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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 5 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 5
In $$\triangle ABC$$
If $$\sin B$$ is $$\displaystyle \frac{12}{m}$$, then $$m$$ is:
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13
0%
12
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11
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10
Explanation
In $$\triangle ABC$$, $$AD \perp BC$$ $$AB = 13$$, $$BD = 5$$, $$DC =16$$
Now, In $$\triangle ABD$$,
$$AB^2 = AD^2 + BD^2$$
$$13^2 = AD^2 + 5^2$$
$$AD^2 = 144$$
$$AD = 12$$
Now, in $$\triangle ADC$$,
$$AC^2 = AD^2 + CD^2$$
$$AC^2 = 12^2 + 16^2$$
$$AC^2 = 144 + 256$$
$$AC = 20$$ cm
Now, $$\sin B = \frac{P}{H} = \frac{AD}{AB} = \frac{12}{13}$$
Find the value of : $$\cot^{2} C\, -\, \displaystyle \frac{1}{\sin^{2} C}$$.
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- 1
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2
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4
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0
Explanation
Given, $$\triangle ABC$$, A perpendicular from B on AC, let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$
In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$
In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$
Now, $$cot^2 C - \frac{1}{\sin^2 C} = cot^2 C - csc^2 C = (\frac{B}{P})^2 - (\frac{H}{P})^2$$
= $$(\frac{8\sqrt{2}}{4})^2 - (\frac{12}{4})^2$$
= $$ 8 - 9$$
= $$-1$$
If $$\sin C$$ is $$\displaystyle \frac{1}{m}$$, then $$m$$ is:
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$$1$$
0%
$$5$$
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$$2$$
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$$3$$
Explanation
Given, $$\triangle ABC$$, A perpendicular from B on AC.
Let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$
In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$
In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$
Now, $$\sin C = \dfrac{P}{H} = \dfrac{BD}{DC} = \dfrac{4}{12} = \dfrac{1}{3}$$
So, option D is correct.
If $$\tan C$$ is $$\displaystyle \frac{3}{m}$$, then $$m$$ is:
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1
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4
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3
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2
Explanation
In $$\triangle ABC$$, $$AD \perp BC$$ $$AB = 13$$, $$BD = 5$$, $$DC =16$$
Now, In $$\triangle ABD$$,
$$AB^2 = AD^2 + BD^2$$
$$13^2 = AD^2 + 5^2$$
$$AD^2 = 144$$
$$AD = 12$$
Now, in $$\triangle ADC$$,
$$AC^2 = AD^2 + CD^2$$
$$AC^2 = 12^2 + 16^2$$
$$AC^2 = 144 + 256$$
$$AC = 20$$ cm
Now, $$\tan C = \frac{P}{B} = \frac{AD}{CD} = \frac{12}{16} = \frac{3}{4}$$
If $$\cos A$$ is $$\displaystyle \frac{3}{m}$$, then m is:
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$$5$$
0%
$$1$$
0%
$$2$$
0%
$$7$$
Explanation
Given, $$\triangle ABC$$, A perpendicular from B on AC, let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$
In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$
In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$
Now, $$\cos A = \dfrac{B}{H} = \dfrac{AD}{AB} = \dfrac{3}{5}$$
$$\Longrightarrow$$ $$m=5$$
If $$ 5 \cot\, \theta\, =\, 12$$, find the value of : $$\csc\, \theta\, +\, \sec \, \theta$$ is $$\displaystyle {3}\cfrac{41}{m}$$, m is
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$$40$$
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$$60$$
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$$50$$
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$$13$$
