$\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}=\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{3}{5}$
$\left(\right.$ let $\left.\sin ^{-1} \frac{4}{5}=\theta \Rightarrow \sin \theta=\frac{4}{5} \Rightarrow \cos \theta=\frac{3}{5}\right)$
$=\frac{\pi}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2014 (Phase 2)
What is $\int \sin ^{-1}(\cos x) \mathrm{dx}$ equal to?

Options:
A $\frac{x \pi}{2}-\frac{x^{2}}{2}+K$
B $\frac{\pi}{2}+\frac{x^{2}}{2}+K$
C $-\frac{x \pi}{2}-\frac{x^{2}}{2}+K$
D $\frac{\pi}{2}-\frac{x^{2}}{2}+K$
Where $\mathrm{K}$ is a constant of integration.

The correct answer is:
$\frac{x \pi}{2}-\frac{x^{2}}{2}+K$

Solution:
Let $I=\int \sin ^{-1}(\cos x) d x$
$=\int \sin ^{-1}\left[\sin \left(\frac{\pi}{2}-x\right)\right] d x=\int\left(\frac{\pi}{2}-x\right) d x$
$=\frac{x \pi}{2}-\frac{x^{2}}{2}+K$
where $K$ is a constant of integration.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2011 (Phase 1)
What is $\sin ^{-1} \sin \frac{3 \pi}{5}$ equal to ?

Options:
A $\frac{3 \pi}{5}$
B $\frac{2 \pi}{5}$
C $\frac{\pi}{5}$
D None of these

The correct answer is:
$\frac{2 \pi}{5}$

Solution:
$\sin ^{-1} \sin \frac{3 \pi}{5}=\sin ^{-1} \sin \left(\pi-\frac{2 \pi}{5}\right)$
$=\sin ^{-1} \sin \frac{2 \pi}{5}=\frac{2 \pi}{5}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2014 (Phase 1)
What is $\sin ^{2} 20^{\circ}+\sin ^{2} 70^{\circ}$ equal to ?

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

The correct answer is:
1

Solution:
$\sin ^{2} 20^{\circ}+\sin ^{2} 70^{\circ}=\sin ^{2} 20^{\circ}+\sin ^{2}\left(90^{\circ}-20^{\circ}\right)$
$=\sin ^{2} 20^{\circ}+\cos ^{2} 20^{\circ}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 2)
What is $\sin ^{2}(3 \pi)+\cos ^{2}(4 \pi)+\tan ^{2}(5 \pi)$ equal to $?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
$\sin ^{2}(3 \pi)+\cos ^{2}(4 \pi)+\tan ^{2}(5 \pi)$
$=\sin ^{2}(3 \pi)+\cos ^{2}(\pi+3 \pi)+\tan ^{2}(5 \pi)$
$=\left(\sin ^{2}(3 \pi)+\cos ^{2}(3 \pi)\right)+\tan ^{2}(2 \times 2 \pi+\pi)$
$=1+\tan ^{2} \pi=\sec ^{2} \pi=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 1)
What is $\sin ^{2} 66 \frac{1^{\circ}}{2}-\sin ^{2} 23 \frac{1}{2}^{\circ}$ equal to ?

Options:
A $\sin 47^{\circ}$
B $\cos 47^{\circ}$
C $2 \sin 47^{\circ}$
D $2 \cos 47^{\circ}$

The correct answer is:
$\cos 47^{\circ}$

Solution:
$\begin{aligned} \sin ^{2} 66 \frac{1^{\circ}}{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\=&\left[\sin \left(90^{\circ}-23 \frac{1^{\circ}}{2}\right)\right]^{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\=& \cos ^{2} 23 \frac{1^{\circ}}{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\=& \cos 2\left(23 \frac{1^{\circ}}{2}\right)=\cos 47^{\circ} \\\left(\because \cos 2 \mathrm{~A}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}\right) \\=\cos \left[2 \times\left(\frac{47}{2}\right)\right]=\cos 47^{\circ} \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 2)
What is $\sin ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~B}$ equal to?

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

The correct answer is:
$1 / 2$

Solution:
$\sin ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~B}=\sin ^{2} 60^{\circ}-\sin ^{2} 30^{\circ}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What is $\int \sin ^{2} x d x+\int \cos ^{2} x d x$ equal to?

Options:
A $x+c$
B $\frac{x^{2}}{2}+c$
C $x^{2}+c$
D None of these
Where $c$ is an arbitary constant.

