datavorous/entrance-exam-dataset huggingface dataset:
Correct order of reactivity of $\mathrm{CH}_3 \mathrm{CHO}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{COCH}_3$ and $\mathrm{CH}_3 \mathrm{COCH}_3$ is

Options:
A $\mathrm{CH}_3 \mathrm{CHO}\gt\mathrm{CH}_3 \mathrm{COCH}_3\gt\mathrm{CH}_3 \mathrm{COC}_2 \mathrm{H}_5$
B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{COCH}_3\gt\mathrm{CH}_3 \mathrm{COCH}_3\gt\mathrm{CH}_3 \mathrm{CHO}$
C $\mathrm{CH}_3 \mathrm{COCH}_3\gt\mathrm{CH}_3 \mathrm{CHO}\gt\mathrm{C}_2 \mathrm{H}_5 \mathrm{COCH}_3$
D $\mathrm{CH}_3 \mathrm{COCH}_3\gt\mathrm{C}_2 \mathrm{H}_5 \mathrm{COCH}_3\gt\mathrm{CH}_3 \mathrm{CHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHO}\gt\mathrm{CH}_3 \mathrm{COCH}_3\gt\mathrm{CH}_3 \mathrm{COC}_2 \mathrm{H}_5$

Solution:
Increasing alkyl group the reactivity decreases.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Correct order of spin only magnetic moment of the following complex ions is:

(Given At. No. Fe: 26, Co: 27)

Options:
A FeF63->CoF63->CoC2O433-
B CoC2O433->CoF63->FeF63-
C FeF63->CoC2O433->CoF63-
D CoF63->FeF63->CoC2O433-

The correct answer is:
FeF63->CoF63->CoC2O433-

Solution:
(1) FeF63-: Fe3+=3d5 (ΔO<P)

Number of unpaired e-=5

μ = n(n+2) BM

∴μ=35 BM

(2) CoF63-: Co3+=3d6ΔO<P

Number of unpaired e-=4

μ = n(n+2) BM

∴μ=24 BM

(3) CoC2O433-: Co3+=3d6ΔO>P

Number of unpaired e-=0

μ = n(n+2) BM

∴μ=0 BM

So, the correct order is FeF63->CoF63->CoC2O433-

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Correct order of stability of carbanion is -

Options:
A \(d > a > c > b\)
B a \( > \) b \( > \) c \( > \) d
C \(d > c > b > a\)
D \(c > b > d > \) a

The correct answer is:
\(d > c > b > a\)

Solution:
As we know compound (d) is aromatic and the compound (a) is anti-aromatic. Hence compound (d) is most stable and compound (a) is least stable among these in compound (b) and (c) carbon atom of that positive charge is $\mathrm{sp}^3$ hybridised they on the basis of angle strain theory compound (c) is more stable than compound (b).

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Correct pair of compounds which gives blue colouration/precipitate and white precipitate, respectively, when their Lassaigne's test is separately done is

Options:
A NH- $\mathrm{NH}_{2}$, HCl and CICH, COOH
B $\mathrm{NH}_{2} \mathrm{CSNH}_{2}$ and $\mathrm{PhCH}_{2} \mathrm{Cl}$
C $\mathrm{NH}_{2} \mathrm{NH}_{2}, \mathrm{COOH}$ and $\mathrm{NH}_{2} \mathrm{CONH}_{2}$
D

The correct answer is:

Solution:
$\begin{array}{ll}\text {Hydrozine, } & \mathrm{NH}_{2} \cdot \mathrm{NH}_{2} \text { does not respond }\end{array}$
Lassaigne's test because it does not contain any carbon and hence, NaCN is not formed. Compound, $\mathrm{NH}_{2}-\mathrm{CS}-\mathrm{NH}_{2}$ contains both $\mathrm{N}$ and $\mathrm{S}$, hence, it will give red colour in Lassaigne test.

