LFM Pure

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jee-main 2022 Q72 Inverse trigonometric equation View
The value of $\cot \left( \sum _ { n = 1 } ^ { 50 } \tan ^ { - 1 } \left( \frac { 1 } { 1 + n + n ^ { 2 } } \right) \right)$ is
(1) $\frac { 25 } { 26 }$
(2) $\frac { 50 } { 51 }$
(3) $\frac { 26 } { 25 }$
(4) $\frac { 52 } { 51 }$
jee-main 2022 Q86 Solve trigonometric equation for solutions in an interval View
Let $S = \left[ - \pi , \frac { \pi } { 2 } \right) - \left\{ - \frac { \pi } { 2 } , - \frac { \pi } { 4 } , - \frac { 3 \pi } { 4 } , \frac { \pi } { 4 } \right\}$. Then the number of elements in the set $A = \{ \theta \in S : \tan \theta ( 1 + \sqrt { 5 } \tan ( 2 \theta ) ) = \sqrt { 5 } - \tan ( 2 \theta ) \}$ is $\_\_\_\_$.
jee-main 2022 Q87 Inverse trigonometric equation View
For $k \in \mathbb { R }$, let the solutions of the equation $\cos \left( \sin ^ { - 1 } \left( x \cot \left( \tan ^ { - 1 } \left( \cos \left( \sin ^ { - 1 } x \right) \right) \right) \right) \right) = k , 0 < | x | < \frac { 1 } { \sqrt { 2 } }$ be $\alpha$ and $\beta$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $x ^ { 2 } - b x - 5 = 0$ are $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } }$ and $\frac { \alpha } { \beta }$, then $\frac { b } { k ^ { 2 } }$ is equal to $\_\_\_\_$ .
jee-main 2023 Q69 Solve trigonometric equation for solutions in an interval View
If $m$ and $n$ respectively are the numbers of positive and negative value of $\theta$ in the interval $[ - \pi , \pi ]$ that satisfy the equation $\cos 2 \theta \cos \frac { \theta } { 2 } = \cos 3 \theta \cos \frac { 9 \theta } { 2 }$, then $m n$ is equal to $\_\_\_\_$.
jee-main 2023 Q70 Inverse trigonometric equation View
If $\sin^{-1}\frac{\alpha}{17} + \cos^{-1}\frac{4}{5} - \tan^{-1}\frac{77}{36} = 0$, $0 < \alpha < 13$, then $\sin^{-1}(\sin\alpha) + \cos^{-1}(\cos\alpha)$ is equal to
(1) $\pi$
(2) 16
(3) 0
(4) $16 - 5\pi$
jee-main 2023 Q71 Inverse trigonometric equation View
Let $y = f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix $y = -\frac{1}{2}$. Then $S = \left\{x \in \mathbb{R} : \tan^{-1}\sqrt{f(x)} + \sin^{-1}\sqrt{f(x)+1} = \frac{\pi}{2}\right\}$:
(1) contains exactly two elements
(2) contains exactly one element
(3) is an infinite set
(4) is an empty set
jee-main 2023 Q76 Solve trigonometric equation for solutions in an interval View
Let $S = \{\theta \in [0, 2\pi): \tan(\pi\cos\theta) + \tan(\pi\sin\theta) = 0\}$. Then $\sum_{\theta \in S} \sin\left(\theta + \frac{\pi}{4}\right)$ is equal to $\_\_\_\_$.
jee-main 2023 Q76 Inverse trigonometric equation View
If the sum of all the solutions of $\tan ^ { - 1 } \left( \frac { 2 x } { 1 - x ^ { 2 } } \right) + \cot ^ { - 1 } \left( \frac { 1 - x ^ { 2 } } { 2 x } \right) = \frac { \pi } { 3 } , - 1 < x < 1 , x \neq 0$, is $\alpha - \frac { 4 } { \sqrt { 3 } }$, then $\alpha$ is equal to $\_\_\_\_$ .