The correct answer is:
$x+c$

Solution:
$\int \sin ^{2} x d x+\int \cos ^{2} x d x=\int\left(\sin ^{2} x+\left(\cos ^{2} x\right) d x\right.$
$\int \mathrm{dx}=\mathrm{x}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2013 (Phase 2)
What is $\int \sin 2 x \ell n(\cot x) d x$ equal to ?

Options:
A 0
B $\pi \ln 2$
C $-\pi \ln 2$
D $\frac{\pi \ell \mathrm{n} 2}{2}$

The correct answer is:
0

Solution:
Let $\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sin 2 \mathrm{x} \ln (\cot \mathrm{x}) \mathrm{d} \mathrm{x}$
$=\left(\because \cot x=\frac{\cos x}{\sin x}\right)$
$=\int_{0}^{\frac{\pi}{2}} \sin 2 \mathrm{x} \ell \mathrm{n}(\cos \mathrm{x}) \mathrm{d} \mathrm{x}-\int_{0}^{\frac{\pi}{2}} \sin 2 \mathrm{x} \ln (\sin \mathrm{x}) \mathrm{dx}$
$=\int_{0}^{\frac{\pi}{2}} \sin \left[2\left(\frac{\pi}{2}+\mathrm{x}\right)\right] \ln \cos \left(\frac{\pi}{2}+\mathrm{x}\right) \mathrm{d} \mathrm{x}$
$=\int_{0}^{\frac{\pi}{2}} \sin 2 \mathrm{x} \ln (\sin \mathrm{x}) \mathrm{dx}$
$=\int_{0}^{\frac{\pi}{2}} \sin 2 \mathrm{x} \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}-\int_{0}^{\frac{\pi}{2}} \sin 2 \mathrm{x} \ln (\sin \mathrm{x}) \mathrm{dx}=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2012 (Phase 2)
What is $\int \sin ^{3} x \cos x d x$ equal to?

Options:
A $\cos ^{4} x+c$
B $\sin ^{4} x+c$
C $\frac{\left(1-\sin ^{2} x\right)^{2}}{4}+c$
D $\frac{\left(1-\cos ^{2} x\right)^{2}}{4}+c$
Where $c$ is the constant of integration.

The correct answer is:
$\frac{\left(1-\cos ^{2} x\right)^{2}}{4}+c$
Where $c$ is the constant of integration.

Solution:
Let $t=\sin x \Rightarrow d t=\cos x d x$
$\int \sin ^{3} x \cos x d x=\int t^{3} d t=\frac{t^{4}}{4}+c=\frac{\sin ^{4} x}{4}+c$
$=\frac{\left(1-\cos ^{2} x\right)^{2}}{4}+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2018 (Phase 2)
What is $\sin 105^{\circ}+\cos 105^{\circ}$ equal to?

Options:
A $\sin 50^{\circ}$
B $\cos 50^{\circ}$
C $\frac{1}{\sqrt{2}}$
D 0

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

Solution:
\sin 105^{\circ}+\cos 105^{\circ} \\
=\frac{\sin \left(\frac{2 x}{2}\right)}{\cos \left(\frac{2 x}{2}\right)}=\frac{\sin x}{\cos x}=\tan x . \\
=\left(\sin 60^{\circ} \cdot \sin 45^{\circ}+\cos 60^{\circ} \cdot \cos 45^{\circ}\right)+\left(\cos 60^{\circ} \cos 45^{\circ}\right. \\
\left.-\sin 60^{\circ} \sin 45^{\circ}\right) \\
=\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 1)
What is $\sin \left[\cot ^{-1}\left\{\cos \left(\tan ^{-1} x\right)\right]\right.$ where $x>0$, equal to?

Options:
A $\sqrt{\frac{\left(x^{2}+1\right)}{\left(x^{2}+2\right)}}$
B $\sqrt{\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)}}$
C $\frac{\left(x^{2}+1\right)}{\left(x^{2}+2\right)}$
D $\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)}$

The correct answer is:
$\sqrt{\frac{\left(x^{2}+1\right)}{\left(x^{2}+2\right)}}$