Explanation
$$5 \cot \theta = 12$$
$$\cot \theta = \cfrac{12}{5}$$
$$\cot \theta = \cfrac{B}{P} = \cfrac{12}{5}$$
Using Pythagoras Theorem,
$$H^2 = P^2 + B^2$$
$$H^2 = 12^2 + 5^2$$
$$H = 13$$
Now, $$\csc \theta + \sec \theta$$
= $$\cfrac{H}{P} + \cfrac{H}{B}$$
= $$\cfrac{13}{12} + \cfrac{13}{5}$$
= $$\cfrac{65 + 156}{60}$$
= $$\cfrac{221}{60}$$
= $$3 \cfrac{41}{60}$$
$$sin\, \angle DBA$$ is $$\displaystyle \frac{4}{m}$$
value of m is
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0
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7
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9
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5
Explanation
In $$\triangle ABC$$
$$AB^2 + BC^2 = AC^2$$ (By Pythagoras theorem)
$$4^2+ 3^2 = AC^2$$
$$AC^2 = 25$$
$$AC =5$$
In $$\triangle ABC$$
$$\angle A + \angle B + \angle C = 180$$
$$\angle A + \angle C = 90$$ (As $$\angle B = 90$$).....(i)
In $$\triangle ABD$$,
$$\angle A + \angle BDA + \angle DBA = 180$$
$$\angle A + \angle DBA = 90$$ (As $$\angle BDA = 90$$).....(i)
From (i) and (ii)
$$\angle C = \angle DBA$$
Similarly, $$\angle DBC = \angle A$$
Now, $$\sin \angle DBA = \sin \angle C = \frac{P}{H} = \frac{AB}{AC}= \frac{4}{5}$$
$$tan\, \angle DBC$$ is $$\displaystyle \frac{m}{4}$$
value of m is
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1
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3
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4
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2
Explanation
In $$\triangle ABC$$
$$AB^2 + BC^2 = AC^2$$ (By Pythagoras theorem)
$$4^2+ 3^2 = AC^2$$
$$AC^2 = 25$$
$$AC =5$$
In $$\triangle ABC$$
$$\angle A + \angle B + \angle C = 180$$
$$\angle A + \angle C = 90$$ (As $$\angle B = 90$$).....(i)
In $$\triangle ABD$$,
$$\angle A + \angle BDA + \angle DBA = 180$$
$$\angle A + \angle DBA = 90$$ (As $$\angle BDA = 90$$).....(i)
From (i) and (ii)
$$\angle C = \angle DBA$$
Similarly, $$\angle DBC = \angle A$$
Now, $$\tan \angle DBC = \tan \angle A = \dfrac{P}{B} = \dfrac{BC}{AB}= \dfrac{3}{4}$$
If $$ \tan A + \cot A = 5$$; find the value if $$\tan^{2}\, A\, +\, \cot ^{2} A$$.
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$$14$$
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$$27$$
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$$23$$
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$$335$$
Explanation
$$\tan A + \cot A = 5$$
Squaring both sides,
$$(\tan A + \cot A)^2 = 25$$
$$\tan^2 A + \cot^2 A + 2 \tan A \cot A = 25$$
$$\tan^2 A + \cot^2 A + 2 = 25$$
$$\tan^2 A + \cot^2 A = 23$$
If $$x = a \cos^{3} \theta \sin^{2} \theta, y = a \sin^{3} \theta \cos^{2} \theta$$ and $$\dfrac {(x^{2} + y^{2})^{p}}{(xy)^{q}}(p, q\epsilon N)$$ is independent of $$\theta$$, then
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$$p = 4$$
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$$p = 5$$
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$$q = 4$$
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$$q = -5$$
Explanation
$$x^{2} + y^{2} = \left (a \cos^{3} \theta \sin^{2}\theta \right )^{2} + \left (a\sin^{3} \theta \cos^{2} \theta \right )^{2} = a^{2} \sin^{4} \theta \cos^{4} \theta$$
$$xy = \left (a\cos^{3}\theta \sin^{2}\theta \right ) \left (a \sin^{3} \theta \cos^{2} \theta \right ) = a^{2} \sin^{5} \theta \cos^{5} \theta$$
Therefore,
$$\dfrac {(x^{2} + y^{2})^{p}}{(xy)^{q}} = \dfrac {a^{2p} (\sin \theta \cos \theta )^{4p}}{a^{2q} (\sin \theta \cos \theta )^{5q}}$$
Is Independent of $$\theta$$ when $$4p = 5q$$
i.e. $$p = 5, q = 4$$
Hence, option $$'B'$$ and $$'C'$$ are correct.
