datavorous/entrance-exam-dataset huggingface dataset:
The value of $\left|\begin{array}{lll}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ \sin ^2 66^{\circ} & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ \tan 135^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|$ is

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

The correct answer is:
$0$

Solution:
Here,
$\left|\begin{array}{ccc}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ \sin ^2 66^{\circ} & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ \tan 135^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|$
Applying $C_1 \rightarrow C_1+C_2+C_3$
$=\left|\begin{array}{lll}\sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \sin ^2 66^{\circ} \tan 135^{\circ} \\ \sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \tan 135^{\circ} \sin ^2 14^{\circ} \\ \sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \sin ^2 14^{\circ} \sin ^2 66^{\circ}\end{array}\right|$
$\begin{aligned} & =\sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} \\ & \qquad\left|\begin{array}{lll}1 & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ 1 & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\end{aligned}$
$=\sin ^2 14^{\circ}+\sin ^2 66^{\circ}-1\left|\begin{array}{ccc}1 & \sin ^2 66 & -1 \\ 1 & -1 & \sin ^2 14^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|$
Applying $R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3$
$\begin{aligned} & =\cos ^2 66^{\circ}+\sin ^2 66^{\circ}-1 \\ & \left|\begin{array}{ccc}0 & 1+\sin ^2 66^{\circ} & -1-\sin ^2 14^{\circ} \\ 0 & -1-\sin ^2 14^{\circ} & \sin ^2 14^{\circ}-\sin ^2 66^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right| \\ & \left(\cos ^2 66^{\circ}+\sin ^2 66^{\circ}-1\right)=0\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
The value of sin⁡9π14sin⁡11π14sin⁡13π14 is equal to

Options:
A 164
B -164
C 18
D -18

The correct answer is:
18

Solution:
sin⁡9π14sin⁡11π14sin⁡13π14=sin⁡π2+π7sin⁡π2+2π7sin⁡π2+3π7
=cos⁡π7cos⁡2π7cos⁡3π7
=cos⁡π7cos⁡2π7cos⁡π-4π7
=-cos⁡π7cos⁡2π7cos⁡4π7
=-sin⁡8π78sin⁡π7=-sin⁡π+π78sin⁡π7
=18

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of $\sin 600^{\circ} \cos 330^{\circ}+\cos 120^{\circ} \sin 150^{\circ}$ is

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

The correct answer is:
-1

Solution:
$\text { } \begin{aligned}
\sin 600^{\circ} \cos 330^{\circ} & +\cos 120^{\circ} \sin 150^{\circ} \\
& =-\sin 60^{\circ} \cos 30^{\circ}-\sin 30^{\circ} \cos 60^{\circ} \\
= & -\left\{\sin \left(60^{\circ}+30^{\circ}\right)\right\}=-1 .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of ∫sin⁡101x.sin99⁡xdx is sin⁡100x sin100⁡xk+5, then k19 is

Options:

The correct answer is:
5

Solution:
I=∫sin⁡100x⋅cos⁡x+cos⁡100x⋅sin⁡xsin99⁡x⋅dx
I=∫sin⁡100x⋅cos⁡x.sin99⁡xdx+∫cos⁡100x⋅sin100⁡x⋅dx III
I=sin⁡100x sin100⁡x100-100100∫cos⁡100xsin100⁡xdx+∫cos⁡100x.sin100⁡x⋅dx
I=sin⁡100x⋅sin100⁡x100+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of $\sin ^{-\cot ^{-1}} \tan ^{-\cos ^{-1} x}$ is equal to

Options:
A $x$
B $\frac{\pi}{2}$
C $1$
D None of these

The correct answer is:
$x$

Solution:
$\begin{aligned}
& \text {Let } \cos ^{-1} x=\theta \Rightarrow x=\cos \theta \Rightarrow \sec \theta=\frac{1}{x} \\
& \Rightarrow \tan \theta=\sqrt{\sec ^2 \theta-1}=\sqrt{\frac{1}{x^2}-1}=\frac{1}{x} \sqrt{1-x^2} \\
& \text {Now } \sin \left(\cot ^{-1} \left(\tan \theta\right)\right)=\sin \left(\cot ^{-1}\left(\frac{1}{x} \sqrt{1-x^2}\right)\right)
\end{aligned}$
Now
Again, putting $x=\sin \theta$
$\begin{aligned}
& \sin \left(\cot ^{-1}\left(\frac{1}{x} \sqrt{1-x^2}\right)\right)=\sin \left(\cot ^{-1}\left(\frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}\right)\right) \\
& =\sin \left(\cot ^{-1}(\cot \theta)\right)=\sin \theta=x
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $\sin \left(\cot ^{-1} x\right)$ is