Topic: Chemistry, General Organic Chemistry, JEE Main
Correct relationship between heat of fusion $\left(\Delta H_{f a r}\right)$, heat of vaporization $\left(\Delta H_{v a p}\right)$ and heat of sublimation $\left(\Delta H_{sab}\right)$ is-

Options:
A $\Delta H_{\text {far }}=\Delta H_{v a p}+\Delta H_{s a b}$-
B $\Delta H_{\text {vap }}=\Delta H_{\text {far }}+\Delta H_{s a b}$
C $\Delta H_{s a b}=\Delta H_{v a p}+\Delta H_{\text {far }}$
D $\Delta H_{s a b}=\Delta H_{v a p}-\Delta H_{\text {far }}$

The correct answer is:
$\Delta H_{s a b}=\Delta H_{v a p}+\Delta H_{\text {far }}$

Solution:
$\Delta H_{s a b}=\Delta H_{v a p}+\Delta H_{\text {fos }}$

Topic: Chemistry, Thermodynamics (C), JEE Main
Correct sequence of CO bond order in given compounds is
(P) Fe(CO)5 (Q) CO (R) H3B←CO (S) [Mn(CO)5]-

Options:
A P>R>S>Q
B S>P>R>Q
C Q>S>P>R
D R>Q>P>S

The correct answer is:
R>Q>P>S

Solution:
Q
:C-≡O+:  B.O. =3
P &S
M⇄C=O  B.O.=2
R
H3B-←C≡O+  B.O.=3

Topic: Chemistry, Coordination Compounds, JEE Main
Correct set up to verify Ohm's law is :

Options:
A

B

C

D

The correct answer is:

Solution:
In ohm's law, we check $V=I R$ where $I$ is the corrent flowing through a resistor and $\mathrm{V}$ is the potential difference across that resistor. Only option (a) fits the above criteria. Remember that ammeter is connected in series with resistance and voltmeter parallel with the resistance.

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2013 (23 Apr Online)
Correct statement about photo-chemical smog is

Options:
A It occurs in humid climate.
B It is a mixture of smoke, fog and SO2.
C It is reducing smog.
D It results from reaction of unsaturated hydrocarbons.

The correct answer is:
It results from reaction of unsaturated hydrocarbons.

Solution:
Photochemical smog occurs in warm, dry and sunny climate.

Photochemical smog results from the action of sunlight on unsaturated hydro carbons and nitrogen oxide.

The common components of photochemical smog are ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate (PAN).

It is oxidising smog.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Correct statement about smog is

Options:
A NO2 is present in classical smog
B Both NO2 and SO2 are present in classical smog
C Photochemical smog has high concentration of oxidizing agents
D Classical smog also has high concentration of oxidizing agents

The correct answer is:
Photochemical smog has high concentration of oxidizing agents

Solution:
A. Classical smog occurs in cool humid climate. It is a mixture of smoke, fog, and sulphur dioxide. Chemically it is a reducing mixture and so it is called reducing smog. So statement A is incorrect.

B. Classical smog is a mixture of smoke, fog and SO2. NO2 is not present in classical smog. So, statement B is incorrect.

C. Photochemical smog occurs in warm, dry and sunny climate. The main components of the photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Photochemical smog has high concentration of oxidising agents and is, therefore, called as oxidising smog. So, statement C is correct.

D. Classical smog is a reducing mixture, so it is called as reducing smog. Hence, statement D is incorrect.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Correct statement about the given chemical reaction is

Options:
A Reaction is possible and compound (A) will be major product.
B -N¨H2 group is ortho and para directive, so product (B) is not possible.
C Reaction is possible and compound (B) will be the major product.
D The reaction will form sulphonated product instead of nitration.

The correct answer is:
Reaction is possible and compound (A) will be major product.

Solution:
Reaction is possible and compound (A) will be major product.

Due to formation of anilinium ion by acid base reaction m-product is form as considerable amount.

Nitration of aniline in strong acidic medium also gives m-nitroaniline because in acidic (strong) medium aniline is present as anilinium ion.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Correct statement is :

Options:
A An average human being consumes more food than air
B An average human being consumes nearly 15 times more air than food
C An average human being consumes equal amount of food and air
D An average human being consumes 100 times more air than food

The correct answer is:
An average human being consumes nearly 15 times more air than food

Solution:
An average human being requires nearly 12-15 times more air than the food. So, even small amounts of pollutants in the air become significant compared to similar levels present in the food. Pollutants can be degradable, like discarded vegetables which rapidly break down by natural processes.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Correct statements among regarding silicones are:
(a) They are polymers with hydrophobic character.
(b) They are biocompatible.
(c) In general, they have high thermal stability and low dielectric strength.
(d) Usually, they are resistant to oxidation and used as greases.