Evaluate : $$\displaystyle \frac{3\cos 53^{\circ} \text{cosec}37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$$
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1
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6
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0
Explanation
$$\displaystyle \frac{3\cos 53^{\circ} cosec 37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$$
= $$\displaystyle \frac{3\cos 53^{\circ} \text{cosec }(90 - 53)^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}(90 - 29)^{\circ})}-3\tan ^{2}45^{\circ}$$
= $$\displaystyle \frac{3\cos 53^{\circ} \sec 53^{\circ}}{(\cos ^{2}29^{\circ}+\sin ^{2}29^{\circ})}-3\tan ^{2}45^{\circ}$$
= $$\displaystyle \frac{3}{1}-3(1)^2$$
= $$0$$
If $$\tan 1^o \tan 2^o ... \tan 89^o=x^2-8$$, then the value of $$ x$$ can be
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$$-1$$
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$$1$$
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$$-3$$
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$$3$$
Explanation
Given that,
$$\tan 1^o \tan 2^o ... \tan 89^o=x^2-8$$
To find out,
The value of $$x$$
We can rewrite the given equation as:
$$x^2-8=(\tan 1^o \tan 89^o)(\tan 2^o \tan 88^o) ....(\tan 44^o \tan 46^o) \tan 45^o$$
We can write $$\tan 89^o$$ as $$\tan (90^o-1^o), \ \tan 88^o$$ as $$\tan (90^o-2^o)$$ and so on up to $$\tan 46^o$$ as $$\tan (90^o-44^o)$$
Hence, the equation becomes:
$$x^2-8=(\tan 1^o \tan (90^o-1^o))(\tan 2^o \tan (90^o-2^o)) ....(\tan 44^o \tan (90^o-44^o)) \tan 45^o$$
We know that, $$\tan(90^o-\theta)=\cot \theta$$
So, $$x^2-8=(\tan 1^o \cot 1^o)(\tan 2^o \cot 2^o)....(\tan 44^o \cot 44^o) \tan 45^o$$
Also, $$\cot \theta=\dfrac{1}{\tan\theta}$$
So, $$x^2-8=\left(\tan 1^o \times \dfrac{1}{\tan 1^o}\right)\left(\tan 2^o \times \dfrac{1}{\tan 2^o}\right)....\left(\tan 44^o \times \dfrac{1}{\tan 44^o}\right) \tan 45^o$$
$$x^2-8=(1)(1)....(1) \times \tan 45^o$$
$$\tan 45^o=1$$
So, $$x^2-8=(1)(1)....(1) \times 1$$
$$\Rightarrow x^2-8=1$$
$$\Rightarrow x^2=9$$
$$\Rightarrow x=\pm 3$$
Hence, if $$\tan 1^o \tan 2^o ... \tan 89^o=x^2-8$$, the value of $$x$$ is $$3$$ or $$-3$$.
$$cot\, \angle DBA$$
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$$\displaystyle \frac{3}{11}$$
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$$\displaystyle \frac{11}{17}$$
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$$\displaystyle \frac{7}{9}$$
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$$\displaystyle \frac{5}{12}$$
Explanation
In $$\triangle ABC$$
$$AB^2 + BC^2 = AC^2$$ (By Pythagoras theorem)
$$12^2+ 5^2 = AC^2$$
$$AC^2 = 169$$
$$AC =13$$
In $$\triangle ABC$$
$$\angle A + \angle B + \angle C = 180$$
$$\angle A + \angle C = 90$$ (As $$\angle B = 90$$).....(i)
In $$\triangle ABD$$,
$$\angle A + \angle BDA + \angle DBA = 180$$
$$\angle A + \angle DBA = 90$$ (As $$\angle BDA = 90$$).....(i)
From (i) and (ii)
$$\angle C = \angle DBA$$
Similarly, $$\angle DBC = \angle A$$
Now, $$\cot \angle DBA = \cot \angle C = \frac{B}{P} = \frac{BC}{AB}= \frac{5}{12}$$
Given: $$\cos A\, =\, \displaystyle \cfrac{5}{13}$$.
