Trigonometric Ratios & Identities · Mathematics · TS EAMCET
MCQ (Single Correct Answer)
If $\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{\beta}{2}\right)$, then $\frac{3+\sin ^2 \beta}{1+3 \sin ^2 \beta}=$
If $P=\sin \frac{2 \pi}{7}+\sin \frac{4 \pi}{7}+\sin \frac{8 \pi}{7}$ and $Q=\cos \frac{2 \pi}{7}+\frac{4 \pi}{7}+\cos \frac{8 \pi}{7}$, then the point $(P, Q)$ lies on the circle of radius
If $\cos \alpha=\frac{l \cos \beta+m}{l+m \cos \beta}$, then $\left(\frac{\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}}\right)^2=$
If $\cos \theta+\sin \theta=\sqrt{2} \cos \theta$ and $0<\theta<\frac{\pi}{2}$, then $\sec 2 \theta+\tan 2 \theta=$
If $x=\log _e 3$, then $\tanh 2 x+\operatorname{sech} 2 x=$
If $\sin A=-\frac{24}{25}, \cos B=\frac{15}{17}, A$ does not belong to 4th quadrant and $B$ does not belong to 1st quadrant, then $(A+B)$ lies in the quadrant
$$ 4 \cos \frac{7 \theta}{2} \cos \frac{3 \theta}{2} \sin 5 \theta= $$
$\cot h^2 x-\tanh ^2 x=$
If $3 \sin \theta+4 \cos \theta=3$ and $\theta \neq(2 n+1) \frac{\pi}{2}$, then $\sin 2 \theta=$
$$ \frac{\cos 15^{\circ} \cos ^2 22 \frac{1^{\circ}}{2}-\sin 75^{\circ} \sin ^2 \cdot 52 \frac{1^{\circ}}{2}}{\cos ^2 15^{\circ}-\cos ^2 75^{\circ}} $$
$16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ}=$
If $5 \sin \theta+3 \cos \left(\theta+\frac{\pi}{3}\right)+3$ lies between $\alpha$ and $\beta$ (including $\alpha, \beta$ also), then $(\alpha-\beta)(\alpha+\beta-6)=$
$$ \frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots \sin 89^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1}= $$
If $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$ and $\alpha+\beta \neq \frac{\pi}{2}$, then $\frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=$
If $\sin A=-\frac{60}{61}, \cot B=-\frac{40}{9}$ and neither $A$ and $B$ is in 4th quadrant, then $6 \cot A+4 \sec B=$
The period of the function $f(x)=\frac{2 \sin \left(\frac{\pi x}{3}\right) \cos \left(\frac{2 \pi x}{5}\right)}{3 \tan \left(\frac{7 \pi x}{2}\right)-5 \sec \left(\frac{5 \pi x}{3}\right)}$ is
If $A+B+C=4 S$, then $\sin (2 S-A)$
$$ +\sin (2 S-B)+\sin (2 S-C)-\sin 2 S= $$
If $1^{\circ}=0.0175$ radians, then the approximate value of $\sec 58^{\circ}$ is
If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$, then $27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha=$
If $\tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \cos ^2 \alpha}$, then $\tan (\alpha+\beta) \cdot \cot \alpha=$
If $\cos A+\cos B+\cos C=0=\sin A+\sin B+\sin C$, then $\cos (A-B)+\cos (B-C)+\cos (C-A)=$
If $\sin x \cdot \cosh y=\cos \theta$ and $\cos x \cdot \sinh y=\sin \theta$, then $\sin ^2 x+\cosh ^2 y=$
The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is
If $\cos \theta=\frac{-3}{5}$ and $\pi<\theta<\frac{3 \pi}{2}$, then $\tan \frac{\theta}{2}+\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2}=$
$$ \sin 6^{\circ}+\sin 54^{\circ}+\sin 126^{\circ}+\cos 156^{\circ}= $$
If $\tan \alpha=\frac{-12}{5}, \cot \beta=\frac{7}{24}, \alpha$ does not belong to second quadrant and $\beta$ does not belong to first quadrant, then $\sqrt{13} \sin \frac{\alpha}{2}+\cos \frac{\beta}{2}+\tan \frac{\alpha}{2} \cot \frac{\beta}{2}=$
$\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14}=$
If $\cot \theta=-\frac{2}{3}$ and $\theta$ does not lie in the 4 th quadrant, then $\frac{(5 \sin \theta+\cos \theta)^2}{\tan \theta+\cot \theta}=$
If $540^{\circ}<\theta<630^{\circ}$ and $\tan \theta=5 / 12$, then
$$ \frac{\cos \frac{\theta}{2}-5 \sin \frac{\theta}{2}}{\sqrt{-(12 \sec \theta+5 \operatorname{cosec} \theta)}}= $$