Solution:
Let $\alpha=\tan ^{-1} \mathrm{x} \Rightarrow \tan \alpha=\mathrm{x}$
then $\cos \alpha=\frac{1}{\sqrt{1+\tan ^{2} \alpha}}=\frac{1}{\sqrt{1+\mathrm{x}^{2}}}$
$\Rightarrow \cos \left(\tan ^{-1} x\right)=\left\{\frac{1}{\sqrt{1+x^{2}}}\right\}$
so, $\cot ^{-1}\left\{\cos \left(\tan ^{-1} x\right)\right\}=\cot ^{-1}\left\{\frac{1}{\sqrt{1+x^{2}}}\right\}$
Let $\cot ^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)=\beta$
$\Rightarrow \cot \beta=\frac{1}{\sqrt{1+\mathrm{x}^{2}}}$
and $\sin \beta=\frac{1}{\sqrt{1+\cot ^{2} \beta}}=\frac{\sqrt{1+\mathrm{x}^{2}}}{\sqrt{\mathrm{x}^{2}+1+1}}=\sqrt{\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}+2}}$
$\Rightarrow \sin \left[\cot ^{-1}\left\{\cos \left(\tan ^{-1}\right)\right\}\right]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2007 (Phase 1)
What is $\sin \left[\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)\right]$ equal to ?

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

The correct answer is:
1

Solution:
$\quad \sin \left[\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)\right]$
$=\sin \left[\sin ^{-1}\left\{\frac{3}{5} \sqrt{1-\frac{16}{25}}+\frac{4}{5} \sqrt{1-\frac{9}{25}}\right\}\right]$
$=\sin \left[\sin ^{-1}\left\{\frac{3}{5} \times \frac{3}{5}+\frac{4}{5} \times \frac{4}{5}\right\}\right]$
$=\sin \left[\sin ^{-1}\left\{\frac{9}{25}+\frac{16}{25}\right\}\right]=\sin \left[\sin ^{-1}(1)\right]$
$=\sin \frac{\pi}{2}=1$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2012 (Phase 2)
What is $\int \sin x \log (\tan x) d x$ equal to?

Options:
A $\cos x \log \tan x+\log \tan (x / 2)+c$
B $-\cos x \log \tan x+\log \tan (x / 2)+c$
C $\cos x \log \tan x+\log \cot (x / 2)+c$
D $-\cos x \log \tan x+\log \cot (x / 2)+c$

The correct answer is:
$-\cos x \log \tan x+\log \tan (x / 2)+c$

Solution:
$\int \sin x \log (\tan x) d x$
$=-\cos x \log \tan x-\int(-\cos x) \frac{1}{\tan x} \cdot \sec ^{2} x d x$
$=-\cos x \log \tan x+\int \frac{1}{\sin x} d x$
$=-\cos x \log (\tan x)+\int \frac{1+\tan ^{2} \frac{x}{2}}{2 \tan \frac{x}{2}} d x$
Let $t=\tan \frac{x}{2}$
$\Rightarrow \frac{d x}{d t}=\frac{2}{1+t^{2}} \Rightarrow d x=\frac{2}{1+t^{2}} \cdot d t$
So, $-\cos x \cdot \log (\tan x)+\int \frac{1+\tan ^{2} \frac{x}{2}}{2 \tan \frac{x}{2}} \cdot d x$
$=-\cos x \cdot \log (\tan x)+\int \frac{1+t^{2}}{2 t} \cdot \frac{2}{1+t^{2}} \cdot d t$
$=-\cos x \log \tan x+\int \frac{1}{t} \cdot d t$
$=-\cos x \log \tan x+\log (t)+c$
$=-\cos x \log \tan x+\log \tan \left(\frac{x}{2}\right)+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2008 (Phase 2)
What is spermatogenesis? Briefly describe the process of spermatogenesis.

Options:

None

Solution:
Spermatogenesis is the process of producing sperms with half the number of chromosomes (haploid) as somatic cells. It occurs in seminiferous tubules. Sperm production begins at puberty continues throughout life with several hundred million sperms being produced each day. Once sperm are formed they move into the epididymis, where they mature and are stored.
During spermatogenesis one spermatogonium produces 4 sperms. Spermatogenesis completes through the following phases - multiplicative phase, growth phase, maturation phase \& spermiogenesis. In multiplicative phase the sperm mother cells divide by mitosis \& produce spermatogonia. The spermatogonia grow in size to form large primary spermatocytes by getting nourishment from sertoli cells in growth phase. Maturation phase involves meiosis I in which primary spermatocytes divide to produce secondary spermatocyte and meiosis II which produces spermatids. Thus each primary spermatocyte gives rise to four haploid spermatids. Spermiogenesis or spermateliosis is process of formation of flagellated spermatozoa from spermatids. Spermiogenesis begins in the seminiferous tubules but usually completed in epididymis.