Options:
A (a),Â (b) and (c) only
B (a) and (b) only
C (a), (b), (c) and (d)
D (a), (b) and (d) only

The correct answer is:
(a), (b), (c) and (d)

Solution:
Silicones with medium length chains behave as viscous oils, jellies and greases. They are chemically inert, i.e., resistant to oxidation. Since silicones are surrounded by non-polar alkyl groups, hence, are hydrophobic in nature. Being biocompatible, they are used in surgical and cosmetic implants. Hence, all options are correct.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Correct statements among regarding silicones are:
(a) They are polymers with hydrophobic character.
(b) They are biocompatible.
(c) In general, they have high thermal stability and low dielectric strength.
(d) Usually, they are resistant to oxidation and used as greases.

Options:
A (a),Â (b) and (c) only
B (a) and (b) only
C (a), (b), (c) and (d)
D (a), (b) and (d) only

The correct answer is:
(a), (b), (c) and (d)

Solution:
Silicones with medium length chains behave as viscous oils, jellies and greases. They are chemically inert, i.e., resistant to oxidation. Since silicones are surrounded by non-polar alkyl groups, hence, are hydrophobic in nature. Being biocompatible, they are used in surgical and cosmetic implants. Hence, all options are correct.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (09 Jan Shift 1)
Correct structure of γ-methylcyclohexane carbaldehyde is

Options:
A

B

C

D

The correct answer is:

Solution:
Aldehydes and ketones are often called by their common names instead of IUPAC names. The common names of most aldehydes are derived from the common names of the corresponding carboxylic acids by replacing the ending –ic of acid with aldehyde. At the same time, the names reflect the Latin or Greek term for the original source of the acid or aldehyde.

The location of the substituent in the carbon chain is indicated by Greek letters α, β, γ, δ, etc. The α-carbon being the one directly linked to the aldehyde group, β carbon the next, and so on.

In the above structure methyl group is attached to the γ Position to the aldehyde group.It has a cyclohexane ring

γ-methyl cyclohexane carbaldehyde

The IUPAC name of the above compound is 3-Methylcyclohexanecarbaldehyde

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Correctly identify the diagram given below:

Options:
A Old dicot root
B Young monocot stem
C Dicot leaf
D Monocot leaf.

The correct answer is:
Dicot leaf

Solution:
Cork lacks stomata but consists of lenticels which carries out gaseous exchange. Passage cells are present in endodermis of root which maintains unidirectional flow of water from cortex to xylem, it also acts as check point which controls type and quantity of ions entering the xylem. The shoot apical meristem lacks quiescent centre, it is present in root apex only.

Topic: Biology, Anatomy of Flowering Plants, NEET
Correlate the hormone levels with its levels for given below days?

(i) 1-5 days
(ii) 12-14 days
(iii) 25-28 days (if the ovum is not fertilised).

Options:
A (i) LH decreases and FSH increases.
(ii) LH increases and FSH decreases.
(iii) LH level maintained and FSH level increases.
B (i) LH increases and FSH decreases.
(ii) LH decreases and FSH increases.
(iii) LH level increases and FSH level maintained.
C (i) LH increases and FSH decreases.
(ii) LH peaks and FSH peaks.
(iii) LH level decreases and FSH level maintained.
D (i) LH peaks and FSH peaks.
(ii) LH increases and FSH decreases.
(iii) LH level decreases and FSH level maintained.

The correct answer is:
(i) LH increases and FSH decreases.
(ii) LH peaks and FSH peaks.
(iii) LH level decreases and FSH level maintained.

Solution:
(i) LH increases and FSH decreases to ensure the maturity of the graffian follicle. This time, the menstrual flow occurs.
(ii) LH peaks and FSH peaks and attain the highest level. Rapid secretion of LH leads to its maximum level during the mid-cycle called LH surge. This induces rupture of Graffian follicle and thereby the release of an ovum (ovulation)
(iii) LH level decreases and FSH level maintained for the initiation of a new cycle.