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

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

Solution:
$\begin{array}{ll} & \sin \left(\cot ^{-1} x\right) \\ & \text { Let } \cot ^{-1} x=\mathrm{t} \\ \therefore \quad & x=\cot \mathrm{t} \\ \therefore \quad & 1+\cot ^2 \mathrm{t}=1+x^2 \\ \therefore \quad & \operatorname{cosec}^2 \mathrm{t}=1+x^2 \\ \therefore \quad & \operatorname{cosec} \mathrm{t}=\sqrt{1+x^2} \\ \therefore \quad & \sin \mathrm{t}=\frac{1}{\sqrt{1+x^2}} \\ \therefore \quad & \mathrm{t}=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \\ \therefore \quad & \sin \left(\cot ^{-1} x\right)=\sin \left(\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right) \\ & =\frac{1}{\sqrt{1+x^2}}\end{array}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of sin⁡cot-1⁡x is

Options:
A 1+x2
B x
C 11+x2
D 1-x2

The correct answer is:
11+x2

Solution:
Let, cot â1 x=A , cot2â¡A=x2
1+cot2â¡A=1+x2=cosec2A
1 sin 2 A = 1 + x 2 â sin A = 1 / 1 + x 2
â sin cot -1 x = 1 / 1 + x 2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of λ such that sum of the squares of the roots of the quadratic equation, x2+3-λ x+2=λ has the least value is:

Options:
A 2
B 49
C 158
D 1

The correct answer is:
2

Solution:
We have, x2+(3-λ)x+2=λ, have roots α & β.
Then, f(λ)=α2+β2
=α+β2-2αβ=3-λ2-22-λ=λ2-6λ+9-4+2λ=λ2-4λ+5
f′λ=2λ-4=0 { for the minimum value of fλ}

As f''(λ)=2>0

⇒λ=2

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The value of $\lambda$ such that the system of equations $2x-y-2 y=2 ; x-2 y+z=-4$,$x+y+\lambda z=4,$ has no solution, is

Options:
A 3
B 1
C 0
D -3

The correct answer is:
-3

Solution:
If the given system of equations has no solution. then $\left|\begin{array}{ccc}2 & -1 & -2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda\end{array}\right|=0$
$\Rightarrow 2(-2 \lambda-1)+1(\lambda-1)-2(1+2)=0$
$\Rightarrow \quad-4 \lambda-2+\lambda-1-6=0$
$\Rightarrow \quad-3 \lambda-9=0$
$\Rightarrow \quad \lambda=-3$

Topic: Mathematics, Determinants, JEE Main
The value of sum of the series \( 3 \cdot{ }^{n} C_{0}+\frac{{ }^{n} C_{1} \cdot 3^{2}}{2}+\frac{{ }^{n} C_{2} \cdot 3^{3}}{3}+\frac{{ }^{n} C_{3} \cdot 3^{4}}{4}+\ldots \frac{{ }^{n} C_{n} \cdot 3^{n+1}}{n+1} \) is

Options:
A \( \frac{4^{n+1}}{n+1} \)
B \( \frac{4^{n+1}-1}{n+1} \)
C \( \frac{4^{n}}{n+1} \)
D \( \frac{4^{n}-1}{n+1} \)

The correct answer is:
\( \frac{4^{n+1}-1}{n+1} \)

Solution:
We know, (1 + x)n = nC0 + nC1·x + nC2·x2 + .... nCn·xn
On integrating, we get, 1+xn+1–1n+1= nC0x+ nC1x22+ nC2x33+ nC3x44+… nCnxn+1n+1
Put x = 3
3·nC0+ nC1⋅322+ nC2⋅333+ nC3⋅344+... nCn⋅3n+1n+1=4n+1–1n+1

Topic: Mathematics, Binomial Theorem, JEE Main
The value of \( \sum_{\mathrm{k}=1}^{99}\left(\mathrm{i}^{\mathrm{k} !}+\omega^{\mathrm{k} !}\right) \) is (where, \( \mathrm{i}=\sqrt{-1} \) and \( \omega \) is non-real cube root of unity)

