datavorous/entrance-exam-dataset huggingface dataset:
Take the breakdown voltage of the zener diode used in the given circuit as 6V. For the input voltage shown in the figure below, the time variation of the output voltage is: (Graphs drawn are schematic and not to the scale)

Options:
A
B
C
D

The correct answer is:

Solution:
A diode in forward biased acts as short circuit, since current can smoothly pass in such case. The potential difference across this is negligible.

In reverse biased, initially, a negligible current can pass through it. When a voltage greater than the breakdown voltageV∘ applied across this, the potential difference remains at the breakdown voltage.

In this question, two diodes are connected in opposite polarity. At a time, one diode is in forward biased and other is reversed biased. For solution, we can ignore the forward-biased diode as its voltage is negligible. If the applied potential difference(input) across the reversed biased diode becomes greater than the breakdown voltageV∘, the potential difference across it will not increase and maintains at breakdown voltage.

In this question, the potential difference is changing direction and there will be one diode in reverse biased always. Hence, every peak will have a cut section and the range of the output graph will be -V∘, +V∘.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Take the mean distance of the moon and the sun from the earth to be $0.4 \times 10^6 \mathrm{~km}$ and $150 \times 10^6 \mathrm{~km}$ respectively. Their masses are $8 \times 10^{22} \mathrm{~kg}$ and $2 \times$ $10^{30} \mathrm{~kg}$ respectively. The radius of the earth is $6400 \mathrm{~km}$. Let $\Delta F_1$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $\Delta \mathrm{F}_2$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $\frac{\Delta F_1}{\Delta F_2}$ is:

Options:
A
$2$

B
$6$

C
$10^{-2}$

D
$0.6$

The correct answer is:
$2$

Solution:
As we know, Gravitational force of attraction,
$\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^2}$
$\mathrm{~F}_1=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^2}$ and $\mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^2}$
$\Delta \mathrm{F}_1=\frac{2 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^3} \Delta \mathrm{r}_1$ and $\Delta \mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^3} \Delta \mathrm{r}_2$
Using $\Delta \mathrm{r}_1=\Delta \mathrm{r}_2=2 \mathrm{R}_{\text {earth }} ; \mathrm{m}=8 \times 10^{22} \mathrm{~kg}$;
$\mathrm{M}_{\mathrm{s}}=2 \times 10^{30} \mathrm{~kg}$
$\mathrm{r}_1=0.4 \times 10^6 \mathrm{~km}^2$ and $\mathrm{r}_2=150 \times 10^6 \mathrm{~km}$
$\frac{\Delta \mathrm{F}_1}{\Delta \mathrm{F}_2}=\left(\frac{8 \times 10^{22}}{2 \times 10^{30}}\right)\left(\frac{150 \times 10^6}{0.4 \times 10^6}\right)^3 \times 1 \cong 2$

Topic: Physics, Gravitation, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Take the mean distance of the moon and the sun from the earth to be 0.4×106 km and 150×106 km, respectively. Their masses are 8×1022 kg and 2×1030 kg, respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest point on the earth, and ΔF2 be the difference in the forces exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1ΔF2 is,

Options:
A 6
B 10-2
C 2
D 0.6

The correct answer is:
2

Solution:
F1=GMemr13,F2=GMeMsr22
ΔF1= (-)2GMemr13 Δr1 , ΔF2=(-)2GMeMsr22 Δr2
ΔF1ΔF2= mΔr1r13 r23MsΔr2=mMsr23r13Δr1Δr2
Using Δr1=Δr2=2Rearth
m=8×1022 kg
Ms= 2×1030 kg
r1=0.4×106 km
r2=150×106 km
We get ΔF1ΔF2=2.