If
$$\displaystyle \frac{\sin A\, -\, \cot A}{2\, \tan\, A}$$ is $$\displaystyle \frac{m}{3744}$$, $$m$$ is:
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695
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595
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295
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395
Explanation
Given, $$\cos A = \frac{5}{13}$$
$$\cos A = \frac{B}{H} = \frac{5}{13}$$
Now, using Pythagoras Theorem,
$$H^2 = P^2 + B^2$$
$$13^2 = P^2 + 5^2$$
$$P = 12$$
Now, $$\displaystyle \frac{\sin A\, -\, \cot A}{2\tan A}$$
= $$\displaystyle \frac{\frac{12}{13}\, -\, \frac{5}{12}}{2 (\frac{12}{5})}$$
= $$\displaystyle \frac{\frac{144 - 65}{156}}{\frac{24}{5}}$$
= $$\displaystyle \frac{\frac{79}{156}}{\frac{24}{5}}$$
= $$\displaystyle \frac{395}{3744}$$
Given : $$sec\, A\, =\, \displaystyle \frac{29}{21}$$, evaluate : $$sin\, A\, -\, \displaystyle \frac{1}{tan\, A}$$ is $$\displaystyle -\frac{m}{580}$$, m is
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0%
130
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215
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209
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524
Explanation
$$\sec A = \frac{29}{21}$$
$$\sec A = \frac{H}{B} = \frac{29}{21}$$
Using Pythagoras Theorem,
$$H^2 = P^2 + B^2$$
$$29^2 = P^2 + 21^2$$
$$841 = 441 + P^2$$$
$$P = 20$$
Now, $$\sin A - \frac{1}{\tan A} = \sin A - \cot A$$
= $$\frac{P}{H} - \frac{B}{P}$$
= $$\frac{20}{29} - \frac{21}{20}$$
= $$\frac{400 - 609}{580}$$
= $$\frac{- 209}{580}$$
If $$x=\csc^2\theta,\ y=\sec^2\theta,$$ and $$z=\dfrac {1}{1-\sin^2\theta \cos^2\theta},$$ then
$$xyz$$ is equal to
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$$x+y+z$$
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$$xy+z$$
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$$x+y-z$$
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$$\displaystyle \frac {x+y}{z}$$
Explanation
Given:
$$x=\csc^2\theta,\ y=\sec^2\theta,$$ and $$z=\dfrac {1}{1-\sin^2\theta \cos^2\theta}$$
Now, $$z=\cfrac { 1 }{ 1-{ \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } $$
Dividing numerator and denominator with $$\sin^2 \theta \cos^2 \theta$$
$$=\dfrac{\dfrac{1}{\sin^2 \theta \cos^2 \theta}}{\dfrac{1}{\sin^2 \theta \cos^2 \theta}-1}$$
$$=\dfrac { { \csc }^{ 2 }\theta. { \sec }^{ 2 }\theta }{ { \csc }^{ 2 }\theta .{ \sec }^{ 2 }\theta -1 } =\dfrac { xy }{ xy-1 } $$ $$...(1)$$
An
d
$$xy={ \csc }^{ 2 }\theta \times { \sec }^{ 2 }\theta $$
$$=\dfrac { 1 }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } $$
$$=\cfrac { { \sin }^{ 2 }\theta+ { \cos }^{ 2 }\theta }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } $$
$$=\cfrac { 1 }{ { \sin }^{ 2 }\theta } +\cfrac { 1 }{ { \cos }^{ 2 }\theta }$$
$$ ={ \csc }^{ 2 }\theta +{ \sec }^{ 2 }\theta $$ $$...(2)$$
From $$(1)$$ and $$(2),$$ we get
$$xyz-z=xy$$
$$\Rightarrow xyz=xy+z$$
or $$xyz=x+y+z$$
If $$tan\, x^{\circ}\, =\, \dfrac{5}{12},\, tan\, y^{\circ}\, =\, \dfrac{3}{4}$$ and $$AB = 48 m$$; find the length of $$CD$$.