Topic: Biology, Human Reproduction
What is standard $\mathrm{N} \equiv \mathrm{N}$ bond enthalpy from following reaction? $\mathrm{N}_{2_{(\mathrm{g})}}+3 \mathrm{H}_{2_{(\mathrm{g})}} \longrightarrow 2 \mathrm{NH}_{3_{(\mathrm{g})}} \Delta \mathrm{H}^{\circ}=-83 \mathrm{KJ}$
$\left(\Delta \mathrm{H}^{\circ}{ }_{(\mathrm{H}-\mathrm{H})}=435 \mathrm{~kJ}, \Delta \mathrm{H}^{\circ}{ }_{(\mathrm{N}-\mathrm{H})}=389 \mathrm{~kJ}\right)$

Options:
A 435 kJ
B $2334 \mathrm{~kJ}$
C $946 \mathrm{~kJ}$
D $1305 \mathrm{~kJ}$

The correct answer is:
$946 \mathrm{~kJ}$

Solution:
(C)
$\mathrm{N}_{2(8)}+3 \mathrm{H}_{2(8)} \longrightarrow 2 \mathrm{NH}_{3(8)} \quad \Delta \mathrm{H}^{\circ}=-83 \mathrm{~kJ}$
$\Delta \mathrm{H}^{0}=\sum \Delta \mathrm{H}^{0}$ (reactant bonds) $-\sum \Delta \mathrm{H}^{0}$ (product bonds)
$\left.\therefore \Delta H^{0}=\left[\Delta H^{0}{ }_{(N=N)}^{0}+3 \Delta H^{0}{ }^{0} H-H\right)\right]-\left[6 \Delta H^{0}{ }^{0}(N-H)\right]$
$\therefore-83=\Delta H_{(N=N)}^{0}+3(435)-6(389)$
$\therefore-83=\Delta H_{(N=N)}^{0}+1305-2334$
$\therefore \Delta H^{0}(N \sim N)=-83-1305+2334=946 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), MHT CET, MHT CET 2020 (19 Oct Shift 2)
What is standard reduction potential of $\mathrm{Cu}^{2+} \mid \mathrm{Cu}_{(\mathrm{s})}$ if $\mathrm{E}^{\circ}$ of following cell is 0.46 ?
$\mathrm{Cu}_{(\mathrm{s})}\left|\mathrm{cu}_{(\mathrm{aq})}^{2+} \| \mathrm{Ag}_{(\mathrm{aq})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})} ; \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.80 \mathrm{v} \mid$

Options:
A $1.56 \mathrm{~V}$
B $0.34 \mathrm{~V}$
C $1.44 \mathrm{~V}$
D $1.26 \mathrm{~V}$

The correct answer is:
$0.34 \mathrm{~V}$

Solution:
$\begin{aligned} & \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\mathrm{Ag}}^{\circ}-\mathrm{E}_{\mathrm{cu}}^{\circ} \\ & 0.46=0.80-\mathrm{E}_{\text {cell }}^{\circ} \\ & \mathrm{E}_{\mathrm{cu}}^{\circ}=0.34 \mathrm{v}\end{aligned}$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2022 (05 Aug Shift 1)
What is stomatal apparatus? Explain the structure of stomata with a labelled diagram.

Options:

None

Solution:
Several minute openings or stomata are found on the epidermis of all the green aerial parts of plants, but are abundant on the lower surface on the leaves as they regulate the process of transpiration. A large number of stomata occur on the upper surface of leaves of aquatic plants. Each stomata is surrounded by two cells known as the guard cells. In the dicotyledons plants these are bean shaped, but in sedges and grasses these are dumb-bell shaped. The guard cell are living. Their outer walls are thin where as inner ones surrounding the aperture are highly thickened. Due to this variation in the thickening, the guard cell may becomes turgid and flaccid, depending upon the supply of water in them, which makes the opening and closing of stomata possible. Some times a few neighbouring epidermal cells in vicinity of guard cells become specialized in their shape and size and contents. These are known as subsidiary cells. The stomatal aperture, guard cells and the surrounding subsidiary cell are together called stomatal apparatus.