jee-main 2024 Q64 Solve trigonometric equation for solutions in an interval View
If $\alpha , - \frac { \pi } { 2 } < \alpha < \frac { \pi } { 2 }$ is the solution of $4 \cos \theta + 5 \sin \theta = 1$, then the value of $\tan \alpha$ is
(1) $\frac { 10 - \sqrt { 10 } } { 6 }$
(2) $\frac { 10 - \sqrt { 10 } } { 12 }$
(3) $\frac { \sqrt { 10 } - 10 } { 12 }$
(4) $\frac { \sqrt { 10 } - 10 } { 6 }$
jee-main 2024 Q64 Solve trigonometric equation for solutions in an interval View
If $2 \tan ^ { 2 } \theta - 5 \sec \theta = 1$ has exactly 7 solutions in the interval $0 , \frac { \mathrm { n } \pi } { 2 }$, for the least value of $\mathrm { n } \in \mathrm { N }$ then $\sum _ { \mathrm { k } = 1 } ^ { \mathrm { n } } \frac { \mathrm { k } } { 2 ^ { \mathrm { k } } }$ is equal to :
(1) $\frac { 1 } { 2 ^ { 15 } } 2 ^ { 14 } - 14$
(2) $\frac { 1 } { 2 _ { 1 } ^ { 14 } } 2 ^ { 15 } - 15$
(3) $1 - \frac { 15 } { 2 ^ { 13 } }$
(4) $\frac { 1 } { 2 ^ { 13 } } 2 ^ { 14 } - 15$
jee-main 2024 Q70 Inverse trigonometric equation View
Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan ^ { - 1 } ( \mathrm { x } ) + \tan ^ { - 1 } ( 2 \mathrm { x } ) = \frac { \pi } { 4 }$ is :
(1) More than 2
(2) 1
(3) 2
(4) 0
jee-main 2024 Q72 Inverse trigonometric equation View
If $a = \sin^{-1}(\sin 5)$ and $b = \cos^{-1}(\cos 5)$, then $a^2 + b^2$ is equal to
(1) $4\pi^2 + 25$
(2) $8\pi^2 - 40\pi + 50$
(3) $4\pi^2 - 20\pi + 50$
(4) 25
jee-main 2024 Q83 Solve trigonometric equation for solutions in an interval View
Let the set of all $a \in R$ such that the equation $\cos 2 x + a \sin x = 2 a - 7$ has a solution be $[ p , q ]$ and $r = \tan 9 ^ { \circ } - \tan 27 ^ { \circ } - \frac { 1 } { \cot 63 ^ { \circ } } + \tan 81 ^ { \circ }$, then $p q r$ is equal to $\_\_\_\_$.
jee-main 2024 Q83 Solve trigonometric equation for solutions in an interval View
The number of solutions of $\sin ^ { 2 } x + \left( 2 + 2 x - x ^ { 2 } \right) \sin x - 3 ( x - 1 ) ^ { 2 } = 0$, where $- \pi \leq x \leq \pi$, is $\_\_\_\_$
jee-main 2024 Q86 Inverse trigonometric equation View
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin ^ { - 1 } x + 3 \cos ^ { - 1 } x = \frac { 2 \pi } { 5 }$, is $\_\_\_\_$
jee-main 2025 Q3 Trigonometric equation with algebraic or logarithmic coupling View
Let $\mathrm{A} = \left\{ x \in (0, \pi) - \left\{ \frac{\pi}{2} \right\} : \log_{(2/\pi)} |\sin x| + \log_{(2/\pi)} |\cos x| = 2 \right\}$ and $\mathrm{B} = \{ x \geqslant 0 : \sqrt{x}(\sqrt{x} - 4) - 3|\sqrt{x} - 2| + 6 = 0 \}$. Then $\mathrm{n}(\mathrm{A} \cup \mathrm{B})$ is equal to:
(1) 4
(2) 8
(3) 6
(4) 2
jee-main 2025 Q18 Solve trigonometric equation for solutions in an interval View
The sum of all values of $\theta \in [ 0,2 \pi ]$ satisfying $2 \sin ^ { 2 } \theta = \cos 2 \theta$ and $2 \cos ^ { 2 } \theta = 3 \sin \theta$ is
(1) $4 \pi$
(2) $\frac { 5 \pi } { 6 }$
(3) $\pi$
(4) $\frac { \pi } { 2 }$
jee-main 2025 Q22 Inverse trigonometric equation View
If for some $\alpha, \beta$; $\alpha \leq \beta$, $\alpha + \beta = 8$ and $\sec^2(\tan^{-1}\alpha) + \operatorname{cosec}^2(\cot^{-1}\beta) = 36$, then $\alpha^2 + \beta$ is $\underline{\hspace{2cm}}$.