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$$50m$$
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$$45m$$
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$$40m$$
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$$55m$$
Explanation
$$\tan x^o = \dfrac{5}{12}$$
$$\tan y^o = \dfrac{3}{4}$$
$$AB = 48m$$
Let $$CD = x\ m$$
From $$\Delta ACD,$$
$$\tan x^o = \dfrac{CD}{AC}$$
$$ \Rightarrow \dfrac{5}{12}=\dfrac{x}{AC}$$
$$\Rightarrow AC = \dfrac{12 x}{5}$$
And from $$\Delta BCD,$$
$$\tan y^o = \dfrac{CD}{BC}$$
$$\Rightarrow \dfrac{3}{4}=\dfrac{x}{BC}$$
$$\Rightarrow BC = \dfrac{4 x}{3}$$
From diagram,
$$AB = AC - BC$$
$$\Rightarrow 48 = \dfrac{12 x}{5} - \dfrac{4x}{3}$$
$$\Rightarrow48 = \dfrac{36 x - 20 x}{15}$$
$$\Rightarrow16 x = 48\times 15$$
$$\Rightarrow x = 45$$
Thus, $$CD = 45$$ m
If, $$\displaystyle \tan \, 9^o = \frac{x}{y}$$ then, value of $$\displaystyle \frac{\sec^2\, 81^o}{1+\cot^2\, 81^o}$$ is
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$$\displaystyle \frac{x^3}{y^3}$$
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$$\displaystyle \frac{x^4}{y^4}$$
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$$\displaystyle \frac{x^5}{y^5}$$
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$$\displaystyle \frac{y^2}{x^2}$$
Explanation
Given,
$$\tan\, 9^{ o }=\dfrac { x }{ y } $$
$$=\dfrac{\sec^2 \, 81^o}{1 + \cot^2 \, 81^o}$$
$$= \dfrac{\sec^2 \, 81^o}{cosec^2 \, 81^o}$$
$$=\dfrac { \dfrac { 1 }{ \cos^{ 2 }81^{ o } } }{ \dfrac { 1 }{ \sin^{ 2 }81^{ o } } \, }$$
$$=\dfrac { \sin^{ 2 }81^{ o } }{ \cos^{ 2 }\, 81^{ o } } $$
$$=\tan^2 \, 81^o$$
$$=[\tan(90^o-9^o)]^2$$
$$= (\cot\, 9^o)^2$$
$$= \cot^2\, 9^o$$
$$=\dfrac{1}{\tan^2 \, 9^o}$$
$$=\dfrac{y^2}{x^2}$$
Hence, option 'D' is correct.
Say yes or no.