Topic: Physics, Gravitation, JEE Main, JEE Main 2018 (15 Apr)
Taking the wavelength of first Balmer line in hydrogen spectrum (n=3 to n=2) as 660 nm , the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :

Options:
A 889.2Â nm
B 488.9Â nm
C 388.9Â nm
D 642.7 nm

The correct answer is:
488.9 nm

Solution:
From Rydberg’s equation,
1λ=R1n2-1n22
1660×10-9=R122-132
1λ=R122-142
λ600×10-9=59×4×16×412
λ=488.9 nm

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Taking the wavelength of first Balmer line in hydrogen spectrum (n=3 to n=2) as 660 nm , the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :

Options:
A 889.2Â nm
B 488.9Â nm
C 388.9Â nm
D 642.7 nm

The correct answer is:
488.9 nm

Solution:
From Rydberg’s equation,
1λ=R1n2-1n22
1660×10-9=R122-132
1λ=R122-142
λ600×10-9=59×4×16×412
λ=488.9 nm

Topic: Physics, Atomic Physics, JEE Main
Taking X-axis along horizontal and Y-axis along vertical, with what minimum speed must a particle be projected from the origin so that it is able to pass through a given point (30 m, 40 m)? Take g = 10 m s-2

Options:
A 60 m s-1
B 30 m s-1
C 50 m s-1
D 40 m s-1

The correct answer is:
30 m s-1

Solution:
Using equation of trajectory for a projectile motion :
y=xtan⁡θ-gx22u21+tan2⁡θ
For the projectile to pass through (30 m, 40 m) :
40 = 30 tan α - g (30) 2 2u 2 1 + tan 2 α
or 900 tan2 α - (6u2 tan α) + (900 + 8u2) = 0
For real value of α. The discriminant of a quadratic equation b2-4ac≥0
(6u2)2 ≥ 3600 (900 + 8u2)
or (u4 - 800 u2) ≥ 9,00,00
or (u2 - 400)2 ≥ (2,50,000)
or u2 - 400 ≥ 500
or u2 ≥ 900 or u ≥ 30 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
tan-11+33+3+sec-18+436+33=

Options:
A π4
B π2
C π3
D π6

The correct answer is:
π3

Solution:
tan-11+33+3+sec-18+436+33

=tan-11+331+3+sec-142+332+3

=tan-113+sec-143

=tan-113+cos-132=π6+π6=π3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2023 (24 Jan Shift 1)
$\tan ^{-1} \frac{a-b}{1+a b}+\tan ^{-1} \frac{b-c}{1+b c}=$

Options:
A $\tan ^{-1} a-\tan ^{-1} b$
B $\tan ^{-1} a-\tan ^{-1} c$
C $\tan ^{-1} b-\tan ^{-1} c$
D $\tan ^{-1} c-\tan ^{-1} a$

The correct answer is:
$\tan ^{-1} a-\tan ^{-1} c$

Solution:
$\begin{aligned} & \tan ^{-1}\left(\frac{a-b}{1+a b}\right)+\tan ^{-1}\left(\frac{b-c}{1+b c}\right) \\ & =\tan ^{-1}(a)-\tan ^{-1}(b)+\tan ^{-1}(b)-\tan ^{-1}(c) \\ & =\tan ^{-1}(a)-\tan ^{-1}(c) .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1}\left[\frac{\cos x}{1+\sin x}\right]=$

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

The correct answer is:
$\frac{\pi}{4}-\frac{x}{2}$

Solution:
$\begin{aligned} & \tan ^{-1}\left[\frac{\cos x}{1+\sin x}\right]=\tan ^{-1}\left[\frac{\sin (\pi / 2-x)}{1+\cos (\pi / 2-x)}\right] \\ & \quad=\tan ^{-1}\left[\frac{2 \sin (\pi / 4-x / 2) \cos (\pi / 4-x / 2)}{2 \cos ^2(\pi / 4-x / 2)}\right] \\ & =\tan ^{-1} \left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right)=\frac{\pi}{4}-\frac{x}{2} .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=$

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{4}$
C $0$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & \tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3 \\ & \quad=\tan ^{-1} 1+\pi+\tan ^{-1}\left(\frac{5}{-5}\right)=\tan ^{-1} 1+\pi-\tan ^{-1} 1=\pi .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=$