Options:
A \( 190+\omega \)
B \( 192+\omega^{2} \)
C \( 190+i \)
D \( 192+i \)

The correct answer is:
\( 190+i \)

Solution:
∑k=199ik!+∑k=199ωk!
∑k=199ik!=i1!+i2!+i3!+i4!+…+i99!
=i-1+i6+1+1+1+…+1
=i-2+96=i+94
∑k=199ωk!=ω1+ω2!+ω3!+ω4!+…+ω99!
=ω+ω2+1+1+1+…+1=96
Sum =i+94+96=i+190

Topic: Mathematics, Complex Number, JEE Main
The value of tan-11-sin⁡x+1+sin⁡x1-sin⁡x-1+sin⁡x ∀x∈0,π2 is equal to

Options:
A x2-π2
B x2+π2
C x2-π
D π2-x2

The correct answer is:
x2-π2

Solution:
tan-11-sin⁡x+1+sin⁡x1-sin⁡x-1+sin⁡x
=tan-11-sin⁡x+1+sin⁡x1-sin⁡x-1+sin⁡x1-sin⁡x+1+sin⁡x1-sin⁡x+1+sin⁡x
=tan-11-sin⁡x+1+sin⁡x+21-sin2x1-sin⁡x-1+sin⁡x=tan-121+cos⁡x-2sin⁡x
=tan-1-2cos2x22sinx2cosx2=tan-1-cotx2=tan-1cotπ-x2
=π2-cot-1cotπ-x2=π2-π-x2=x2-π2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of ∫tan-1sin⁡x+1cos⁡x3+2sin⁡x-cos2xdx is (where c is the constant of integration)

Options:
A tan-1sin⁡x+c
B tan-1sin⁡x2+c
C tan-1sin⁡x+122+c
D tan-1sin⁡x22+c

The correct answer is:
tan-1sin⁡x+122+c

Solution:
Let, Ι=∫tan-1sin⁡x+1cos⁡x3+2sin⁡x-1-sin2xdx
Substituting, 1+sin⁡x=t⇒cos⁡xdx=dt
Ι=∫tan-1tdt2+t-12+2t-1
Ι=∫tan-1tdt1+t2
Ι=tan-1t22+c
Ι=tan-1sin⁡x+122+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of $\frac{\tan x}{\tan 3 x}$ whenever defined never lie between

Options:
A $1 / 3$ and 3
B $1 / 4$ and 4
C $1 / 5$ and 5
D 5 and 6

The correct answer is:
$1 / 3$ and 3

Solution:
$\begin{aligned}
& y=\frac{\tan x}{\tan 3 x}=\frac{\tan x}{3 \tan x-\tan ^3 x} \\
& \text { Let } \quad 1-3 \tan ^2 x \\
& y=\frac{1-3 \tan ^2 x}{3-\tan ^2 x}=\frac{\frac{1}{3}-\tan ^2 x}{1-\frac{1}{3} \cdot \tan ^2 x} \\
&
\end{aligned}$
Hence, $y$ should never lie between $\frac{1}{3}$ and 3 whenever defined.

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$ is

Options:
A $\frac{5 \pi}{6}$
B $\frac{\pi}{2}$
C $\frac{2 \pi}{3}$
D $\frac{3 \pi}{4}$

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

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

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of tan-1⁡1+x2+ 1-x21+x2- 1-x2, x<12, x≠0, is equal to:

Options:
A π4+12cos-1⁡x2
B π4-cos-1⁡x2
C π4-12cos-1⁡x2
D π4+cos-1⁡x2

The correct answer is:
π4+12cos-1⁡x2

Solution:
Assume that ,
x2=cos⁡2θ;θ=12cos-1x2
∴tan- 1⁡1+cos⁡2θ+ 1-cos⁡2θ1+cos⁡2θ- 1-cos⁡2θ=tan- 1⁡2cos2θ+ 2sin2θ2cos2θ- 2sin2θ=tan- 1⁡cosθ+ sinθcosθ- sinθ
=tan-1⁡1+tan⁡θ1-tan⁡θ
=tan-1⁡tan⁡π4+θ
=π4+12cos-1⁡x2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2017 (08 Apr Online)
The value of $\tan ^{-1} 2+\tan ^{-1} 3$ is