Topic: Biology, Anatomy of Flowering Plants
What is sum to the 100 terms of the series $9+99+999+\ldots ?$

Options:
A $\frac{10}{9}\left(10^{100}-1\right)-100$
B $\frac{10}{9}\left(10^{99}-1\right)-100$
C $100\left(100^{10}-1\right)$
D $\frac{9}{100}\left(10^{100}-1\right)$

The correct answer is:
$\frac{10}{9}\left(10^{100}-1\right)-100$

Solution:
Let $\mathrm{S}=9+99+999+\ldots$
$=\left(10^{1}-1\right)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots$
$=\left(10+10^{2}+10^{3}+\ldots\right)-(1+1+1+\ldots 100$ times $)$
$=\frac{10\left(10^{100}-1\right)}{10-1}-100 \quad\left(\because 10,10^{2}, 10^{3} \ldots \ldots .\right.$ is GP. with
$\left.\mathrm{a}=10, \mathrm{r}=10, \mathrm{~S}_{100}=\frac{a\left(r^{100}-1\right)}{r-1}\right)$
$=\frac{10}{9}\left(10^{100}-1\right)-100$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
What is $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}$ equal to ?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
\quad & \frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}} \\ \text { 147. (a) } & \sin ^{2} 20^{\circ}+\sin ^{2} 70^{\circ}=\sin ^{2} 20^{\circ}+\sin ^{2}\left(90^{\circ}-20^{\circ}\right) \\=\sin ^{\circ} 20^{\circ}+\cos ^{2} 20^{\circ}=& 1 \\ \text { 148. (d) }\left(1-\sin ^{2} \theta\right)\left(1+\tan ^{2} \theta\right) &=\cos ^{2} \theta \cdot \sec ^{2} \theta \\=\cos ^{2} \theta \cdot \frac{1}{\cos ^{2} \theta}=1 \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 2)
What is $\frac{1-\tan 2^{0} \cot 62^{\circ}}{\tan 152^{\circ}-\cot 88^{\circ}}$ equal to?

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

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

Solution:
$\begin{aligned} L=\frac{1-\tan 2^{\circ} \cot 62^{\circ}}{\tan 152^{\circ}-\cot 88^{\circ}} &=\frac{1-\tan 2^{\circ} \cot (90-28)^{\circ}}{\tan (180-28)^{\circ}-\cot (90-2)^{\circ}} \\ \Rightarrow L=\frac{1-\tan 2^{\circ} \tan 28^{\circ}}{-\tan 28^{\circ}-\tan 2^{\circ}}=-\left[\frac{1-\tan 2^{\circ} \tan 28^{\circ}}{\tan 2^{\circ}+\tan 28^{\circ}}\right] \\ \Rightarrow L=-\frac{1}{\tan (2+28)^{\circ}}=-\frac{1}{\tan 30^{\circ}}=-\sqrt{3} \\\left[\cdots \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\right] \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
What is $\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)$ equal to ?

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

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

Solution:
$\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)$
$=\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)=\tan ^{-1}(1)=\frac{\pi}{4}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2013 (Phase 2)
What is $\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{3}{5}\right)$ equal to?

Options:
A 0
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

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

Solution:
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{3}{5}\right)$
We know, $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
So, $\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{3}{5}}{1-\left(\frac{1}{4}\right)\left(\frac{3}{5}\right)}\right)$
$=\tan ^{-1}\left(\frac{\frac{5+12}{20}}{1-\frac{3}{20}}\right)$
$=\tan ^{-1}\left(\frac{\frac{17}{20}}{\frac{17}{20}}\right)=\tan ^{-1}(1)=\frac{\pi}{4}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2018 (Phase 1)
What is tan $15^{\circ}$ equal to ?

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

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

Solution:
$\tan 15^{\circ}=\sqrt{\frac{1-\cos 30^{\circ}}{1+\cos 30^{\circ}}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}$
$=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2-\sqrt{3})^{2}}{1}}=2-\sqrt{3}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 2)
What is $\int \tan ^{-1}(\sec x+\tan x) \mathrm{dx}$ equal to?

Options:
A $\frac{\pi x}{4}+\frac{x^{2}}{4}+c$
B $\frac{\pi x}{2}+\frac{x^{2}}{4}+c$
C $\frac{\pi x}{4}+\frac{\pi x^{2}}{4}+c$
D $\frac{\pi x}{4}-\frac{x^{2}}{4}+c$

The correct answer is:
$\frac{\pi x}{4}+\frac{x^{2}}{4}+c$

Solution:
$\int \tan ^{-1}(\sec x+\tan x) d x$
$=\int \tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) d x$ $\quad \because \sec x+\tan x=\frac{1+\sin x}{\cos x}$
$=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) \mathrm{d} x$
$=\frac{\pi x}{4}+\frac{x^{2}}{4}+c .$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2017 (Phase 2)
What is $\int \tan ^{2} x \sec ^{4} x d x$ equal to?

