JEE Main Trigonometric Ratio and Identites Practice Questions With Solutions

If the value of $\frac{3\cos {36}^{\circ }+5\sin {18}^{\circ }}{5\cos {36}^{\circ }-3\sin {18}^{\circ }}$\frac{3 cos ⁡ 36^{\circ} + 5 sin ⁡ 18^{\circ}}{5 cos ⁡ 36^{\circ} - 3 sin ⁡ 18^{\circ}} is $\frac{a\sqrt{5}-b}{c}$\frac{a \sqrt{5} - b}{c}, where $a,b,c$a , b , c are natural numbers and $\gcd (a,c)=1$gcd \left(\right. a , c \left.\right) = 1, then $a+b+c$a + b + c is equal to :

If $\sin x=-\frac{3}{5}$sin ⁡ x = - \frac{3}{5}, where $\pi <x<\frac{3\pi }{2}$\pi < x < \frac{3 \pi}{2}, then $80({\tan }^{2}x-\cos x)$80 \left(\right. tan^{2} ⁡ x - cos ⁡ x \left.\right) is equal to

Suppose $\theta \in [0,\frac{\pi }{4}]$\theta \in \left[\right. 0 , \frac{\pi}{4} \left]\right. is a solution of $4\cos \theta -3\sin \theta =1$4 cos ⁡ \theta - 3 sin ⁡ \theta = 1. Then $\cos \theta$cos ⁡ \theta is equal to :

The number of solutions, of the equation $e^{\sin x}-2e^{-\sin x}=2$e^{sin ⁡ x} - 2 e^{- sin ⁡ x} = 2, is :

For $\alpha ,\beta \in (0,\pi /2)$\alpha , \beta \in \left(\right. 0 , \pi / 2 \left.\right), let $3\sin (\alpha +\beta )=2\sin (\alpha -\beta )$3 sin ⁡ \left(\right. \alpha + \beta \left.\right) = 2 sin ⁡ \left(\right. \alpha - \beta \left.\right) and a real number $k$k be such that $\tan \alpha =k\tan \beta$tan ⁡ \alpha = k tan ⁡ \beta. Then, the value of $k$k is equal to

$96\cos \frac{\pi }{33}\cos \frac{2\pi }{33}\cos \frac{4\pi }{33}\cos \frac{8\pi }{33}\cos \frac{16\pi }{33}$96 cos ⁡ \frac{\pi}{33} cos ⁡ \frac{2 \pi}{33} cos ⁡ \frac{4 \pi}{33} cos ⁡ \frac{8 \pi}{33} cos ⁡ \frac{16 \pi}{33} is equal to :

The value of $36(4{\cos }^2{9}^{\circ }-1)(4{\cos }^2{27}^{\circ }-1)(4{\cos }^2{81}^{\circ }-1)(4{\cos }^2{243}^{\circ }-1)$36 \left(\right. 4 cos^{2} ⁡ 9^{\circ} - 1 \left.\right) \left(\right. 4 cos^{2} ⁡ 27^{\circ} - 1 \left.\right) \left(\right. 4 cos^{2} ⁡ 81^{\circ} - 1 \left.\right) \left(\right. 4 cos^{2} ⁡ 243^{\circ} - 1 \left.\right) is :

If $\tan {15}^{\circ }+\frac{1}{\tan {75}^{\circ }}+\frac{1}{\tan {105}^{\circ }}+\tan {195}^{\circ }=2a$tan ⁡ 15^{\circ} + \frac{1}{tan ⁡ 75^{\circ}} + \frac{1}{tan ⁡ 105^{\circ}} + tan ⁡ 195^{\circ} = 2 a, then the value of $\left(a+\frac{1}{a}\right)$\left(\right. a + \frac{1}{a} \left.\right) is :

The set of all values of $\lambda$\lambda for which the equation ${\cos }^{2}2x-2{\sin }^{4}x-2{\cos }^{2}x=\lambda$cos^{2} 2 x - 2 sin^{4} x - 2 cos^{2} x = \lambda has a real solution $x$x, is :