Options:
A $\left(\frac{3 \pi}{4}\right)^c$
B $\left(\frac{\pi}{2}\right)^c$
C $\left(\frac{\pi}{4}\right)^c$
D $\left(\frac{\pi}{6}\right)^c$

The correct answer is:
$\left(\frac{3 \pi}{4}\right)^c$

Solution:
$\begin{aligned} & \tan ^{-1} 2+\tan ^{-1} 3 \\ & =\tan ^{-1}\left[\frac{2+3}{1-(2)(3)}\right]=\tan ^{-1}\left(\frac{5}{1-6}\right)=\tan ^{-1}(-1)=\left(\frac{3 \pi}{4}\right)^c\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ is

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

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

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

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)$ is

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

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

Solution:
$\begin{aligned} & \tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right) \\ & =\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2} \cdot \frac{1}{5}}\right) \\ & =\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{7}{9}\right) \\ & =\tan ^{-1}\left(\frac{\frac{1}{8}+\frac{7}{9}}{1-\frac{1}{8} \cdot \frac{7}{9}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{72}{65}}{\frac{72}{9}}\right) \\ & =\tan ^{-1}(1) \\ & =\frac{\pi}{4}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of tan-1cos15π4-1sinπ4 is equal to

Options:
A -π4
B -π8
C -5π12
D -4π9

The correct answer is:
-π8

Solution:
Given,

tan-1cos15π4-1sinπ4

=tan-1cos4π-π4-1sinπ4

=tan-112-112

=tan-11-2 =-tan-12-1

=-π8

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The value of tan2tan-135+sin-1513 is equal to:

Options:
A -18169
B 22021
C -29176
D 15163

The correct answer is:
22021

Solution:
We have,

tan2tan-135+sin-1513

=tantan-135+tan-135+sin-1513

=tantan-135+351-925+sin-1513

=tantan-1158+tan-1512

=tan-1158+5121-158·512=tan-122021

=tantan-122021=22021

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The value of $\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]$ is

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

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

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

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $\tan 7 \frac{1}{2}^{\circ}$. is 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:
We have $\tan A=\frac{\sin A}{\cos A}=\frac{2 \sin A \cos A}{2 \cos ^2 A}=\frac{\sin 2 A}{1+\cos ^2 A}$
Putting
$A=7 \frac{1}{2}^{\circ} \Rightarrow \tan 7{\frac{1^{\circ}}{2}}^{\circ}=\frac{\sin 15^{\circ}}{1+\cos 15^{\circ}}$
On simplification, we get
$\tan 7 \frac{1}{2}^{\circ}=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of tan9o-tan27o-tan63o+tan81o is _____.

Options:

The correct answer is:
4

Solution:
Given,

tan9o-tan27o-tan63o+tan81o

=cot81°+tan81o-tan27°+cot27o

=tan9°+cot9°-tan27°+cot27°

=sin9°cos9°+cos9°sin9°-sin27°cos27°+cos27°sin27°

=2sin18o-2sin54o

=2×45-1-2×45+1

=815-1-15+1

=85+1-5-15-15+1

=825-1=4

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of tan9o-tan27o-tan63o+tan81o is _____.

Options:

The correct answer is:
4

Solution:
Given,

tan9o-tan27o-tan63o+tan81o

=cot81°+tan81o-tan27°+cot27o

=tan9°+cot9°-tan27°+cot27°

=sin9°cos9°+cos9°sin9°-sin27°cos27°+cos27°sin27°

=2sin18o-2sin54o

=2×45-1-2×45+1

=815-1-15+1

=85+1-5-15-15+1

=825-1=4

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The value of tan⁡63o-cot⁡63o is equal to

Options:
A 25+110-25
B 25+110+25
C 5-1410-25
D 5-1410+25

The correct answer is:
25+110-25

Solution:
sin⁡63ocos⁡63o-cos⁡63osin⁡63o=sin263o-cos2⁡63ocos⁡63osin⁡63o
=-cos⁡126o12sin⁡126o=-2cot⁡126o
=-2cotπ2+36∘=2tan36∘=2sin36∘cos36∘
=25+110-25

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

Options:
A $\frac{6}{17}$
B $\frac{7}{16}$
C $\frac{16}{7}$
D $\frac{17}{6}$

The correct answer is:
$\frac{17}{6}$

Solution:
$\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)$
$=\tan \tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right)=\tan \tan ^{-1}\left(\frac{17}{6}\right)=\frac{17}{6}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of \( \tan \left\{\cos ^{-1}\left(-\frac{2}{7}\right)-\frac{\pi}{2}\right\} \) is

