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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 1 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 1
The value of $$\displaystyle\,\frac{2}{tan\,30^{\circ}}$$ is ________
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3.46
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3.56
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34.6
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3.48
Explanation
$$\dfrac{2}{\tan30^{\circ}} = \tfrac{2}{\frac{1}{\sqrt{3}}} = 2 \sqrt{3} = 3.46$$
The value of $$\tan\theta \times \cot\theta =$$ _____
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$$2$$
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$$0$$
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$$1$$
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$$-1$$
Explanation
To find: Value of $$\tan\theta \times \cot\theta$$
We know that $$\cot\theta = \dfrac{1}{\tan\theta}$$
So, $$\tan\theta \times \dfrac{1}{\tan\theta} = 1$$
Which of the following is equal to $$\sin x \sec x$$?
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$$\tan x$$
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$$\cot x$$
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$$\cos x \tan x$$
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$$\cos x \text{cosec}\, x$$
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$$\cot x \text{cosec}\, x$$
Explanation
$$\sin { x }\cdot \sec { x } $$
$$=\sin { x } \times\dfrac { 1 }{ \cos x } $$
$$=\dfrac { \sin { x } }{ \cos x } $$
$$=\tan { x } $$
The value of $$\sin\theta\times \text{cosec}\,\theta =$$ _____
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$$2$$
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$$1$$
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$$-1$$
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$$0$$
Explanation
$$\sin\theta \times cosec\;\theta $$
$$=\sin \theta \times \dfrac{1}{\sin \theta}$$
$$=1$$
Hence, the answer is $$1.$$
Who published the trigonometry in 1595?
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William Rowan Hamilton
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Hipparchus
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Bartholomaeus Pitiscus
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Newton
Explanation
Trigonometry was first published by Bartholomaeus Pitiscus in 1595.
So, option C is correct.
Choose the correct alternative answer for the following question.
$$cosec \ 45^\circ= ?$$
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$$\dfrac12$$
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$$\sqrt{2}$$
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$$\dfrac{\sqrt{3}}{2}$$
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$$\dfrac{2}{\sqrt{3}}$$
Explanation
$$\sin 45^{\circ}=\dfrac {1}{\sqrt2}$$
$$cosec \ 45^\circ = \sqrt{2}$$
Hence, the correct answer is $$\sqrt{2}$$.
If $$ \sqrt{3} \cos A = \sin A $$, then the value of $$ \cot A $$ is:
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$$ \sqrt{3} $$
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$$ 1 $$
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$$\dfrac{1}{\sqrt{3}} $$
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$$ 2$$
$$\tan {45^ \circ } =?$$
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$$1$$
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$$\sqrt3$$
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$$0$$
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can't be determined
Explanation
$$\tan 45^{\circ}=\dfrac{\sin 45^{\circ}}{\cos 45^{\circ}}$$
$$=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}}$$
$$=1$$
So the relation is $$\text{True}$$
The value of $$\cos 1^o.\cos 2^o.\cos 3^o.....\cos 179^o$$ is equal to
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$$-1$$
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$$0$$
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$$1$$
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$$\dfrac{1}{\sqrt{2}}$$
Explanation
$$A = \cos1^0\cdot\cos2^0\cdot\cos3^0\cdots\cos{90}^0\cdots\cos{179}^0$$
$$\because \cos{90}^0 = 0$$
$$\therefore A = 0$$
Hence, option B.