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{4}$
C $0$
D None of these

The correct answer is:
None of these

Solution:
$\begin{gathered} \text{So, } \tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3 \\ \Rightarrow \tan ^{-1} 1+\pi+\tan ^{-1}\left(\frac{5}{-5}\right)=\tan ^{-1} 1+\pi-\tan ^{-1} 1=\pi\end{gathered}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} 2+\cot ^{-1}(-3)+\cot ^{-1} \frac{1}{3}+\tan ^{-1}\left(-\frac{1}{2}\right)=$

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

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

Solution:
$\begin{aligned} & \tan ^{-1} 2+\cot ^{-1}(-3)+\cot ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(-\frac{1}{2}\right) \\ & =\tan ^{-1}(2)-\tan ^{-1}\left(\frac{1}{2}\right)+\pi-\cot ^{-1}(3)+\cot ^{-1}\left(\frac{1}{3}\right) \\ & =\tan ^{-1}(2)-\tan ^{-1}\left(\frac{1}{2}\right)+\pi-\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}(3) \\ & =\tan ^{-1}\left(\frac{2-\frac{1}{2}}{1+2 \cdot \frac{1}{2}}\right)+\pi+\tan ^{-1}(3)-\tan ^{-1}\left(\frac{1}{3}\right) \\ & =\tan ^{-1}\left(\frac{3}{2}\right)+\pi+\tan ^{-1}\left(\frac{3-\frac{1}{3}}{1+3 \cdot \frac{1}{3}}\right) \\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\pi+\tan ^{-1}\left(\frac{8}{6}\right) \\ & =\pi+\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{4}{3}\right) \\ & =\pi+\tan ^{-1}\left(\frac{3}{4}\right)+\cot ^{-1}\left(\frac{3}{4}\right)=\pi+\pi / 2 \\ & =3 \pi / 2\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} 2+\tan ^{-1} 3=$

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

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

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

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
tan-1(-2)-tan-1(3) is equal to

Options:
A 3π4
B -π6
C π6
D -3π4

The correct answer is:
-3π4

Solution:
Given that: tan-1(-2)-tan-1(3)

As we know that, tan-1(-x)=-tan-1x

=-tan-1(2)-tan-1(3)

=-(tan-1(2)+tan-1(3))

As we know that,tan-1(x)+tan-1(y) =π+tan-1(x+y1-xy) when xy>1 .

=-π-tan-1(2+31-2×3)

=-π-tan-1(-1)

=-π+tan-11

=-3π4

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ has the value

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

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

Solution:
$\begin{aligned} & \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8} \\ & =\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} \times \frac{1}{5}}\right)+\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \times \frac{1}{8}}\right) \\ & =\tan ^{-1}\left(\frac{4}{7}\right)+\tan ^{-1}\left(\frac{3}{11}\right) \\ & =\tan ^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7} \times \frac{3}{11}}\right)\end{aligned}$
$=\tan ^{-1}\left(\frac{65}{65}\right)=\tan ^{-1}(1)=\frac{\pi}{4}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$ is equal to

Options:
A $\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$
B $\frac{1}{2} \sin ^{-1}\left(\frac{3}{5}\right)$
C $\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)$
D $\tan ^{-1}\left(\frac{1}{2}\right)$

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

Solution:
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\tan ^{-1}\left\{\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right\}$
$=\tan ^{-1}\left\{\frac{9+8}{36-2}\right\}=\tan ^{-1}\left(\frac{1}{2}\right)$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1}\left(\frac{1}{11}\right)+\tan ^{-1}\left(\frac{2}{12}\right)=$

Options:
A $\tan ^{-1}\left(\frac{33}{132}\right)$
B $\tan ^{-1}\left(\frac{1}{2}\right)$
C $\tan ^{-1}\left(\frac{132}{33}\right)$
D None of these