Options:
A \( \frac{2}{3 \sqrt{5}} \)
B \( \frac{2}{3} \)
C \( \frac{1}{\sqrt{5}} \)
D \( \frac{4}{\sqrt{5}} \)

The correct answer is:
\( \frac{2}{3 \sqrt{5}} \)

Solution:
Let E=tan⁡cos-1⁡-27-π2
=tan⁡π-cos-1⁡ 27-π2

=tan⁡π2-cos-1⁡ 27

=tan⁡ sin-1⁡27

Since, sin-127=tan-1271-272
Now, E=tan⁡tan-1⁡235

=2 35

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are ℓ1 and ℓ2, respectively. The actual length of the metal wire is:

Options:
A T1ℓ2-T2ℓ1 T1-T2
B T1ℓ1-T2ℓ2 T1-T2
C ℓ1+ℓ22
D T1 T2ℓ1ℓ2

The correct answer is:
T1ℓ2-T2ℓ1 T1-T2

Solution:
Y=FLAΔL
⇒Y=T1ℓ0 Aℓ1-ℓ0=T2ℓ0 Aℓ2-ℓ0
1=T1ℓ2-ℓ0T2ℓ1-ℓ0
T2ℓ1-T2ℓ0=T1ℓ2-T1ℓ0
T1-T2ℓ0=T1ℓ2-T2ℓ1
ℓ0=T1ℓ2-T2ℓ1 T1-T2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The value of the acceleration due to gravity is g1 at a height h=R2 (R= radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface the earth. The ratio dR equals:

Options:
A 49
B 59
C 13
D 79

The correct answer is:
59

Solution:
Given that

gn=gd

GM(R+h)2=GMR3(R-d)

GM(R+R/2)2=GMR3(R-d)

4GM9R2=GMR21-dR

49=1-dR

dR=1-49=59

d=59R

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The value of the definite integral Î=â«-11lnâ¡2-sin3x2+sin3xdx is equal to

Options:
A ln⁡4
B ln⁡2
C 0
D ln⁡12

The correct answer is:
0

Solution:
Let, ln⁡2-sin3x2+sin3x=fx
Then, f-x=ln⁡2+sin3x2-sin3x
=-ln⁡2-sin3x2+sin3x
=-fx
Hence, ∫-11fxdx=0

Topic: Mathematics, Definite Integration, JEE Main
The value of the definite integral ∫-π4π4dx1+excosxsin4x+cos4x is equal to :

Options:
A -π2
B π22
C -π4
D π2

The correct answer is:
π22

Solution:
I=∫-π4π4dx1+excosxsin4x+cos4x …1

Using ∫abfxdx=∫abfa+b-xdx

I=∫-π/4π/4dx1+e-xcosxsin4x+cos4x ...2

Add 1 and 2

2I=∫-π4π4dxsin4x+cos4x

⇒2I=2∫0π4dxsin4x+cos4x

⇒I=∫0π41+tan2xsec2xtan4x+1dx

⇒I=∫0π41+1tan2xsec2xtanx-1tanx2+2dx

Put tanx-1tanx=t

⇒1+1tan2xsec2xdx=dt

So, I=∫-∞0dtt2+2=12tan-1t2-∞0

⇒I=0-12-π2=π22

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The value of the definite integral ∫π/245π/24dx1+tan2x3 is

Options:
A π3
B π6
C π12
D π18

The correct answer is:
π12

Solution:
Let I=∫π/245π/24dx1+tan2x3

⇒I=∫π/245π/24(cos2x)1/3(cos2x)1/3+(sin2x)1/3dx ...i

⇒I=∫π/245π/24cos2π4-x13cos2π4-x13+sin2π4-x13dx ∫abfxdx=∫abfa+b-xdx

So, I=∫π/245π/24(sin2x)1/3(sin2x)1/3+(cos2x)1/3dx ...ii

Adding the equation (i)+(ii), we get

Hence, 2I=∫π/245π/24dx

⇒2I=xπ245π24

⇒2I=5π24-π24

⇒2I=4π24⇒I=π12

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The value of the definite integral I=â«lnâ¡32lnâ¡23lnâ¡2-tan7â¡x2+tan7â¡xdx is equal to