Topic: Biology, Human Reproduction, NEET
Corresponding to the process shown in figure, the heat given to the gas in the process ABCA is 0.2x J. Find value of x.

Options:

The correct answer is:
5

Solution:
Cycle is clockwise. Hence heat will be given to the gas, as work done will be positive.
Q=W= area of cycle = 1 J

Topic: Physics, Thermodynamics, JEE Main
Cortisol is secreted from

Options:
A
pancreas

B
thyroid

C
adrenal

D
thymus

The correct answer is:
adrenal

Solution:
The zona fasciculata region of adrenal cortex secretes cortisol, corticosterone and cortisone. These hormones are involved in maintaining glucose homeostasis.
Pancreas is composite gland which secretes insulin glucagon and somatostatin.
Thyroid gland secretes thyroxin or tetraiodothyronine $\left(\mathrm{T}_4\right)$ and triodothyronine $\left(\mathrm{T}_3\right)$ which regulate the metabolic rate of the body and maintain basal metabolic rate. It also secretes calcitonin.
Thymus secretes thymosin, involved in providing cell mediated immunity to the body.

Topic: Biology, Chemical Coordination and Integration
$\cos ^{-1} \sqrt{1-x}+\sin ^{-1} \sqrt{1-x}=$

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

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

Solution:
$\cos ^{-1} \sqrt{1-x}+\sin ^{-1} \sqrt{1-x}=\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}=\frac{\pi}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$ is equal to

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

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

Solution:
$\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}=\frac{\pi}{3}+\frac{2 \pi}{6}=\frac{2 \pi}{3}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$$
\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=
$$

Options:
A $\cos ^{-1}\left(\frac{24}{25}\right)$
B $\cos ^{-1}\left(\frac{33}{65}\right)$
C $\cos ^{-1}\left(\frac{5}{13}\right)$
D $\cos ^{-1}\left(\frac{3}{5}\right)$

The correct answer is:
$\cos ^{-1}\left(\frac{33}{65}\right)$

Solution:
$$
\begin{aligned}
& \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right) \\
& =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right) \\
& =\tan ^{-1}\left[\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1-\left(\frac{3}{12}\right)\left(\frac{5}{12}\right)}\right] \\
& =\tan ^{-1}\left(\frac{36+20}{48-15}\right)=\tan ^{-1}\left(\frac{56}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right)
\end{aligned}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=$

Options:
A $\tan ^{-1} \frac{27}{11}$
B $\sin ^{-1} \frac{11}{27}$
C $\cos ^{-1} \frac{11}{27}$
D None of these

The correct answer is:
$\tan ^{-1} \frac{27}{11}$

Solution:
$\begin{aligned} & \cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=\tan ^{-1}\left[\frac{\sqrt{\left(1-\frac{16}{25}\right)}}{\frac{4}{5}}\right]+\tan ^{-1} \frac{3}{5} \\ & {\left[\text { Since } \cos ^{-1} x=\tan ^{-1} \frac{\sqrt{\left(1-x^2\right)}}{x}\right]} \\ & =\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4} \cdot \frac{3}{5}}\right)=\tan ^{-1}\left(\frac{27}{11}\right)\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cos ^{-1}\left(\frac{15}{17}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)=$

Options:
A $\frac{\pi}{2}$
B $\cos ^{-1}\left(\frac{171}{221}\right)$
C $\frac{\pi}{4}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & \cos ^{-1}\left(\frac{15}{17}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right) \quad=\cos ^{-1}\left(\frac{15}{17}\right)+\cos ^{-1}\left(\frac{1-1 / 25}{1+1 / 25}\right) \\ & =\cos ^{-1}\left(\frac{15}{17}\right)+\cos ^{-1}\left(\frac{12}{13}\right) \\ & =\cos ^{-1}\left(\frac{15}{17} \times \frac{12}{13}-\sqrt{1-\left(\frac{15}{17}\right)^2} \sqrt{1-\left(\frac{12}{13}\right)^2}\right) \\ & =\cos ^{-1}\left(\frac{140}{221}\right)\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
cos-1(cos(-5))+sin-1(sin(6))-tan-1(tan(12)) is equal to :
(The inverse trigonometric functions take the principal values)