In $$\triangle ABC$$, $$\angle B= {90}^{\circ}$$, $$\sin{C} = \dfrac{3}{5}$$, then $$\cos{A} =$$ ______
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$$\dfrac{3}{5}$$
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$$\dfrac{4}{5}$$
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$$\dfrac{5}{4}$$
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$$\dfrac{5}{3}$$
Explanation
In a right angled triangle, $$A+B+C=180^\circ$$
$$B=90^\circ$$
So, $$A+C=90^\circ$$
$$A=90^\circ-C$$
$$\cos A=\cos(90^\circ-A)$$
$$\Rightarrow \cos A=\sin C=\dfrac35$$
$$\Rightarrow \cos A=\cfrac35$$
If sin $$A=\dfrac{1}{2}$$ and cos $$B=\dfrac{1}{2}$$, then the value of $$(A+B)$$ is equal to :
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$$0^{\circ}$$
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$$60^{\circ}$$
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$$90^{\circ}$$
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$$30^{\circ}$$
Explanation
$$\sin { A } =\dfrac { 1 }{ 2 } ,\cos { B } =\dfrac { 1 }{ 2 } $$
$$\Rightarrow \sin { A } =\sin { 30° } ,\cos { B } =\cos { 60° } $$
$$\Rightarrow A=30°,B=60°$$
$$\therefore A+B=30°+60°=90°$$
Hence, the answer is $$90°.$$
If cot $$\theta =\dfrac{7}{8}$$, then the value of $$tan^{2}\theta$$ equals to :
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$$\dfrac{8}{7}$$
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$$\dfrac{49}{64}$$
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$$\dfrac{64}{49}$$
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$$\dfrac{7}{8}$$
Explanation
$$\cot { \theta } =\dfrac { 7 }{ 8 } $$
$$\Rightarrow \tan { \theta } =\dfrac { 1 }{ \cot { \theta } } \\\ \ \ \ \ \ \ \ \ \ \ \ =\dfrac { 8 }{ 7 } $$
$$\Rightarrow \tan ^{ 2 }{ \theta } =\dfrac { 64 }{ 49 } $$
Hence, the correct option is (C)
The value of (sin $$45^{\circ}+cos 45^{\circ}$$) is :
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1
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$$\dfrac{1}{\sqrt{2}}$$
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$$\dfrac{\sqrt{3}}{2}$$
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$$\sqrt{2}$$
Explanation
$$\sin { 45° } +\cos { 45° } $$
$$=\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 2 } } $$
$$=\dfrac { 2 }{ \sqrt { 2 } } $$
$$=\sqrt { 2 } $$
Hence, the answer is $$=\sqrt { 2 } .$$
If $$2 \cos\theta+ \sin\theta=1$$, then the value of $$4 \cos\theta + 3 \sin\theta$$ is equal to
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$$3$$
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$$-5$$
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$$\dfrac {7}{5}$$
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$$-4$$
$$\cos ^{2} 60^{\circ}+\sin ^{2} 30^{\circ} = \displaystyle \frac{1}{a}$$
Then $$a=$$
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$$2$$
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$$3$$
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$$1$$
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$$\dfrac{1}{2}$$
Explanation
$$\cos^2 60^{\circ} + \sin^2 30^{\circ}$$
$$=\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2$$
$$=\cfrac{1}{4} + \cfrac{1}{4}$$
$$=\cfrac{2}{4}$$
$$=\cfrac{1}{2}$$
$$\therefore \dfrac1a = \dfrac12$$
$$\Rightarrow a = 2$$
If $$\sec 4A = cosec (A-20^{\small\circ})$$, where $$4A$$ is an acute angle, find the value of $$A$$.
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$$A = 32^{\small\circ}$$
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$$A = 22^{\small\circ}$$
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$$A = 41^{\small\circ}$$
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$$A = 16^{\small\circ}$$
Explanation
Hence
$$\sin(A-20^{0})=\cos(4A)$$
$$\sin(A-20^{0})=\sin(90^{0}-4A)$$
Hence
$$A-20^{0}=90^{0}-4A$$
$$5A=110^{0}$$
$$A=22^{0}$$
In the given figure, $$BC=15cm$$ and $$\sin{B}=\dfrac{4}{5}$$, What is the value of $$AB$$?
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$$25cm$$
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$$20cm$$
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$$5cm$$
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$$4cm$$
Explanation
$$\sin B= \dfrac{AC}{AB} =\dfrac{\sqrt{AB^2 - BC^2}}{AB}$$
$$ \dfrac{4}{5} AB= \sqrt{AB^2 - 15^2}$$
$$\Rightarrow 16AB^2 = 25AB^2 - (25)(15)^2$$
$$AB^2 = \dfrac{25\times 15 \times 15}{9}$$
$$\Rightarrow \boxed {AB = 25}$$
Choose the correct option. Justify your choice.
$$\displaystyle 9{ \sec }^{ 2 }A-9{ \tan }^{ 2 }A=$$
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$$1$$
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$$9$$
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$$8$$
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$$0$$
Explanation
We know,
$$1= \sec ^2A-\tan^2A$$
$$\therefore 9(\sec^2A-\tan^2A)$$
$$=9\times 1=9$$
If $$sin({ 90 }^{ 0 }-\theta )=\dfrac { 3 }{ 7 } $$, then $$cos\theta $$
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$$\cfrac { 7 }{ 3 } $$
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$$\cfrac { 3 }{ 7 } $$
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$$\cfrac { -7 }{ 3 } $$
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$$\cfrac { -3 }{ 7 } $$
Explanation
$$\sin (90^o - \theta) = \dfrac{3}{7}$$
We know,
$$\sin (90^o - \theta) = \cos \theta$$
Hence, $$\cos \theta = \dfrac{3}{7}$$
If $$\tan \theta= \cot \theta$$, then the value of $$\sec \theta$$ is :
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$$2$$
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$$1$$
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$$\dfrac{2}{\sqrt{3}}$$
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$$\sqrt{2}$$
Explanation
$$\tan { \theta } =\cot { \theta } $$
$$\Rightarrow \tan ^{ 2 }{ \theta } =1$$
$$\Rightarrow \tan { \theta } =1$$
$$\Rightarrow \theta =45°$$
$$\therefore \sec { 45° } =\sqrt { 2 } $$
Hence, the answer is $$\sqrt { 2 }.$$
If $$\sqrt{3}\,=\,1.732$$, find (correct to two decimal places) the value of each of the following :
$$sin\,60^{\circ}$$ is 0.87
State true or false.