If $A+B+C+D=2 \pi$, then $\cos A-\cos B+\cos C-\cos D=$
If $\cosh x=\frac{4}{3}$, then $3 \cosh x+3^2 \cosh 2 x+3^3 \cosh 3 x=$
$$ \text { If } \frac{2 \sin \theta}{1+\cos \theta+\sin \theta}=y, \text { then } \frac{1-\cos \theta+\sin \theta}{1+\sin \theta}= $$
If $\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7}=\frac{\sin \frac{8 \pi}{7}}{8 \sin \frac{\pi}{7}}$, then $\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}=$
If $f(\theta)=\cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\theta-\frac{2 \pi}{3}\right)$, then $f\left(\frac{\pi}{5}\right)=$
If $\sin A=\frac{-7}{25}, \cos B=\frac{8}{17}, A$ does not lie in the 3rd quadrant and $B$ does not lie in the 1st quadrant, then $8 \tan A-5 \cot B=$
If $\sin \theta-\cos \theta=\frac{1}{\sqrt{3}}$, then $\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta)=$
If $a \tan \alpha+b \tan \beta=(a+b) \tan \left(\frac{\alpha+\beta}{2}\right)$ and $\alpha-\beta \neq 2 n \pi$ then $\frac{\cos \beta}{\cos \alpha}=$
If $\frac{5 \sinh 2 x}{7+6 \cosh 2 x}=\frac{3}{2}$, then $3 \tanh ^2 x+20 \tanh x=$
If $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B) =\frac{1}{2}$ and $0
$$ \frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}}= $$
If $A+B+C=\frac{\pi}{2}$, then $\sqrt{2} \cos \left(\frac{\pi}{4}-A\right)$
$$ +\sqrt{2} \cos \left(\frac{\pi}{4}-B\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-C\right)+1= $$
If $\sinh x=\tan A$, then $|\tanh x|=$
$$ \frac{\sinh (x+y)+\sinh (x-y)}{\cosh (x+y)-\cosh (x-y)}= $$
Let $\alpha$ be the period of $3 \sin \frac{\pi x}{3}-\cos \frac{\pi x}{2}+\tan \frac{\pi x}{4}, \beta$ be the period of $\sin ^2\left(\frac{\pi}{7}+\frac{x}{4}\right)-\sin ^2\left(\frac{\pi}{7}-\frac{x}{4}\right)$, and $\gamma$ be the period of $\cos ^4 x+\sin ^4 x$. Then, $\frac{\alpha \gamma}{\beta}=$
If $\theta$ does not lie in the second quadrant and $\tan \theta=\frac{-3}{4}$, then $\tan \frac{\theta}{2}+\sin 2 \theta=$
$$ \cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ}= $$
If $|\sin \alpha-\cos \alpha|=\frac{3}{4}$, then $|\sec 2 \alpha-\tan 2 \alpha|=$
If $\frac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{1}{\sin 46^{\circ} \sin 47^{\circ}}+\ldots$ up to 45 terms $=\frac{1}{\sin x^{\circ}}$, then $\sin \left(\frac{\pi}{2} x\right)=$
If $\sinh x=\frac{-1}{2}$, then $\tanh 2 x=$
If $\cos x+\cos y=p, \sin x+\sin y=q$, then $\cos \left(\frac{x-y}{2}\right)=$
If $A+B+C=\frac{3 \pi}{2}$, then $4 \sin A \sin B \sin C+\cos 2 A+\cos 2 B+\cos 2 C=$
$$ \frac{e^{4 x}+e^{-4 x}+14}{4\left(e^x-e^{-x}\right)^2}= $$
If $\tanh x=\frac{1}{2}$, then $\sinh 2 x-\operatorname{sech} 2 x=$
If $A$ and $B(A>B)$ are acute angles, $\sin (A-B)=\frac{16}{65}$ and $\sin B=\frac{5}{13}$, then $\tan A+\cot A=$
If $\tan A=\frac{2}{3}$, then $\sin 4 A=$
$$ \frac{\sqrt{2} \cos 45^{\circ}+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} $$
If $\theta=\frac{\pi}{12}$ and $x=\log \left(\cot \left(\frac{\pi}{4}+\theta\right)\right)$, then $\cosh x=$
$2 \cosh (x+y) \sinh (x-y)+\sinh 2 y=$
If $\cos x-\sin x=\sqrt{a} \sin x$, then $a \sin x+\cos x-\sin x=$
$$ \text { Match the items of List-I to the items of List-II } $$
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| $$ \text { List-I } $$ |
$$ \text { List-II } $$ |
||
|---|---|---|---|
| A. | The period of $\sin ^2 x$ is | I. | $$ \frac{2 \pi}{3} $$ |
| B. | $$ \begin{aligned} &\text { Maximum value of }\\ &\frac{\pi}{3}(\sqrt{3} \cos 3 x+\sin 3 x) \end{aligned} $$ |
II. | $$ 12 \pi $$ |