kyotsu-test 2010 QCourse2-IV-Q2 Geometric problem using trigonometric function graphs View
Q2 Suppose that the curve $y=2\cos 2x$ and the curve $y=4\cos x+k$ have a common tangent at $x=a\left(0(1) We set $f(x)=2\cos 2x$ and $g(x)=4\cos x+k$. Since we have assumed that $y=f(x)$ and $y=g(x)$ have a common tangent at $x=a$, we see that
$$f'(a)=g'(a), \quad f(a)=g(a).$$
Since $f'(a)=g'(a)$ and $0Hence the coordinates of the tangent point are $\left(\frac{\pi}{\mathbf{O}},-\mathbf{Q}\right)$, and the equation of the common tangent line is
$$y=-\mathbf{R}\sqrt{\mathbf{S}}\left(x-\frac{\pi}{\mathbf{T}}\right)-\mathbf{U}.$$
(2) We are to find the area $S$ of the region bounded by these two curves over the range $-\frac{\pi}{2}\leqq x\leqq\frac{\pi}{2}$.
Since both of these curves are symmetric with respect to the $y$-axis, by putting $b=\mathbf{V}$ and $c=\frac{\pi}{\mathbf{O}}$ we have
$$S=\mathbf{W}\int_{b}^{c}(2\cos 2x-4\cos x-k)\,dx$$
By calculating this, we obtain
$$S=\mathbf{X}\cdot\pi-\mathbf{Y}\cdot\mathbf{Z}.$$
taiwan-gsat 2023 Q18 3 marks Evaluate trigonometric expression given a constraint View
On the coordinate plane, $O$ is the origin, and points $A(1,0)$ and $B(-2,0)$ are given. There are also two points $P$ and $Q$ in the upper half-plane satisfying $\overline{AP} = \overline{OA}$, $\overline{BQ} = \overline{OB}$, $\angle POQ$ is a right angle. Let $\angle AOP = \theta$.
The length of line segment $\overline{OP}$ is which of the following options? (Single choice question, 3 points)
(1) $\sin\theta$
(2) $\cos\theta$
(3) $2\sin\theta$
(4) $2\cos\theta$
(5) $\cos 2\theta$
taiwan-gsat 2024 Q4 5 marks Solve trigonometric equation for solutions in an interval View
How many real numbers $x$ satisfy $\sin\left(x + \frac{\pi}{6}\right) = \sin x + \sin\frac{\pi}{6}$ and $0 \leq x < 2\pi$?
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5 or more
taiwan-gsat 2025 Q1 6 marks Solve trigonometric equation for solutions in an interval View
On the coordinate plane, the graph of the function $y = \sin x$ is symmetric about $x = \frac{\pi}{2}$, as shown in the figure. Find the value of $\theta$ in the range $0 < \theta \leq \pi$ that satisfies $\sin \theta = \sin\left(\theta + \frac{\pi}{5}\right)$.
(1) $\frac{\pi}{5}$
(2) $\frac{2\pi}{5}$
(3) $\frac{3\pi}{5}$
(4) $\frac{4\pi}{5}$
(5) $\pi$
taiwan-gsat 2025 Q20 6 marks Applied trigonometric modeling View
Continuing from question 19, where $f ( x ) = a \sin ( b x )$ with $f(0) = f(12) = 0$ and $f(2) = 4$, a person wants to sunbathe when the UVI value is between $4 \sqrt { 2 }$ and $4 \sqrt { 3 }$ (inclusive). The time during which he can sunbathe is set as $t$ hours after sunrise. Find the maximum possible range of $t$.
turkey-yks 2013 Q23 Solve trigonometric equation for solutions in an interval View
Given that $0 < \mathrm { x } < \pi$,
$$\sin ^ { 4 } x = \cos ^ { 4 } x$$
What is the sum of the $\mathbf { x }$ values that satisfy this equality?
A) $\frac { 3 \pi } { 2 }$
B) $\frac { 4 \pi } { 3 }$
C) $\frac { 5 \pi } { 4 }$
D) $\pi$
E) $2 \pi$
turkey-yks 2014 Q24 Evaluate trigonometric expression given a constraint View
ABCD is a square $| \mathrm { AB } | = 3$ units $| \mathrm { BE } | = | \mathrm { CF } | = 1$ unit $m ( \widehat { F A E } ) = x$
According to the given information above, what is the value of $\cot \mathrm { x }$?
A) $\frac { 6 } { 5 }$
B) $\frac { 8 } { 5 }$
C) $\frac { 7 } { 6 }$
D) $\frac { 9 } { 7 }$
E) $\frac { 11 } { 8 }$