Options:
A ln⁡4
B ln⁡2
C 0
D ln⁡12

The correct answer is:
0

Solution:
If fx=ln⁡2-tan7⁡x2+tan7⁡x, then
f-x=ln⁡2+tan7⁡x2-tan7⁡x
=-ln⁡2-tan7⁡x2+tan7⁡x
=-fx

So, fx is an odd function
Hence, ∫ln⁡32ln⁡23fxdx=0

Topic: Mathematics, Definite Integration, JEE Main
The value of the determinant $\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$ is

Options:
A $a+b+c$
B $(a+b+c)^2$
C $0$
D $1+a+b+c$

The correct answer is:
$0$

Solution:
$\Delta=\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|=(a+b+c)\left|\begin{array}{lll}1 & 1 & b+c \\ 1 & 1 & c+a \\ 1 & 1 & a+b\end{array}\right| \quad\left(C_2 \rightarrow C_2+C_3\right)$
$=0, \quad\left(C_1 = C_2\right)$.

Topic: Mathematics, Determinants, JEE Main
The value of the determinant ∆=13+325515+265103+65155 is equal to

Options:
A 152-253
B 253-152
C 35
D -152+73

The correct answer is:
152-253

Solution:
Taking 5 common from C2 and C3 , we get
∆=5213+32115+26523+6535
Applying C1→C1-13C3-3C2, we get
∆=5-321052035=5-35-6
=518-253=152-253

Topic: Mathematics, Determinants, JEE Main
The value of the determinant $\left|\begin{array}{ccc}31 & 37 & 92 \\ 31 & 58 & 71 \\ 31 & 105 & 24\end{array}\right|$ is

Options:
A -2
B 0
C 81
D None of these

The correct answer is:
0

Solution:
$\left|\begin{array}{ccc}31 & 37 & 92 \\ 31 & 58 & 71 \\ 31 & 105 & 24\end{array}\right|=\left|\begin{array}{ccc}31 & 37 & 92 \\ 0 & 21 & -21 \\ 0 & 47 & -47\end{array}\right|$ ;$\begin{aligned} & R_3 \rightarrow R_3-R_2 \\ & R_2 \rightarrow R_2-R_1\end{aligned}$
$=\left|\begin{array}{ccc}31 & 129 & 92 \\ 0 & 0 & -21 \\ 0 & 0 & -47\end{array}\right|=0$; (by $C_2 \rightarrow C_2+C_3$ ).

Topic: Mathematics, Determinants, JEE Main
The value of the determinant
$\left|\begin{array}{lll}265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{array}\right|$ is

Options:
A 1000
B 779
C 679
D 0

The correct answer is:
0

Solution:
Applying $\mathrm{C}_{1}-\mathrm{C}_{2}$ and $\mathrm{C}_{2}-\mathrm{C}_{3},$ we get
$\begin{aligned}
\text { Det. } &=\left|\begin{array}{ccc}
25 & 21 & 219 \\
15 & 27 & 198 \\
21 & 17 & 181
\end{array}\right|=\left|\begin{array}{ccc}
4 & 21 & 9 \\
-12 & 27 & -72 \\
4 & 17 & 11
\end{array}\right| \\
=&\left[\begin{array}{llc}
4 & 21 & 9 \\
0 & 90 & -45 \\
0 & -4 & 2
\end{array} \mid\left[\mathrm{By} \mathrm{R}_{2}+3 \mathrm{R}_{1}, \mathrm{R}_{3}-\mathrm{R}_{1}\right]\right.\\
=4(180-180)=0
\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
The value of the determinant \( \left|\begin{array}{lll}10 ! & 11 ! & 12 ! \\ 11 ! & 12 ! & 13 ! \\ 12 ! & 13 ! & 14 !\end{array}\right| \) is

Options:
A \( 2(10 ! 11 !) \)
B \( 2(10 ! 13 !) \)
C \( 2(10 ! 11 ! 12 !) \)
D \( 2(11 ! 12 ! 13 !) \)

The correct answer is:
\( 2(10 ! 11 ! 12 !) \)

Solution:
Let ∆=10!11!12!11!12!13!12!13!14!
=10! 11! 12!11111×1211212×1311313×14
Applying R2→R2-R1 and R3→R3-R1
=10! 11! 12!11111×1201240250
=10! 11! 12!50-48
=2⋅10! 11! 12!

