MCQ 511 Mark
$\tan ^2 \theta \sin ^2 \theta$ is equal to
- ✓
$\tan ^2 \theta-\sin ^2 \theta$
- B
$\tan ^2 \theta+\sin ^2 \theta$
- C
$\frac{\tan ^2 \theta}{\sin ^2 \theta}$
- D
$\sin ^2 \theta \cot ^2 \theta$
AnswerCorrect option: A. $\tan ^2 \theta-\sin ^2 \theta$
$=\sin ^2 \theta\left(\frac{1-\cos ^2 \theta}{\cos ^2 \theta}\right)$
$=\frac{\sin ^2 \theta}{\cos ^2 \theta}-\sin ^2 \theta$
$=\tan ^2 \theta-\sin ^2 \theta$
View full question & answer→MCQ 521 Mark
If $\cos \theta=\frac{\sqrt{3}}{2}$ and $\theta$ is acute, then $(3 \tan \theta$ $\left.-\tan ^2 \theta\right)$ is egual to
- ✓
$\frac{8}{3 \sqrt{3}}$
- B
$\frac{1}{2 \sqrt{3}}$
- C
$\frac{\sqrt{3}}{6}$
- D
AnswerCorrect option: A. $\frac{8}{3 \sqrt{3}}$
(a) ; We have $\cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=30^{\circ}\left(; \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)$Now, $3 \tan \theta-\tan ^3 \theta$
$
=3 \tan 30^{\circ}-\tan ^3 30^{\circ}
$
$
=3 \times \frac{1}{\sqrt{3}}-\left(\frac{1}{\sqrt{3}}\right)^3=\frac{3}{\sqrt{3}}-\frac{1}{3 \sqrt{3}}=\frac{9-1}{3 \sqrt{3}}=\frac{8}{3 \sqrt{3}}
$
View full question & answer→MCQ 531 Mark
If $\theta$ is scute and $\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3$, then $\theta$ is equal to is equal to
- ✓
$60^{\circ}$
- B
$30^{\circ}$
- C
$90^{\circ}$
- D
AnswerCorrect option: A. $60^{\circ}$
We have $\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3$
$\Rightarrow \cos ^2 \theta=3\left(\cot ^2 \theta-\cos ^2 \theta\right)$
$\Rightarrow \cos ^2 \theta=3 \cot ^2 \theta-3 \cos ^2 \theta$
$\Rightarrow 4 \cos ^2 \theta=3 \cot ^2 \theta-4 \cos ^2 \theta$
$=\frac{3 \cos ^2 \theta}{\sin ^2 \theta}$
$\Rightarrow \sin ^2 \theta=\frac{3}{4}$
$\Rightarrow \sin \theta=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta=60^{\circ}$
View full question & answer→MCQ 541 Mark
If $4 \cos \theta=11 \sin \theta$, then the value of $\frac{11 \cos \theta-7 \sin \theta}{11 \cos \theta+7 \sin \theta}$ is
- ✓
$93 / 149$
- B
$94 / 149$
- C
$91 / 149$
- D
$97 / 149$
AnswerCorrect option: A. $93 / 149$
(a) : Given $4 \cos \theta=11 \sin \theta$
$\Rightarrow \cos \theta=\frac{11}{4} \sin \theta$
$\begin{aligned} & \text { Now } \frac{11 \cos \theta-7 \sin \theta}{11 \cos \theta+7 \sin \theta}=\frac{11 \times \frac{11}{4} \sin \theta-7 \sin \theta}{11 \times \frac{11}{4} \sin \theta+7 \sin \theta} \\ & =\frac{\sin \theta\left(\frac{121}{4}-7\right)}{\sin \theta\left(\frac{121}{4}+7\right)}=\frac{121-28}{121+28}=\frac{93}{149}\end{aligned}$
View full question & answer→MCQ 551 Mark
$2\left(\cos ^4 60^{\circ}+\sin ^4 30^{\circ}\right)-\left(\tan ^2 60^{\circ}+\cot ^2 45^{\circ}\right)$ $+3 \sec ^2 30^{\circ}$ is equal to
- A
$1 / 2$
- B
- C
$3 \times 4$
- D
$1 / 4$
Answer$\begin{aligned} & \text {(d): } 2\left(\cos ^4 60^{\circ}+\sin ^4 30^{\circ}\right)-\left(\tan ^2 60^{\circ}+\cot ^2 45^{\circ}\right) \\ & +3 \sec ^2 30^{\circ} \\ & =2\left[\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^4\right]-\left[(\sqrt{3})^2+(1)^2\right]+3\left(\frac{2}{\sqrt{3}}\right)^2 \\ & =2\left[\frac{1}{16}+\frac{1}{16}\right]-[3+1]+3 \times \frac{4}{3}=2\left[\frac{2}{16}\right]-4+4=\frac{4}{16}=\frac{1}{4} \\ & \end{aligned}$
View full question & answer→MCQ 561 Mark
The values of .......... are equal.