The correct answer is:
None of these

Solution:
$\tan ^{-1}\left(\frac{1}{11}\right)+\tan ^{-1}\left(\frac{2}{12}\right)$
$\begin{aligned} & \tan ^{-1}\left(\frac{\frac{1}{11}+\frac{2}{12}}{1-\frac{1}{11} \times \frac{2}{12}}\right)=\tan ^{-1}\left(\frac{12+22}{130}\right) \\ = & \tan ^{-1}\left(\frac{34}{130}\right)=\tan ^{-1}\left(\frac{17}{65}\right) .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} \frac{c_1 x-y}{c_1 y+x}+\tan ^{-1} \frac{c_2-c_1}{1+c_2 c_1}+\tan ^{-1} \frac{c_3-c_2}{1+c_3 c_2}+\ldots+\tan ^{-1} \frac{1}{c_n}=$

Options:
A $\tan ^{-1} \frac{y}{x}$
B $\tan ^{-1} y x$
C $\tan ^{-1} \frac{x}{y}$
D $\tan ^{-1}(x-y)$

The correct answer is:
$\tan ^{-1} \frac{x}{y}$

Solution:
$\begin{aligned} & \tan ^{-1}\left(\frac{c_1 x-y}{c_1 y+x}\right)+\tan ^{-1}\left(\frac{c_2-c_1}{1+c_2 c_1}\right)+\tan ^{-1}\left(\frac{c_3-c_2}{1+c_3 c_2}\right)+ \\ & \ldots \ldots+\tan ^{-1} \frac{1}{c_n} \\ & \tan ^{-1}\left(\frac{\frac{x}{y}-\frac{1}{c_1}}{1+\frac{x}{y} \cdot \frac{1}{c_1}}\right)+\tan ^{-1}\left(\frac{\frac{1}{c_1}-\frac{1}{c_2}}{1+\frac{1}{c_1 c_2}}\right)+\tan ^{-1}\left(\frac{\frac{1}{c_2}-\frac{1}{c_3}}{1+\frac{1}{c_2 c_3}}\right)+\ldots \ldots+\tan ^{-1} \frac{1}{c_n} \\ & =\tan ^{-1} \frac{x}{y}-\tan ^{-1} \frac{1}{c_1}+\tan ^{-1} \frac{1}{c_1}-\tan ^{-1} \frac{1}{c_2}+\tan ^{-1} \frac{1}{c_2}-\tan ^{-1} \frac{1}{c_3}+\ldots+\tan ^{-1} \frac{1}{c_{n-1}}-\tan ^{-1} \frac{1}{c_n}+\tan ^{-1} \frac{1}{c_n} \\ & =\tan ^{-1}\left(\frac{x}{y}\right) .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \tan 4^{\circ} \tan 89^{\circ}=$

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

The correct answer is:
1

Solution:
$\tan 1^{\circ} \tan 2^{\circ} \ldots \cdot \tan 89^{\circ}$
$=\left(\tan 1^{\circ} \tan 89^{\circ}\right)\left(\tan 2^{\circ} \tan 88^{\circ}\right) \ldots \ldots .=1 \times 1 \times 1 \ldots=1 .$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=$

Options:
A 0
B $3 \pi$
C $\frac{-\pi}{6}$
D $\frac{\pi}{6}$

The correct answer is:
0

Solution:
$\begin{aligned} & \tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \\ & =\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{6}\right)\right]+\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right] \\ & =\tan ^{-1}\left[-\tan \frac{\pi}{6}\right]+\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right] \\ & =\tan ^{-1}\left[\tan \left(-\frac{\pi}{6}\right)\right]+\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=-\frac{\pi}{6}+\frac{\pi}{6}=0\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} \frac{1}{\sqrt{x^2-1}}=$

Options:
A $\frac{\pi}{2}+\operatorname{cosec}^{-1} x$
B $\frac{\pi}{2}+\sec ^{-1} x$
C $\operatorname{cosec}^{-1} x$
D $\sec ^{-1} x$