Topic: Mathematics, Determinants, JEE Main
The value of the expression 1+cosecπ4+cosecπ8+cosecπ16 is equal to

Options:
A cotπ8
B cotπ16
C cotπ32
D cosec2π16

The correct answer is:
cotπ32

Solution:
We know that
cosecÎ¸+cotâ¡Î¸=1+cosâ¡Î¸sinâ¡Î¸=2cos2Î¸22sinÎ¸2cosÎ¸2=cotÎ¸2
âcosecÎ¸=cotÎ¸2-cotâ¡Î¸
âcosecÏ4=cotÏ8-cotÏ4
cosecÏ8=cotÏ16-cotÏ8
cosecÏ16=cotÏ32-cotÏ16
Adding them, we get,
cosecÏ4+cosecÏ8+cosecÏ16=cotÏ32-cotÏ4
â1+cosecÏ4+cosecÏ8+cosecÏ16=cotÏ32

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of the expression 2(sin1Â°+sin2Â°+sin3Â°+........+sin89Â°)2(cos1Â°+cos2Â°+cos3Â°+........+cos44Â°)+1 is equal to

Options:
A 2
B 12
C 12
D 0

The correct answer is:
2

Solution:
2sin1o+sin89o+sin2o+sin88o+.......+sin44o+sin46o+sin45o2cos1o+cos2o+cos3o+.......+cos44o+1
=22sin45ocos44o+cos43o+.......+cos1o+sin45o2cos1o+cos2o+cos3o+.......+cos44o+1
=2sin45o2cos1o+cos2o+.......+cos44o+12cos1o+cos2o+.......+cos44o+1
=2⋅12
=2

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of the expression cot -1 1 2 + cot -1 9 2 + cot -1 2 5 2 + cot -1 4 9 2 + ....... upto n terms is

Options:
A tan-1⁡2n
B tan-1⁡2n-1
C tan-1⁡n
D tan-1⁡2n-tan-1⁡1

The correct answer is:
tan-1⁡2n

Solution:
Given expression = tan -1 2 + tan -1 2 9 + tan -1 2 2 5 + tan -1 2 4 9 + ........
General term=22n-12=24n2-4n+1=21+4nn-1=2n-2n-21+2n2n-2
Tn=tan-1⁡2n-tan-1⁡2n-2
∴ Sum of the series =tan-1⁡2-tan-1⁡0+tan-1⁡4-tan-1⁡2+tan-1⁡6-tan-1⁡4+…tan-1⁡2n-tan-1⁡2n-2

=tan-1⁡2n-tan-1⁡0=tan-1⁡2n

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of the expression ∑k=027k⋅Ck 2713k2327-k is equal to

Options:

The correct answer is:
9

Solution:
Given expression can be rewritten as∑k=127k⋅27kCk-1 261313k-12326-(k-1)
=9∑k=127Ck-1 2613k-12326-(k-1)
=9∑k-1=026Ck-1 262326-(k-1)13k-1
=923+1326
=9×1=9

Topic: Mathematics, Binomial Theorem, JEE Main
The value of the expression ∑r=0∞12r∑k=0∞13k∑P=0∞23P is

Options:
A 1
B 2
C 12
D 43

The correct answer is:
1

Solution:
âr=0â12r
=1+12+122....â
=a1-r=11-12=2
(Applying formula of the sum of infinite G.P.)
Similarly, âK=0â13K=11-13=32
& âP=0â23P=11-23=3
Hence, the expression simplifies to 2Ã323=1

Topic: Mathematics, Sequences and Series, JEE Main
The value of the integral ∫ 0 1 1−x 1+x dx is

Options:
A -1
B 1
C âÏ 2 â1
D π 2 +1

The correct answer is:
​π 2 −1

Solution:
I=â«01[1âx1+xÃ1âx1âx]ââdxââââââââââââââââ(rationalising the denominator)

= â« 0 1 1âx 1â x 2 dx = â« 0 1 1 1â x 2 dxâ â« 0 1 xâdx 1â x 2

âââI= [ sin â1 x ] 0 1 â 1 2 â« 0 1 2x 1â x 2 dx

=[ Ï 2 â0 ]+ 1 2 [ 2 1â x 2 ] 0 1

= Ï 2 +[ 0â1 ] = Ï 2 â1

Topic: Mathematics, Definite Integration, JEE Main