- ✓
$\sin 60^{\circ}$ and $\cos 30^{\circ}$
- B
$\sin 60^{\circ}$ and $\cos 45^{\circ}$
- C
$\sin 45^{\circ}$ and $\tan 45^{\circ}$
- D
AnswerCorrect option: A. $\sin 60^{\circ}$ and $\cos 30^{\circ}$
$\sin 60^{\circ}$ and $\cos 30^{\circ}$
View full question & answer→MCQ 571 Mark
If $x=a \operatorname{cosec} \theta$ and $y=b \cot \theta$ then $b^2 x^2-a^2 y^2=$.
- A
$a b$
- B
$a^2-b^2$
- ✓
$a^2 b^2$
- D
$a^2+b^2$
AnswerCorrect option: C. $a^2 b^2$
$a^2 b^2$
View full question & answer→MCQ 581 Mark
If $\tan 7 \theta \cdot \tan 3 \theta=1$ then the value of $\theta$ is
- A
$10^{\circ}$
- ✓
$9^{\circ}$
- C
$18^{\circ}$
- D
$0^{\circ}$
AnswerCorrect option: B. $9^{\circ}$
$9^{\circ}$
View full question & answer→MCQ 591 Mark
$\cot ^2 \theta-\operatorname{cosec}^2 \theta=\ldots \ldots \ldots . .\left(0<\theta<90^{\circ}\right)$
View full question & answer→MCQ 601 Mark
$2 \sin ^2 30^{\circ} \cdot \cot 30^{\circ}-3 \cos ^2 60^{\circ} \sec ^2 30^{\circ}=$
- A
$\frac{\sqrt{3}-1}{2}$
- ✓
$\frac{\sqrt{3}-2}{2}$
- C
$0$
- D
AnswerCorrect option: B. $\frac{\sqrt{3}-2}{2}$
$\frac{\sqrt{3}-2}{2}$
View full question & answer→MCQ 611 Mark
The simplest form of $\frac{\cos \left(90^{\circ}- A \right) \sin \left(90^{\circ}- A \right)}{\tan \left(90^{\circ}- A \right)}$ is
- ✓
$\sin ^2 A$
- B
$\cos ^2 A$
- C
$\sin A$
- D
AnswerCorrect option: A. $\sin ^2 A$
$\sin ^2 A$
View full question & answer→MCQ 621 Mark
$\tan \alpha=1$, then $\alpha=$
- A
$0^{\circ}$
- ✓
$45^{\circ}$
- C
$30^{\circ}$
- D
$60^{\circ}$
AnswerCorrect option: B. $45^{\circ}$
$45^{\circ}$
View full question & answer→MCQ 631 Mark
$\cos A =4 \sin A$ then, $\tan A =$
- ✓
$\frac{1}{4}$
- B
$\frac{5}{4}$
- C
- D
$\frac{4}{5}$
AnswerCorrect option: A. $\frac{1}{4}$
$\frac{1}{4}$
View full question & answer→MCQ 641 Mark
$\frac{\operatorname{cosec}^4 \theta-\cot ^4 \theta}{\operatorname{cosec}^2 \theta+\cot ^2 \theta}=$
View full question & answer→MCQ 651 Mark
If $A$ and $B$ are complementary angles then $\sin A \cdot \sec B =$
View full question & answer→MCQ 661 Mark
$\sin ^2 35+\cos ^2 \theta=1$ then $\theta=$
View full question & answer→MCQ 671 Mark
If $\theta$ is an acute angle and $b \sin \theta=a \cos \theta$ then $\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=$
AnswerCorrect option: B. $\frac{a^2-b^2}{a^2+b^2}$
$\frac{a^2-b^2}{a^2+b^2}$
View full question & answer→MCQ 681 Mark
For $0<\theta<90^{\circ}$, when value of $\theta$ increases from $0^{\circ}$ to $90^{\circ}$ the value of ......... increases.