$$\tan48^{\small\circ}\tan23^{\small\circ}\tan42^{\small\circ}\tan67^{\small\circ} = 1$$
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Yes
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No
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Ambiguous
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Data insufficient
Explanation
$$\quad LHS = \tan48^{\small\circ}\tan23^{\small\circ}\tan42^{\small\circ}\tan67^{\small\circ} $$
$$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times\tan(90^{\small\circ} - 48^{\small\circ})\times\tan(90^{\small\circ} - 23^{\small\circ})$$
$$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times cot 48^{\small\circ}\times cot 23^{\small\circ}$$
$$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times\displaystyle\frac{1}{\tan48^{\small\circ}}\times\displaystyle\frac{1}{\tan23^{\small\circ}} = 1$$
$$\therefore \quad LHS = RHS$$
$$\displaystyle \left (\frac{\sin\, 50^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, 62^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$
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$$0$$
0%
$$\dfrac{1}{2}$$
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$$1$$
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$$-2$$
Explanation
$$\displaystyle \left (\frac{\sin\, 50^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, 62^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$
= $$\displaystyle \left (\frac{\sin\, (90 - 40)^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, (90 - 28)^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$
= $$\displaystyle \left (\frac{\cos\, 40^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\cos\, 28^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$$
= $$1 + 1 - 2 (1)^2$$
= $$0$$
If $$\displaystyle \cfrac {\cos\alpha}{\cos\beta}=a, \cfrac {\sin\alpha}{\sin\beta}=b$$, then
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$$\sin^2\beta=\displaystyle \frac {a^2-1}{a^2-b^2}$$
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$$\cos^2\beta=\displaystyle \frac {a^2-1}{a^2-b^2}$$
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$$\cos^2\alpha=\displaystyle \frac {(b^2-1)a^2}{b^2-a^2}$$
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$$\sin^2\alpha=\displaystyle \frac {(b^2-1)a^2}{b^2-a^2}$$
Explanation
Given, $$\cfrac{\cos\alpha}{\cos\beta} = a, \cfrac{\sin\alpha}{\sin\beta} = b $$
$$\Rightarrow \cos^2\alpha = a^2 \cos^2\beta $$ and $$ \sin^2\alpha = b^2 \sin^2\beta$$
Adding both, $$ a^2 \cos^2\beta + b^2 \sin^2\beta = 1$$
$$\Rightarrow a^2 (1 - \sin^2\beta) + b^2 \sin^2\beta = 1$$
$$\Rightarrow \sin^2\beta = \dfrac{a^2 - 1}{a^2 - b^2} \Rightarrow \cos^2\beta = 1 - \dfrac{a^2 - 1}{a^2 - b^2} = \dfrac{1 - b^2}{a^2 - b^2}$$
also we have, $$ \cos^2\alpha = a^2 \cos^2\beta $$
$$\Rightarrow \cos^2\alpha = \dfrac{a^2(1 - b^2)}{a^2 - b^2} $$
$$\Rightarrow \sin^2\alpha = 1 - \dfrac{a^2(1 - b^2)}{a^2 - b^2} = \dfrac{b^2(a^2 - 1)}{a^2 - b^2} $$
Choose the correct option
$$\quad \displaystyle\frac{2\tan30^{\small\circ}}{1+(\tan30^{0})^{2}}$$
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$$\sin60^{\small\circ}$$
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$$\cos60^{\small\circ}$$
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$$\tan60^{\small\circ}$$
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$$\sin30^{\small\circ}$$
Explanation
$$\quad \displaystyle\frac{2\tan30^{\small\circ}}{1+\tan30^{\small\circ}} = \displaystyle\frac{2\left(\displaystyle\frac{1}{\sqrt3}\right)}{1+\left(\displaystyle\frac{1}{\sqrt3}\right)^2} = \displaystyle\frac{\displaystyle\frac{2}{\sqrt3}}{1+\displaystyle\frac{1}{3}}$$
$$\quad = \displaystyle\frac{2}{\sqrt3}\times\displaystyle\frac{3}{4} = \displaystyle\frac{\sqrt3}{2} = \sin60^{\small\circ}$$
Find the value of $$\displaystyle \left( \frac{3 \cos 43^{\circ}}{\sin 47^{\circ}} \right)^2 -\frac{\cos 37 ^{\circ}. \text{cosec} 53^{\circ}}{\tan 5^{\circ}. \tan 25^{\circ}. \tan 45^{\circ}. \tan 65^{\circ} \tan 85^{\circ}} $$
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$$7$$
0%
$$0$$
0%
$$1$$
0%
$$8$$
Explanation
$$\left( \dfrac{3 \cos 43^{\circ}}{\sin 47^{\circ}} \right)^2 -\dfrac{\cos 37 ^{\circ}. \text{cosec} 53^{\circ}}{\tan 5^{\circ}. \tan 25^{\circ}. \tan 45^{\circ}. \tan 65^{\circ} \tan 85^{\circ}} = ?$$
$$=\left( \dfrac { 3\cos43^{ \circ } }{ \sin47^{ \circ } } \right) ^{ 2 }-\dfrac { \cos37^{ \circ }\times \text{cosec}53^{ \circ } }{ \tan5^{ \circ }\times \tan25^{ \circ }\times \tan45^{ \circ }\times \tan65^{ \circ }\times \tan85^{ \circ } } $$
$$=\left( \dfrac { 3\sin47^{ \circ } }{ \sin47^{ \circ } } \right) ^{ 2 }-\dfrac { \cos37^{ \circ } }{ \sin53^{ \circ } } \times \dfrac { 1 }{ \tan5^{ \circ }\times \tan25^{ \circ }\times (1)\times cot25^{ \circ }\times cot5^{ \circ } } $$
$$=3^2 - \dfrac{\sin 53^{\circ}}{\sin 53^{\circ}} \times \dfrac{1}{\displaystyle \dfrac{\tan 5^{\circ}}{\tan 5^{\circ}} \times \dfrac{\tan 25^{\circ}}{\tan 25^{\circ}}}$$
$$=9-1=8$$
Hence, option 'D' is correct.