The correct answer is:
$\operatorname{cosec}^{-1} x$

Solution:
\(\begin{aligned}
& \tan ^{-1} \frac{1}{\sqrt{x^2-1}},|x|>1 \\
& \text {Put } x=\operatorname{cosec} \theta \Rightarrow \theta=\operatorname{cosec}^{-1} x \\
& \therefore \tan ^{-1} \frac{1}{\sqrt{x^2-1}}=\tan ^{-1} \frac{1}{\sqrt{\operatorname{cosec}^2 \theta-1}} \\
& =\tan ^{-1}\left(\frac{1}{\cot \theta}\right)=\tan ^{-1}(\tan \theta) \\
& =\theta=\operatorname{cosec}-1 x=\frac{\pi}{2}-\sec ^{-1} x \\
& {\left[\operatorname{cosec}^{-1} x+\sec ^{-1} x=\frac{\pi}{2}\right]}
\end{aligned}\)

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8}$, then $x$ equals

Options:
A $-1$
B $1$
C $0$
D None of these

The correct answer is:
$-1$

Solution:
$\begin{aligned} & \left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8} \\ & \Rightarrow \quad\left(\tan ^{-1} x+\cot ^{-1} x\right)^2-2 \tan ^{-1} x\left(\frac{\pi}{2}-\tan ^{-1} x\right)=\frac{5 \pi^2}{8}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{\pi^2}{4}-2 \times \frac{\pi}{2} \tan ^{-1} x+2\left(\tan ^{-1} x\right)^2=\frac{5 \pi^2}{8} \\ & \Rightarrow 2\left(\tan ^{-1} x\right)^2-\pi \tan ^{-1} x-\frac{3 \pi^2}{8}=0 \\ & \Rightarrow \tan ^{-1} x=-\frac{\pi}{4}, \frac{3 \pi}{4} \Rightarrow \tan ^{-1} x=-\frac{\pi}{4} \Rightarrow x=-1 \\ & \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

Options:
A -1
B 1
C 0
D $\pi \sqrt{\frac{5}{8}}$

The correct answer is:
-1

Solution:
Given equation is
$\begin{gathered}
\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8} \\
\left(\tan ^{-1} x+\cot ^{-1} x\right)^2-2 \tan ^{-1} x \cdot \cot ^{-1} x=\frac{5 x^2}{8} \\
{\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]}
\end{gathered}$
$\begin{aligned} & \Rightarrow \quad \frac{\pi^2}{4}-2 \tan ^{-1} x\left(\pi / 2-\tan ^{-1} x\right)=\frac{5 \pi^2}{8} \\ & \Rightarrow-2 \cdot \frac{\pi \tan ^{-1} x}{2}+2\left(\tan ^{-1} x\right)^2=\frac{5 \pi^2}{8}-\frac{\pi^2}{4} \\ & \Rightarrow \quad\left(\tan ^{-1} x\right)^2-\frac{\pi \tan ^{-1} x}{2}=\frac{3 \pi^2}{16}\end{aligned}$
$\begin{array}{cc}\therefore \quad & \tan ^{-1} x=\frac{\frac{\pi}{2} \pm \sqrt{\frac{\pi^2}{4}+\frac{3 \pi^2}{4}}}{2} \\ \Rightarrow \quad & \tan ^{-1} x=\frac{\frac{\pi}{2} \pm \pi}{2} \\ & =\frac{3 \pi}{4},-\frac{\pi}{4}\end{array}$
$\begin{array}{ll}\Rightarrow & x=\tan \left(\frac{3 \pi}{4}\right), \tan \left(-\frac{\pi}{4}\right) \\ \Rightarrow & x=-1,-1\end{array}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1} x+\cot ^{-1}(x+1)=$

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

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

Solution:
$\begin{aligned} & \tan ^{-1} x+\cot ^{-1}(x+1)=\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x+1}\right) \\ & =\tan ^{-1}\left[\frac{x+\frac{1}{x+1}}{1-\frac{x}{x+1}}\right]=\tan ^{-1}\left(x^2+x+1\right)\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ is