AnswerCorrect option: D. $\sin \theta$
$\sin \theta$
View full question & answer→MCQ 691 Mark
If $\cos \theta+\cos ^2 \theta=1$ then $\sin ^2 \theta+\sin ^4 \theta=\ldots \ldots \ldots$
View full question & answer→MCQ 701 Mark
$\cos ^4 \theta-\sin ^4 \theta=\ldots \ldots \ldots .$.
- A
$2 \cos ^2 \theta+1$
- B
$2 \sin ^2 \theta+1$
- ✓
$2 \cos ^2 \theta-1$
- D
$2 \sin ^2 \theta-1$
AnswerCorrect option: C. $2 \cos ^2 \theta-1$
$2 \cos ^2 \theta-1$
View full question & answer→MCQ 711 Mark
$\frac{1+\tan ^2 A }{1+\cot ^2 A }=$
- A
$\sec ^2 A$
- B
$\cot ^2 A$
- ✓
$\tan ^2 A$
- D
AnswerCorrect option: C. $\tan ^2 A$
$\tan ^2 A$
View full question & answer→MCQ 721 Mark
If $\tan \theta+\cot \theta=2$ then $(\tan \theta)^{2019}+(\cot \theta)^{2020}=\ldots \ldots \ldots$
View full question & answer→MCQ 731 Mark
$2 \cos 30^{\circ} \times \sin 30^{\circ}=\sin \theta$ then find the value of $\theta$.
- ✓
$60^{\circ}$
- B
$45^{\circ}$
- C
$30^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: A. $60^{\circ}$
$60^{\circ}$
View full question & answer→MCQ 741 Mark
$\frac{3 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}$
- A
$\frac{\sqrt{3}}{2}$
- B
$\frac{2 \sqrt{3}}{3}$
- ✓
$\frac{3 \sqrt{3}}{2}$
- D
AnswerCorrect option: C. $\frac{3 \sqrt{3}}{2}$
$\frac{3 \sqrt{3}}{2}$
View full question & answer→MCQ 751 Mark
$\frac{\sin ^2 60^{\circ}}{1-\cos ^2 60^{\circ}}+\frac{\operatorname{cosec}^2 60^{\circ}}{1+\cot ^2 60^{\circ}}$
View full question & answer→MCQ 761 Mark
$\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}=$
- A
$\cos 60^{\circ}$
- B
$\sin 60^{\circ}$
- ✓
$\tan 60^{\circ}$
- D
$\sin 30^{\circ}$
AnswerCorrect option: C. $\tan 60^{\circ}$
$\tan 60^{\circ}$
View full question & answer→MCQ 771 Mark
When $A =\ldots \ldots \ldots, \sin 2 A =2 \sin A$ is true.
- ✓
$0^{\circ}$
- B
$30^{\circ}$
- C
$45^{\circ}$
- D
$60^{\circ}$
AnswerCorrect option: A. $0^{\circ}$
$0^{\circ}$
View full question & answer→MCQ 781 Mark
$\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\ldots \ldots \ldots .$.
- A
$\tan 90^{\circ}$
- B
- C
$\sin 45^{\circ}$
- ✓
$0$
View full question & answer→MCQ 791 Mark
$\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}=$
- ✓
$\sin 60^{\circ}$
- B
$\cos 60^{\circ}$
- C
$\tan 60^{\circ}$
- D
$\sin 30^{\circ}$
AnswerCorrect option: A. $\sin 60^{\circ}$
$\sin 60^{\circ}$
View full question & answer→MCQ 801 Mark
If $\sin ^2 \theta=\frac{1}{2}$, then value of $\tan ^2 \theta=$___________.
- A
$\frac{1}{\sqrt{3}}$
- B
$\sqrt{3}$
- C
- ✓
View full question & answer→