Options:
A 3π+1
B 3π-11
C 4π-11
D 4π-9

The correct answer is:
4π-11

Solution:
cos-1(cos(-5))+sin-1sin(6)-tan-1(tan(12))

=2π-5+(-2π+6)-(12-4π)

=4π-11

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (01 Sep Shift 2)
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=$

Options:
A $\frac{7 \pi}{6}$
B $\frac{5 \pi}{6}$
C $\frac{\pi}{6}$
D None of these

The correct answer is:
$\frac{5 \pi}{6}$

Solution:
$\begin{aligned} & \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left\{\cos \left(\pi+\frac{\pi}{6}\right)\right\} \\ & =\cos ^{-1}\left(-\cos \frac{\pi}{6}\right)=\pi-\cos ^{-1} \cos \frac{\pi}{6}=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=$

Options:
A $\frac{7 \pi}{6}$
B $\frac{5 \pi}{6}$
C $\frac{\pi}{6}$
D None of these

The correct answer is:
$\frac{5 \pi}{6}$

Solution:
$\begin{aligned} & \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left\{\cos \left(\pi+\frac{\pi}{6}\right)\right\} \\ & \quad \cos ^{-1}\left(-\cos \frac{\pi}{6}\right)=\pi-\cos ^{-1} \cos \frac{\pi}{6}=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=$

Options:
A $\frac{3}{8}$
B $\frac{1}{8}$
C $\frac{\sqrt{3}}{8}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{16}$

Solution:
We have,
$$
\begin{aligned}
\cos 20^{\circ} \cos & 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} \\
& =\cos 20^{\circ} \cos 40^{\circ}\left(\frac{1}{2}\right) \cos 80^{\circ} \\
& =\frac{1}{2}\left(\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}\right) \\
& =\frac{1}{2}\left[\cos 20^{\circ} \cos 2\left(20^{\circ}\right) \cos 2^2\left(20^{\circ}\right)\right] \\
& =\frac{1}{2} \cdot \frac{\sin 2^3 \times 20^{\circ}}{2^3 \sin 20^{\circ}} \\
{[} & \left.\because \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^n A=\frac{\sin 2^n A}{2^n \sin A}\right] \\
& =\frac{1}{16} \frac{\sin 160^{\circ}}{\sin 20^{\circ}}=\frac{1}{16} \frac{\sin \left(180^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}} \\
& =\frac{1}{16} \frac{\sin 20^{\circ}}{\sin 20^{\circ}}=\frac{1}{16} \quad\left[\because \sin \left(180^{\circ}-\theta\right)=\sin \theta\right]
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
$\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}=$

Options:
A $1 / 2$
B $1 / 4$
C $1 / 6$
D $1 / 8$

The correct answer is:
$1 / 8$

Solution:
$\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}=\frac{\sin 2^3 20^{\circ}}{2^3 \sin 20^{\circ}}=\frac{\sin 160^{\circ}}{8 \sin 20^{\circ}}=\frac{1}{8}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$$
\cos ^2\left(\frac{\pi}{4}-\beta\right)-\sin ^2\left(\alpha-\frac{\pi}{4}\right)=
$$

Options:
A $\sin (\alpha+\beta) \sin (\alpha-\beta)$
B $\cos (\alpha+\beta) \cos (\alpha-\beta)$
C $\sin (\alpha-\beta) \cos (\alpha+\beta)$
D $\sin (\alpha+\beta) \cos (\alpha-\beta)$

The correct answer is:
$\sin (\alpha+\beta) \cos (\alpha-\beta)$

Solution:
$\begin{aligned}
\cos ^2\left(\frac{\pi}{4}-\beta\right)- & \sin ^2\left(\alpha-\frac{\pi}{4}\right) \\
& =\cos \left(\frac{\pi}{4}-\beta+\alpha-\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}-\beta-\alpha+\frac{\pi}{4}\right) \\
& =\cos (\alpha-\beta) \cos \left(\frac{\pi}{2}-\overline{\alpha+\beta}\right)=\cos (\alpha-\beta) \sin (\alpha+\beta)
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)=$