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

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

Solution:
$\begin{aligned} & \tan ^{-1} \frac{x}{y}-\tan ^{-1}\left(\frac{x-y}{x+y}\right)=\tan ^{-1} \frac{x}{y}-\tan ^{-1}\left(\frac{1-y / x}{1+y / x}\right) \\ & =\tan ^{-1} \frac{x}{y}-\left(\tan ^{-1} 1-\tan ^{-1} \frac{y}{x}\right) \\ & =\tan ^{-1} \frac{x}{y}+\tan ^{-1} \frac{y}{x}-\frac{\pi}{4} \\ & =\tan ^{-1} \frac{x}{y}+\cot ^{-1} \frac{x}{y}-\frac{\pi}{4}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}=$

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

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

Solution:
We know that
$\begin{gathered}\tan \left(20^{\circ}+40^{\circ}\right)=\frac{\tan 20^{\circ}+\tan 40^{\circ}}{1-\tan 20^{\circ} \tan 40^{\circ}} \\ \quad \Rightarrow \sqrt{3}=\frac{\tan 20^{\circ}+\tan 40^{\circ}}{1-\tan 20^{\circ} \tan 40^{\circ}} \\ \quad \Rightarrow \sqrt{3}-\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}=\tan 20^{\circ}+\tan 40^{\circ} \\ \quad \Rightarrow \tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}=\sqrt{3} .\end{gathered}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}=$

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

The correct answer is:
3

Solution:
$\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}$
$=\frac{\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} \tan 60^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}}$
Here numerator $=\left(\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}\right)$
$\begin{aligned}
& =\frac{\sin 20^{\circ}}{2}\left(2 \sin 40^{\circ} \sin 80^{\circ}\right) \\
& =\frac{\sin 20^{\circ}}{2}\left(\cos 40^{\circ}-\cos 120^{\circ}\right) \\
& =\frac{1}{2} \sin 20^{\circ}\left(1-2 \sin ^2 20^{\circ}+\frac{1}{2}\right) \\
& =\frac{1}{2} \sin 20^{\circ}\left(\frac{3}{2}-2 \sin ^2 20^{\circ}\right)=\frac{\sin 60^{\circ}}{4}=\frac{\sqrt{3}}{8}
\end{aligned}$
Now, we take denominator $=\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}$
$\begin{aligned}
& =\frac{\sin 2^3 20^{\circ}}{2^3 \sin 20^{\circ}}=\frac{\sin 160^{\circ}}{8 \sin 20^{\circ}}=\frac{\sin 20^{\circ}}{8 \sin 20^{\circ}}=\frac{1}{8} \\
& \therefore \text { Hence } \tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ}=\frac{\sqrt{3} / 8}{1 / 8}
\end{aligned}$
Therefore $\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}=\sqrt{3} \cdot \sqrt{3}=3$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\tan \left(2 \cos ^{-1} \frac{3}{5}\right)=$

Options:
A $\frac{7}{25}$
B $\frac{24}{25}$
C $-\frac{24}{7}$
D $\frac{8}{3}$

The correct answer is:
$-\frac{24}{7}$

Solution:
$\begin{aligned} & \tan \left(2 \cos ^{-1} \frac{3}{5}\right)=\tan \left[\cos ^{-1}\left(2 \cdot \frac{9}{25}-1\right)\right] \\ & =\tan \left(\cos ^{-1}\left(-\frac{7}{25}\right)\right)=\tan \left(\tan ^{-1}\left[\sqrt{\frac{1-\frac{49}{625}}{-\frac{7}{25}}}\right]\right)=-\frac{24}{7} \\ & \text { Therefore } \tan \left(2 \cos ^{-1} \frac{3}{5}\right)=-\frac{24}{7} .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\tan ^2 \theta+\cot ^2 \theta$ is

Options:
A $\geq 2$
B $\leq 2$
C $\geq-2$
D None of these

The correct answer is:
$\geq 2$

Solution:
We know that $\left(x-\frac{1}{x}\right)^2 \geq 0 \Rightarrow x^2+\frac{1}{x^2}-2 \geq 0$ Put $x=\tan \theta \Rightarrow \tan ^2 \theta+\cot ^2 \theta \geq 2$.

