M.C.Q (1 Marks)

MCQ 511 Mark

$\frac{2\tan^230^\circ\sec^252^\circ\sin^238^\circ}{\text{cosec}^270^\circ-\tan^220^\circ}=?$

A
$2$
B
$\frac{1}{2}$
✓
$\frac{2}{3}$
D
$\frac{3}{2}$

Answer

Correct option: C.

$\frac{2}{3}$

$\frac{2\tan^230^\circ\sec^252^\circ\sin^238^\circ}{\text{cosec}^270^\circ-\tan^220^\circ}$
$=\frac{2\times\big(\frac{1}{\sqrt3}\big)^2\times\frac{1}{\cos^252^\circ}\times\sin^2(90^\circ-52^\circ)}{\text{cosec}^2(90^\circ-20^\circ)-\tan^220^\circ}$
$=\frac{2\times\frac{1}{3}\times\frac{1}{\cos^252^\circ}\times\cos^252^\circ}{\sec^220^\circ-\tan^220^\circ}$
$=\frac{\frac{2}{3}}1{}$
$=\frac{2}{3}$

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MCQ 521 Mark

$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ=?$

A
$0$
✓
$\frac{1}{4}$
C
$4$
D
$1$

Answer

Correct option: B.

$\frac{1}{4}$

$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ$
$=\Big(\frac{1}{2}\Big)^2+4(1)^2-(2)^2$
$=\frac{1}{4}+4-4$
$=\frac{1}4{}$

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MCQ 531 Mark

If $\tan\theta=\frac{1}{\sqrt7}$ then $\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=? $

A
$\frac{-2}{3}$
B
$\frac{-3}{4}$
C
$\frac{2}{3}$
✓
$\frac{3}{4}$

Answer

Correct option: D.

$\frac{3}{4}$

$\tan\theta=\frac{1}{\sqrt7}$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\sqrt7$
Now, $\sec^2\theta=(1+\tan^2\theta)$
$1+\Big(\frac{1}{\sqrt7}\Big)^2=1+\frac{1}{7}=\frac{8}{7}$
And, $\text{cosec}^2\theta=(1+\cot^2\theta)$
$=1+\big(\sqrt7\big)^2=1+7=8$
$\therefore\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}=\frac{48}{64}=\frac{3}{4}$

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MCQ 541 Mark

If $\cos\text{A}+\cos^2\text{A}=1$ then $\sin^2\text{A}+\sin^4\text{A}=?$

✓
$1$
B
$2$
C
$4$
D
$3.$

Answer

Correct option: A.

$1$

$\cos\text{A}+\cos^2\text{A}=1$
$\Rightarrow\cos\text{A}=1-\cos^2\text{A}$
$\Rightarrow\cos\text{A}=\sin^2\text{A}$
Now, $\sin^2\text{A}+\sin^4\text{A}=\cos\text{A}+\cos^2\text{A}=1$

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MCQ 551 Mark

If A and B are acute angles such that $\sin\text{A}=\cos\text{B}$ then $(\text{A}+\text{B})=?$

A
$45^\circ$
B
$60^\circ$
✓
$90^\circ$
D
$180^\circ$

Answer

Correct option: C.

$90^\circ$

$\sin\text{A}=\cos\text{B}$
$\Rightarrow\sin\text{A}=\sin(90^\circ-\text{B})$
$\Rightarrow\text{A}=90^\circ-\text{B}$
$\Rightarrow\text{A}+\text{B}=90^\circ$

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MCQ 561 Mark

$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}=?$

A
$\frac32$
B
$\frac23$
C
$2$
✓
$3$

Answer

Correct option: D.

$3$

$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}$
$=\frac{2\sin^263^\circ+2\sin^227+1}{3\cos^217^\circ+3\cos^273^\circ-2}$
$=\frac{2\sin^263^\circ+2\cos^263^\circ+1}{3\cos^217^\circ+3\sin^217^\circ-2}$
$=\frac{2\times1+1}{3\times1-2}$
$=\frac{2+1}{3-2}$
$=3$

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MCQ 571 Mark

If $\text{2x}=\sec\text{A}$ and $\frac{2}{\text{x}}=\tan\text{A}$ then $2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=?$

✓
$\frac{1}{2}$
B
$\frac{1}{4}$
C
$\frac{1}{8}$
D
$\frac{1}{16}$

Answer

Correct option: A.

$\frac{1}{2}$

We know that
$\sec^2\text{A}-\tan^2\text{A}=1$
$\Rightarrow(\text{2x})^2-\Big(\frac{2}{\text{x}}\Big)^2=1$
$\Rightarrow\text{4x}^2-\frac{4}{\text{x}^2}=1$
$\Rightarrow4\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=1$
$\Rightarrow\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}4{}$
$\Rightarrow2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}2{}$

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Maths M.C.Q (1 Marks)