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True
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False
Explanation
$$\sin 60^{\circ} = \dfrac{\sqrt{3}}{2} = \dfrac{1.732}{2} = 0.87$$
The value of cosec $$30^{\circ}$$ + cot $$45^{\circ}$$ is:
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2
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1
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3
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-1
Explanation
cosec $$30^{\circ}$$ + cot $$45^{\circ}$$
$$=2+1$$
$$=3$$
Hence, the answer is $$3.$$
The value of $$5cosec^{2}\theta -5cot^{2 }\theta$$ is equal to:
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5
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1
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0
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-5
Explanation
Using the identity:
$$cosec^{2}\theta -\cot^{2 }\theta=1$$
We get,
$$5cosec^{2}\theta -5\cot^{2 }\theta$$ = 5($$cosec^{2}\theta -\cot^{2 }\theta)=5$$
If $$\theta =45^{\circ},$$ then the value of $$cosec^{2}\theta$$ is
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$$\dfrac{1}{\sqrt{2}}$$
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$$1$$
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$$\dfrac{1}{2}$$
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$$2$$
Explanation
$$\theta =45°$$
$$\Rightarrow cosec^{ 2 }45°={ \left( \sqrt { 2 } \right) }^{ 2 }=2$$
Hence, the answer is $$2.$$
For an acute angle $$\theta$$ in a right angled triangle, $$\dfrac{1}{\sin{\theta}} =$$
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$$\cos{\theta}$$
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$$\cot{\theta}$$
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$$\sec{\theta}$$
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$$\text{cosec}\,{\theta}$$
Explanation
$$\sin \theta=\dfrac{opposite}{hypotenuse}$$
$$\cos \theta = \dfrac{adjacent}{hypotenuse}$$
$$\dfrac{1}{\sin \theta}=\dfrac{1}{\dfrac{opposite}{hypotenuse}}=\dfrac{hypotenuse}{opposite}=\text{cosec}\ \theta$$
$$\therefore\ \dfrac{1}{\sin \theta}=\text{cosec}\ \theta$$
What is the value of $$\displaystyle \sin { { 30 }^{ o } } \times co\sec{ 30 }^{ o }$$?
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$$1$$
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$$0$$
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$$\displaystyle \frac { 1 }{ 2 } $$
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$$2$$
Explanation
$$\displaystyle \sin { { 30 }^{ o }\times co\sec{ 30 }^{ o } } =\cfrac { 1 }{ 2 } \times \cfrac { 2 }{ 1 } =\cfrac { 1 }{ 1 } =1$$
The value of $$\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }$$ is :
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$$0$$
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$$1$$
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$$2$$
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$$\displaystyle \frac { 1 }{ 2 } $$
Explanation
$$\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }=1\times 1=1$$
$$\dfrac{1}{\cos{\theta}} =$$_____ , for an acute angle $$\theta$$ in a right angled triangle.
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$$\text{cosec}\,{\theta}$$
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$$\sec{\theta}$$
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$$\cot{\theta}$$
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None of these
Explanation
$$\sin \theta=\dfrac{opposite}{hypotenuse}$$
$$\cos \theta = \dfrac{adjacent}{hypotenuse}$$
$$\dfrac{1}{\cos \theta}=\dfrac{1}{\dfrac{adjacent}{hypotenuse}}=\dfrac{hypotenuse}{adjacent}=\sec \theta$$
$$\therefore\ \dfrac{1}{\cos \theta}=\sec \theta$$
The value of $$\tan{\theta}\cdot\tan{(90-\theta)}$$ is equal to
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$$\sin^2{\theta}$$
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$$1$$
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$$\cos^2{\theta}$$
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$$0$$
Explanation
$$ \tan\theta \cdot \tan(90-\theta) = \tan\theta \times \cot\theta = 1 $$
$$\because \tan(90-\theta) =\cot \theta $$
If $$\sec{2A}=\csc{(A-42^\circ)}$$ where $$2A$$ is acute angle then value of $$A$$ is
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$$44^\circ$$
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$$22^\circ$$
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$$21^\circ$$
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$$66^\circ$$
Explanation
$$\sec{2A}=\csc{(A-42^\circ)}$$
$$\csc{(90-2A)}=\csc{(A-42^\circ)}$$
$$90-2A=A-42$$
$$3A=132$$
$$A=44^\circ$$
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