JEE Main Mathematics — Trigonometry previous year questions with solutions.
Given that the inverse trigonometric function assumes principal values only. Let $x$, $y$ be any two real numbers in $[-1,1]$ such that $\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$. Then, the minimum value of $x^2+y^2+2 x y \sin \alpha$ is
If $\alpha ,-\frac{\pi }{2}<\alpha <\frac{\pi }{2}$ is the solution of $4\mathrm{cos}\theta +5\mathrm{sin}\theta =1$, then the value of $\mathrm{tan}\alpha$ is
If $2{\mathrm{sin}}^{3}x+\mathrm{sin}2x\mathrm{cos}x+4\mathrm{sin}x-4=0$ has exactly $3$ solutions in the interval $[0,\frac{n\pi }{2}⌉,n\in N$, then the roots of the equation ${x}^{2}+nx+(n-3)=0$ belong to :
If the value of $\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$ is $\frac{a \sqrt{5}-b}{c}$, where $a, b, c$ are natural numbers and $\operatorname{gcd}(a, c)=1$, then $a+b+c$ is equal to :
The integral $\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$ is equal to
Let $y=f(x)$ represent a parabola with focus $(-\frac{1}{2},0)$ and directrix $y=-\frac{1}{2}$. Then $S={x\in \mathbb{R}:{\mathrm{tan}}^{-1}(\sqrt{f(x)}+{\mathrm{sin}}^{-1}(\sqrt{f(x)+1}))=\frac{\pi }{2}}:$
Let $S$ be the set of all solutions of the equation ${\mathrm{cos}}^{-1}(2x)-2{\mathrm{cos}}^{-1}(\sqrt{1-{x}^{2}})=\pi ,x\in [-\frac{1}{2},\frac{1}{2}]$. Then $\underset{x\in S}{\sum }2{\mathrm{sin}}^{-1}({x}^{2}-1)$ is equal to
$96\mathrm{cos}\frac{\pi }{33}\mathrm{cos}\frac{2\pi }{33}\mathrm{cos}\frac{4\pi }{33}\mathrm{cos}\frac{8\pi }{33}\mathrm{cos}\frac{16\pi }{33}$ is equal to
The value of $\mathrm{tan}{9}^{{}^{o}}-\mathrm{tan}{27}^{{}^{o}}-\mathrm{tan}{63}^{{}^{o}}+\mathrm{tan}{81}^{{}^{o}}$ is _____.
Let $(a,b)\subset (0,2\pi )$ be the largest interval for which ${\mathrm{sin}}^{-1}(\mathrm{sin}\theta )-{\mathrm{cos}}^{-1}(\mathrm{sin}\theta )>0,\theta \in (0,2\pi )$, holds . If $\alpha {x}^{2}+\beta x+{\mathrm{sin}}^{-1}({x}^{2}-6x+10)+{\mathrm{cos}}^{-1}({x}^{2}-6x+10)=0$ and $\alpha -\beta =b-a$, then $\alpha$ is equal to;
For $x\in (-1,1]$, the number of solutions of the equation ${\mathrm{sin}}^{-1}x=2{\mathrm{tan}}^{-1}x$ is equal to
If ${\mathrm{sin}}^{-1}\frac{\alpha }{17}+{\mathrm{cos}}^{-1}\frac{4}{5}-{\mathrm{tan}}^{-1}\frac{77}{36}=0,0<\alpha <13$, then ${\mathrm{sin}}^{-1}(\mathrm{sin}\alpha )+{\mathrm{cos}}^{-1}(\mathrm{cos}\alpha )$ is equal to
Let ${a}_{1}=1,{a}_{2},{a}_{3},{a}_{4},\ldots .$. be consecutive natural numbers. Then ${\mathrm{tan}}^{-1}(\frac{1}{1+{a}_{1}{a}_{2}})+{\mathrm{tan}}^{-1}(\frac{1}{1+{a}_{2}{a}_{3}})$ $+\ldots ..+{\mathrm{tan}}^{-1}(\frac{1}{1+{a}_{2021}{a}_{2022}})$ is equal to
If $\mathrm{tan}15^{\circ}+\frac{1}{\mathrm{tan}75^{\circ}}+\frac{1}{\mathrm{tan}105^{\circ}}+\mathrm{tan}195^{\circ}=2a$, then the value of $(a+\frac{1}{a})$ is :
The value of $36(4{\mathrm{cos}}^{2}{9}^{\circ }-1)(4{\mathrm{cos}}^{2}{27}^{\circ }-1)(4{\mathrm{cos}}^{2}{81}^{\circ }-1)(4{\mathrm{cos}}^{2}{243}^{\circ }-1)$ is
The general solution of sin²θ - 2cosθ + 1/4 = 0 is:
Let $S={x\in R:0<x<1\text{and}2{\mathrm{tan}}^{-1}(\frac{1-x}{1+x})={\mathrm{cos}}^{-1}(\frac{1-{x}^{2}}{1+{x}^{2}})}$. If $n(S)$ denotes the number of elements in $S$ then :
${\mathrm{tan}}^{-1}(\frac{1+\sqrt{3}}{3+\sqrt{3}})+{\mathrm{sec}}^{-1}\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}=$
The set of all values of $\lambda$ for which the equation ${\mathrm{cos}}^{2}2x-2{\mathrm{sin}}^{4}x-2{\mathrm{cos}}^{2}x=\lambda$
If $S={x\in \mathbb{R}:{\mathrm{sin}}^{-1}(\frac{x+1}{\sqrt{{x}^{2}+2x+2}})-{\mathrm{sin}}^{-1}(\frac{x}{\sqrt{{x}^{2}+1}})=\frac{\pi }{4}}$ then $\underset{x\in S}{\sum }(\mathrm{sin}(({x}^{2}+x+5)\frac{\pi }{2})-\mathrm{cos}(({x}^{2}+x+5)\pi ))$ is equal to $_________.$
If the sum of all the solutions of ${\mathrm{tan}}^{-1}(\frac{2x}{1-{x}^{2}})+{\mathrm{cot}}^{-1}(\frac{1-{x}^{2}}{2x})=\frac{\pi }{3}$, $-1<x<1,x\neq 0$, is $\alpha -\frac{4}{\sqrt{3}}$, then $\alpha$ is equal to _____ .
The number of elements in the set $S={\theta \in [0,2\pi ]:3{\mathrm{cos}}^{4}\theta -5{\mathrm{cos}}^{2}\theta -2{\mathrm{sin}}^{6}\theta +2=0}$ is
Let $S={x\in (-\frac{\pi }{2},\frac{\pi }{2}):{9}^{1-{\mathrm{tan}}^{2}x}+{9}^{{\mathrm{tan}}^{2}x}=10}$ and $\beta =\underset{x\in S}{\sum }{\mathrm{tan}}^{2}(\frac{x}{3})$, then $\frac{1}{6}(\beta -14{)}^{2}$ is equal to
${\mathrm{sin}}^{-1}(\mathrm{sin}\frac{2\pi }{3})+{\mathrm{cos}}^{-1}(\mathrm{cos}\frac{7\pi }{6})+{\mathrm{tan}}^{-1}(\mathrm{tan}\frac{3\pi }{4})$ is equal to