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

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

Solution:
$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)$
$=\cos \left(\frac{\pi}{6}+\theta+\frac{\pi}{6}-\theta\right) \cos \left(\frac{\pi}{6}+\theta-\frac{\pi}{6}+\theta\right)$
$=\cos \frac{2 \pi}{6} \cos 2 \theta=\frac{1}{2} \cos 2 \theta$
$\left[\cos ^2 A-\sin ^2 B=\cos (A+B) \cos (A-B)\right]$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}$ is equal to

Options:
A $0$
B $\frac{1}{2}$
C $-\frac{1}{4}$
D $\frac{3}{2}$

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

Solution:
$\begin{aligned} & \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ} \\ & =\cos ^2 76^{\circ}+1-\sin ^2 16^{\circ}-\frac{1}{2} \cdot 2 \cos 76^{\circ} \cos 16^{\circ} \\ & =1+\cos \left(76^{\circ}+16^{\circ}\right) \cos \left(76^{\circ}-16^{\circ}\right) \\ & \quad-\frac{1}{2}\left(\cos 92^{\circ}+\cos 60^{\circ}\right) \\ & =1+\frac{1}{2} \cos 92^{\circ}-\frac{1}{2} \cos 92^{\circ}-\frac{1}{4} \\ & =1-\frac{1}{4}=\frac{3}{4}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
$\cos \left[2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right]=$

Options:
A $\frac{2 \sqrt{6}}{5}$
B $-\frac{2 \sqrt{6}}{5}$
C $\frac{1}{5}$
D $-\frac{1}{5}$

The correct answer is:
$-\frac{2 \sqrt{6}}{5}$

Solution:
$\begin{gathered}\cos \left(\cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}+\cos ^{-1} \frac{1}{5}\right)=\cos \left(\frac{\pi}{2}+\cos ^{-1} \frac{1}{5}\right) \\ =-\sin \left(\cos ^{-1} \frac{1}{5}\right)=-\sin \left(\sin ^{-1} \sqrt{\frac{24}{25}}\right)=-\frac{2 \sqrt{6}}{5}\end{gathered}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cos 36^{\circ}-\cos 72^{\circ}$ is equal to

Options:
A $1$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{1}{8}$

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

Solution:
$\cos 36^{\circ}-\cos 72^{\circ}=-2 \sin \left(\frac{36^{\circ}+72^{\circ}}{2}\right)$
$\sin \left(\frac{72^{\circ}-36^{\circ}}{2}\right)$
$\begin{aligned} & =-2 \sin 54^{\circ} \sin \left(-18^{\circ}\right) \\ & =2 \sin 54^{\circ} \sin 18^{\circ} \\ & =2 \times \frac{\sqrt{5}+1}{4} \times \frac{\sqrt{5}-1}{4}=2 \times \frac{5-1}{4 \times 4}=\frac{4}{8} \\ & =\frac{1}{2}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
$\int \cos ^3 x e^{\log (\sin x)^2} d x=$

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

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

Solution:
$\begin{aligned}
& \text { Let } I=\int \cos ^3 x e^{\log (\sin x)^2} d x \\
& =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x \\
& =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x
\end{aligned}$
Put $\sin x=t \Rightarrow \cos x d x=d t$
$\therefore \mathrm{I}=\int\left(\mathrm{t}^2-\mathrm{t}^4\right) \mathrm{dt}=\frac{\mathrm{t}^3}{3}-\frac{\mathrm{t}^5}{5}+\mathrm{c}=\frac{\sin ^3 \mathrm{x}}{3}-\frac{\sin ^5 \mathrm{x}}{5}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \cos ^{3} x \cdot e^{\log (\sin x)} d x$ is equal to

Options:
A $-\frac{\sin ^{4} x}{4}+c$
B $-\frac{\cos ^{4} x}{4}+c$
C $\frac{e^{\sin x}}{4}+c$
D None of these

The correct answer is:
$-\frac{\cos ^{4} x}{4}+c$

Solution:
Let $I=\int \cos ^{3} x e^{\log \sin x} d x=\int \cos ^{3} x \sin x d x$
Put $x=t \quad \Rightarrow-\sin x d x=d t$
$$
I=-\int t^{3} d t=-\frac{t^{4}}{4}+c=-\frac{\cos ^{4} x}{4}+c
$$

Topic: Mathematics, Indefinite Integration, JEE Main