Let $$x=(1+\sin A)(1-\sin B)(1+\sin C), y=(1-\sin A)(1-\sin B)(1-\sin C)$$ and if $$x=y$$, then
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$$x=\cos A \cos B \cos C$$
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$$y=\sin A \sin B \sin C$$
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$$y=-\cos A \cos B \cos C$$
0%
$$x=-\sin A \sin B \sin C$$
Explanation
$$P=\left( 1+\sin A \right) \left( 1+\sin B \right) \left( 1+\sin C \right) =\left( 1-\sin A \right) \left( 1-\sin B \right) \left( 1-\sin C \right) $$......As $$x=y$$ (given)
$$ \Rightarrow { P }^{ 2 }=\left( 1+\sin A \right) \left( 1+\sin B \right) \left( 1+\sin C \right) \times \left( 1-\sin A \right) \left( 1-\sin B \right) \left( 1-\sin C \right) $$
$$ \Rightarrow { P }^{ 2 }=\left( 1-\sin ^{ 2 }{ A } \right) \left( 1-\sin ^{ 2 }{ B } \right) \left( 1-\sin ^{ 2 }{ C } \right) =\cos ^{ 2 }{ A } \cos ^{ 2 }{ B } \cos ^{ 2 }{ C } $$
$$ \therefore P=\pm \cos { A } \cos { B } \cos { C } $$
If $$\cos A + \cos^2A = 1$$ then $$\sin^2A + \sin^4A =1$$.Is it true or false?
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True
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False
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Ambiguous
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Data insufficient
Explanation
$$cosA+cos^{2}A=1$$
$$cosA=1-cos^{2}A$$
$$cosA=sin^{2}A$$
Hence
$$sin^{4}A+sin^{2}A$$
$$cos^{2}A+cosA$$
$$=1$$
Hence True.
Is LHS=RHS?
$$\displaystyle\frac{cosec\theta + \cot\theta}{cosec\theta - \cot\theta} = 1+ 2\cot^2\theta + 2cosec\theta\cot\theta$$
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Yes
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No
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Can't say
0%
Ambiguous
If $$\tan A = \displaystyle\dfrac{3}{4}$$ and $$A+B = 90^{\small\circ}$$, then what is the value of $$\cot B$$?
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0%
$$\displaystyle\dfrac{1}{2}$$
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$$-\displaystyle\dfrac{2}{5}$$
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$$\displaystyle\dfrac{3}{4}$$
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$$-\displaystyle\dfrac{7}{5}$$
Explanation
Given,
$$A+B=90^{ \circ }$$
$$=>A=90^{ \circ }-B$$
Given,
$$\tan A=\dfrac{3}{4}$$
$$=>\cot B=\dfrac{3}{4}$$
Is LHS=RHS?
$$\displaystyle\frac{\tan^2\theta}{1+\tan^2\theta}+\displaystyle\frac{\cot^3\theta}{1+\cot^2\theta} = \sec\theta sin\theta - 2 cosec\theta\cos\theta$$
Say true or false.