Topic: Mathematics, Trigonometric Equations, JEE Main
tan2tan-115+sec-152+2tan-118 is equal to:

Options:
A 1
B 2
C 14
D 54

The correct answer is:
2

Solution:
Given,

tan2tan-115+sec-152+2tan-118

=tan2tan-115+tan-118+tan-112

=tan2tan-115+181-140+tan-112

=tan2tan-113+tan-112

=tantan-12×131-132+tan-112

=tantan-134+tan-112

=tantan-134+121-38

=tantan-15458

=tantan-12

=2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (26 Jul Shift 1)
$\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]=$

Options:
A $\frac{17}{7}$
B $-\frac{17}{7}$
C $\frac{7}{17}$
D $-\frac{7}{17}$

The correct answer is:
$-\frac{7}{17}$

Solution:
$\begin{aligned} & \tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]=\tan ^{-1}\left[\tan ^{-1} \frac{\frac{2}{5}}{1-\frac{1}{25}}-\tan ^{-1}(1)\right] \\ & \quad=\tan \left[\tan ^{-1} \frac{5}{12}-\tan ^{-1}(1)\right]=\tan \tan ^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12}}\right)=-\frac{7}{17}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\int \tan ^2 x d x$ is equal to

Options:
A $\tan x+x+c$
B $\tan x-x+c$
C $\sec x+x+c$
D $\sec x-x+c$

The correct answer is:
$\tan x-x+c$

Solution:
$\int \tan ^2 x d x=\int\left(\sec ^2 x-1\right) d x=\tan x-x+c$

Topic: Mathematics, Indefinite Integration, JEE Main
$\tan 3 A-\tan 2 A-\tan A=$

Options:
A $\tan 3 A \tan 2 A \tan A$
B $-\tan 3 A \tan 2 A \tan A$
C $\tan A \tan 2 A-\tan 2 A \tan 3 A-\tan 3 A \tan A$
D None of these

The correct answer is:
$\tan 3 A \tan 2 A \tan A$

Solution:
$\begin{aligned} \tan 3 A= & \frac{\tan A+\tan 2 A}{1-\tan A \tan 2 A} \\ & \Rightarrow \tan 3 A-\tan 2 A-\tan A=\tan 3 A \tan 2 A \tan A\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\int \tan (3 x-5) \sec (3 x-5) d x=$

Options:
A $\sec (3 x-5)+c$
B $\frac{1}{3} \sec (3 x-5)+c$
C $\tan (3 x-5)+c$
D None of these

The correct answer is:
$\frac{1}{3} \sec (3 x-5)+c$

Solution:
Put $t=3 x-5 \Rightarrow d t=3 d x$, therefore
$\begin{aligned}
\int \tan (3 x-5) \sec (3 x-5) d x & =\frac{1}{3} \int \tan t . \sec t d t \\
=\frac{\sec t}{3}+c & =\frac{\sec (3 x-5)}{3}+c .
\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=$

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

The correct answer is:
4

Solution:
$\begin{aligned}
\tan 9^{\circ}-\tan 27^{\circ} & -\tan 63^{\circ}+\tan 81^{\circ} \\
& =\tan 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ}+\cot 9^{\circ} \\
& =\left(\tan 9^{\circ}+\cot 9^{\circ}\right)-\left(\tan 27^{\circ}+\cot 27^{\circ}\right) \\
& =\frac{\left.\cos 9^{\circ}-9^{\circ}\right)}{\sin 9^{\circ} \cos 9^{\circ}}-\frac{\cos \left(27^{\circ}-27^{\circ}\right)}{\sin 27^{\circ} \cdot \cos 27^{\circ}}=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \\
& =2\left\{\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}\right\}=2 \cdot \frac{2 \cdot \cos 36^{\circ} \cdot \sin 18^{\circ}}{\sin 18^{\circ} \cdot \sin 54^{\circ}}=4
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main