Options:
A $\frac{\sec ^{5} x}{5}+\frac{\sec ^{3} x}{3}+c$
B $\frac{\tan ^{5} x}{5}+\frac{\tan ^{3} x}{3}+c$
C $\frac{\tan ^{5} x}{5}+\frac{\sec ^{3} x}{3}+c$
D $\frac{\sec ^{5} x}{5}+\frac{\tan ^{3} x}{3}+c$

The correct answer is:
$\frac{\tan ^{5} x}{5}+\frac{\tan ^{3} x}{3}+c$

Solution:
Let $I=\int \tan ^{2} x \sec ^{4} x d x$
Let $\tan x=t$
$\Rightarrow \sec ^{2} x d x=d t$
$\begin{aligned} \therefore \quad & \mathrm{I}=\int \tan ^{2} x \cdot \sec ^{2} x \cdot \sec ^{2} x \cdot d x \\ &=\int \tan ^{2} x\left(1+\tan ^{2} x\right) \sec ^{2} x \cdot d x \end{aligned}$
$\begin{aligned} \therefore \quad I &=\int \mathrm{t}^{2}\left(1+t^{2}\right) d t=\int\left(t^{2}+t^{4}\right) d t \\ &=\frac{t^{5}}{5}+\frac{t^{3}}{3}+c=\frac{\tan ^{5} x}{5}+\frac{\tan ^{3} x}{3}+c \end{aligned}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2009 (Phase 1)
What is $\tan ^{4} \mathrm{~A}-\sec ^{4} \mathrm{~A}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}$ equal to?

Options:
A $\overline{0}$
B 1 $[2013-I]$
C 2
D $-1$

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

Solution:
Consider $\tan ^{4} \mathrm{~A}-\sec ^{4} \mathrm{~A}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}$
$=\left(\tan ^{2} \mathrm{~A}\right)^{2}-\left(\sec ^{2} \mathrm{~A}\right)^{2}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} .$
$=\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)\left(\tan ^{2} \mathrm{~A}-\sec ^{2} \mathrm{~A}\right)$
$+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}$
$=-\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
What is tan $\left(7 \frac{1}{2}\right)^{\circ}$ equal to?

Options:
A $\sqrt{6}+\sqrt{3}-\sqrt{2}+2$
B $\sqrt{6}+\sqrt{3}+\sqrt{2}+2$
C $\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
D $\sqrt{6}+\sqrt{3}+\sqrt{2}-2$

The correct answer is:
$\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

Solution:
$\tan \left(7 \frac{1}{2}\right)^{-}=\frac{\sin \left(7 \frac{1}{2}\right)^{\circ}}{\cos \left(7 \frac{1}{2}\right)}^{\prime}$
Multiply and divide by $2 \sin \left(7 \frac{1}{2}\right)$; we get
$\frac{2 \sin ^{2}\left(7 \frac{1}{2}\right)^{\circ}}{2 \sin \left(7 \frac{1}{2}\right)^{\circ} \cos \left(7 \frac{1}{2}\right)^{\circ}}=\frac{2 \sin ^{2}\left(\frac{15}{2}\right)^{\circ}}{2 \sin \left(\frac{15}{2}\right)^{\circ} \cos \left(\frac{15}{2}\right)^{\circ}}$
$\left(\because \cos 2 \theta=1-2 \sin ^{2} \theta\right.$ and $\left.\sin 2 \theta=2 \sin \theta \cos \theta\right)$
$=\frac{1-\cos \left(2 \times \frac{15}{2}\right)^{\circ}}{\sin \left(2 \times \frac{15}{2}\right)^{\circ}}$
$=\frac{1-\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{1-\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}=\sqrt{2}(\sqrt{3}+1)-(2+\sqrt{3})$
$=\sqrt{6}+\sqrt{2}-2-\sqrt{3}=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 1)
What is $\tan \left(\frac{\pi}{12}\right)$ equal to?

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

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

Solution:
$\quad \tan \frac{\pi}{12}=\tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{\tan \pi / 3-\tan \pi / 4}{1+\tan \pi / 3 \tan \pi / 4}$
$=\frac{\sqrt{3}-1-3+\sqrt{3}}{(1)^{2}-(\sqrt{3})^{2}}=\frac{2(\sqrt{3}-2)}{-2}=2-\sqrt{3}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)