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0%
True
0%
False
0%
Ambiguous
0%
Data insufficient
Explanation
$$\dfrac{cot^{3}x}{1+cot^{2}x}$$
$$=\dfrac{tan^{2}(x).cot^{3}x}{1+tan^{2}(x)}$$
$$=\dfrac{cot(x)}{1+tan^{2}(x)}$$
Hence the above expression becomes,
$$\dfrac{tan^{2}x+cotx}{1+tan^{2}(x)}$$
$$=\dfrac{tan^{3}x+1}{(tanx)(tan^{2}x+1)}$$
$$=\dfrac{cos^{3}(x)(sin^{3}x+cos^{3}x)}{cos^{3}x.sinx}$$
$$=\dfrac{cos^{3}x+sin^{3}x}{sinx}$$
RHS
$$=\dfrac{1-2sin^{2}xcos^{2}x}{sinx.cosx}$$
$$=\dfrac{(sin^{2}x+cos^{2}x)^{2}-2sin^{2}x cos^{2}x}{sinx.cosx}$$
$$=\dfrac{sin^{4}x+cos^{4}x}{sinx.cosx}$$
Hence $$LHS\neq RHS$$
Evaluate the following
$$(i)\quad \displaystyle\frac{1+\tan^230^{\small\circ}}{1-\tan^230^{\small\circ}}+cosec^260^{\small\circ} - \cos^245^{\small\circ} + \sin^245^{\small\circ} + \displaystyle\frac{1+\cot^260^{\small\circ}}{1-\cot^260^{\small\circ}}$$
$$(ii)\quad 4(\sin^430^{\small\circ} + \cos^460^{\small\circ}) - 3(\cos^245^{\small\circ} - \sin^290^{\small\circ})$$
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$$(i)\quad \displaystyle\frac{16}{3} \\ (ii)\quad 4$$
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$$(i)\quad \displaystyle\frac{19}{3} \\ (ii)\quad 7$$
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$$(i)\quad \displaystyle\frac{16}{7} \\ (ii)\quad 11$$
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$$(i)\quad \displaystyle\frac{17}{5} \\ (ii)\quad 5$$
Explanation
$$\dfrac{1+tan^{2}30^{0}}{1-tan^{2}30^{0}}+cosec^{2}60^{0}-cos^{2}45^{0}+sin^{2}45^{0}+\dfrac{1+cot^{2}60^{0}}{1-cot^{2}60^{0}}$$
$$=\frac{3+1}{3-1}+\frac{4}{3}-\frac{1}{2}+\frac{1}{2}+\frac{3+1}{3-1}$$
$$=2(2)+\frac{4}{3}$$
$$=\frac{16}{3}$$
$$4(sin^{4}30^{0}+cos^{4}30^{0})-3(cos^{2}45^{0}-sin^{2}90^{0})$$
$$=4(\frac{1+9}{16})-3(\frac{1}{2}-1)$$
$$=\frac{10}{4}+\frac{3}{2}$$
$$=\frac{8}{2}$$
$$=4$$
If $$\sin\theta - \cos\theta = 0$$, then the value of $$(\sin^4\theta + \cos^4\theta)$$ is
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0%
$$1$$
0%
$$\displaystyle\frac{3}{4}$$
0%
$$\displaystyle\frac{1}{2}$$
0%
$$\displaystyle\frac{1}{4}$$
Explanation
From the above equation, we get
$$\cos\theta=\sin\theta$$
Hence
$$\theta=45^{0}$$
Therefore
$$\sin^{4}\theta+\cos^{4}\theta$$
$$=\sin^{4}45^{0}+\cos^{4}45^{0}$$
$$=2\times(\dfrac{1}{\sqrt{2}})^{4}$$
$$=2\times\dfrac{1}{4}$$
$$=\dfrac{1}{2}$$
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