| C. | The period of $\sin \frac{x}{3}+\cos \frac{x}{2}$ is | III. | $$ \frac{\pi}{2} $$ |
| D. | Intersection points of $y=|\sin x|$ and $y=1$ in $(0, \pi)$ | IV. | $$ \frac{3\pi}{2} $$ |
| V | $$ \pi $$ |
||
$$ \text { The correct match is } $$
If $\cot \left(\frac{A}{2}\right)=\sqrt{\frac{1+a}{1-a}} \cdot \cot \left(\frac{\theta}{2}\right)$, then $\cos \theta=$
If $\sin \theta \cosh \alpha=\tan x, \cos \theta \sinh \alpha=\sec x$, then $\cos 2 \theta \cosh 2 \alpha=$
For $n \in \mathbf{N}$, if $f(n)=(\cos n x)(\sec x)^n$ and $g(n)=(\sin n x)(\sec x)^n$, then $f(2020)-f(2019)+(\tan x) g(2019)=$
$\theta$ and $\alpha$ lie in $Q_3$. If $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in harmonic progression, then $\cos \theta \sec \frac{\alpha}{2}=$
The ratio of the maximum and minimum values attained by the function $f(x)=1+2 \sin x+3 \cos ^2 x, 0 \leq x \leq \frac{2 \pi}{3}$ is
$$ \text { Match the items of List-I with those of List-II } $$
| $$ \text { List-I } $$ |
$$ \text { List-II } $$ |
||
|---|---|---|---|
| A. | $$ \text { If } A=\left[\begin{array}{ccc} \cos ^2 37^{\circ} & \cos ^2 53^{\circ} & \cot 135^{\circ} \\ \sin ^2 76^{\circ} & \sin 270^{\circ} & \sin ^2 14^{\circ} \\ \cos 180^{\circ} & \cos ^2 28^{\circ} & \cos ^2 62^{\circ} \end{array}\right] \text {, then } 3-|A|= $$ |
I. | -4 |
| B. | If the period of $\frac{\cos (6 x-4)-\sec (3-4 x)}{\cot (5 x+3)+\sin (3 x+4)}$ is $\frac{2 k \pi}{5}$, then $k=$ | II. | 2 |
| C. | $$ \text { The maximum value of } \cos ^2\left(\frac{\pi}{4}-x\right)+(\sin x-\cos x)^2 \text { is } $$ |
III. | 3 |
| D. | $$ \text { If } x+y+z=0^{\circ}, \text { then } \frac{\sin 2 x+\sin 2 y+\sin 2 z}{\sin (-x) \sin (-y) \sin (-z)} $$ |
IV. | 4 |
| V. | 5 | ||
$$ \text { The correct match is } $$
The period of $\cos (3 x+5)+7$ is
If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$, then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$
For some $a, b, c \in \mathbf{R}$, if $\sin 5 \theta=a \cos ^4 \theta \sin \theta+b \cos ^2 \theta \sin ^3 \theta+c \sin ^5 \theta$, then $a b c=$
The period of $\frac{\sin x}{\cos 3 x}+\frac{\sin 3 x}{\cos 9 x}+\frac{\sin 9 x}{\cos 27 x}+\frac{\sin 27 x}{\cos 81 x}$ is
If $\alpha=\frac{\sin ^3 x}{\cos ^2 x}, \beta=\frac{\cos ^3 x}{\sin ^2 x}$ and $\sin x+\cos x=k$, then $\alpha \sin x+\beta \cos x+3=$
If $A+B+C=60^{\circ}$, then $\cos \left(30^{\circ}-A\right)+\cos \left(30^{\circ}-B\right)+\cos \left(30^{\circ}-C\right)+\sin (A+B+C)=$
Let $a$ be maximum value of $(3 \cos \theta-4 \sin \theta)$ and $\theta \neq \frac{n \pi}{2}$. If $\alpha=a \sin ^2 \theta \cdot \cos ^3 \theta$ and $\beta=a \sin ^3 \theta \cdot \cos ^2 \theta$, then $\sqrt{\frac{\left(\alpha^2+\beta^2\right)^5}{(\alpha \beta)^4}}=$
If $A$ does not belong to the first quadrant, $B$ does not belong to the second quadrant, $\sin A=\frac{11}{61}$ and $\cos B=\frac{-7}{25}$, then $A-B$ and $A+B$ lie respectively in the quadrants
If $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x =\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$, then a possible value of $\sec x$ is
$$ \begin{aligned} \sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8} & +\sin ^4 \frac{5 \pi}{8} +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{7 \pi}{8}= \end{aligned} $$
Assertion (A) If $A=15^{\circ}, B=17^{\circ}$ and $C=13^{\circ}$, then $\cot 2 A+\cot 2 B+\cot 2 C=\cot 2 A \cot 2 B \cot 2 C$
Reason (R) In a $\triangle P Q R$,
$$ \tan \frac{P}{2} \tan \frac{Q}{2}+\tan \frac{Q}{2} \tan \frac{R}{2}+\tan \frac{P}{2} \tan \frac{R}{2}=1 $$
The correct option among the following is