$$\displaystyle \frac{11}{17}$$

MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 5

$\triangle ABC$ If

$\sin B$ is

$\displaystyle \frac{12}{m}$ , then

$m$ is:

0%
13
0%
12
0%
11
0%
10

Explanation

$\triangle ABC$ ,

$AD \perp BC$

$AB = 13$ ,

$BD = 5$ ,

$DC =16$

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 :

$\cot^{2} C\, -\, \displaystyle \frac{1}{\sin^{2} C}$ .

0%
- 1
0%
2
0%
4
0%
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,

$cot^2 C - \frac{1}{\sin^2 C} = cot^2 C - csc^2 C = (\frac{B}{P})^2 - (\frac{H}{P})^2$
=

$(\frac{8\sqrt{2}}{4})^2 - (\frac{12}{4})^2$
=

$8 - 9$
=

$-1$

$\sin C$ is

$\displaystyle \frac{1}{m}$ , then

$m$ is:

0%
$1$
0%
$5$
0%
$2$
0%
$3$

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.

$\tan C$ is

$\displaystyle \frac{3}{m}$ , then

$m$ is:

0%
1
0%
4
0%
3
0%
2

Explanation

$\triangle ABC$ ,

$AD \perp BC$

$AB = 13$ ,

$BD = 5$ ,

$DC =16$

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,

$\tan C = \frac{P}{B} = \frac{AD}{CD} = \frac{12}{16} = \frac{3}{4}$

$\cos A$ is

$\displaystyle \frac{3}{m}$ , then m is:

0%
$5$
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,

$\cos A = \dfrac{B}{H} = \dfrac{AD}{AB} = \dfrac{3}{5}$

$\Longrightarrow$

$m=5$

$5 \cot\, \theta\, =\, 12$ , find the value of :

$\csc\, \theta\, +\, \sec \, \theta$ is

$\displaystyle {3}\cfrac{41}{m}$ , m is

0%
$40$
0%
$60$
0%
$50$
0%
$13$

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

0%
0
0%
7
0%
9
0%
5

Explanation

$\triangle ABC$

$AB^2 + BC^2 = AC^2$ (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

0%
1
0%
3
0%
4
0%
2

Explanation

$\triangle ABC$

$AB^2 + BC^2 = AC^2$ (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,

$\tan \angle DBC = \tan \angle A = \dfrac{P}{B} = \dfrac{BC}{AB}= \dfrac{3}{4}$

$\tan A + \cot A = 5$ ; find the value if

$\tan^{2}\, A\, +\, \cot ^{2} A$ .

0%
$14$
0%
$27$
0%
$23$
0%
$335$

Explanation

$\tan A + \cot A = 5$
Squaring both sides,

$(\tan A + \cot A)^2 = 25$

$\tan^2 A + \cot^2 A + 2 \tan A \cot A = 25$

$\tan^2 A + \cot^2 A + 2 = 25$

$\tan^2 A + \cot^2 A = 23$

$x = a \cos^{3} \theta \sin^{2} \theta, y = a \sin^{3} \theta \cos^{2} \theta$ and

$\dfrac {(x^{2} + y^{2})^{p}}{(xy)^{q}}(p, q\epsilon N)$ is independent of

$\theta$ , then

0%
$p = 4$
0%
$p = 5$
0%
$q = 4$
0%
$q = -5$

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 :

$\displaystyle \frac{3\cos 53^{\circ} \text{cosec}37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$

0%
1
0%
3
0%
6
0%
0

Explanation

$\displaystyle \frac{3\cos 53^{\circ} cosec 37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$

$\displaystyle \frac{3\cos 53^{\circ} \text{cosec }(90 - 53)^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}(90 - 29)^{\circ})}-3\tan ^{2}45^{\circ}$

$\displaystyle \frac{3\cos 53^{\circ} \sec 53^{\circ}}{(\cos ^{2}29^{\circ}+\sin ^{2}29^{\circ})}-3\tan ^{2}45^{\circ}$

$\displaystyle \frac{3}{1}-3(1)^2$

$0$

$\tan 1^o \tan 2^o ... \tan 89^o=x^2-8$ , then the value of

$x$ can be

0%
$-1$
0%
$1$
0%
$-3$
0%
$3$

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

$-3$ .

$cot\, \angle DBA$

0%
$\displaystyle \frac{3}{11}$
0%
$\displaystyle \frac{11}{17}$
0%
$\displaystyle \frac{7}{9}$
0%
$\displaystyle \frac{5}{12}$

Explanation

$\triangle ABC$

$AB^2 + BC^2 = AC^2$ (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:

0%
695
0%
595
0%
295
0%
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

0%
130
0%
215
0%
209
0%
524

Explanation

$\sec A = \frac{29}{21}$

$\sec A = \frac{H}{B} = \frac{29}{21}$

Using Pythagoras Theorem,

$H^2 = P^2 + B^2$

$29^2 = P^2 + 21^2$

$841 = 441 + P^2$

$P = 20$

$Now,$

$\sin A - \frac{1}{\tan A} = \sin A - \cot A$

$=$

$\frac{P}{H} - \frac{B}{P}$

$=$

$\frac{20}{29} - \frac{21}{20}$

$=$

$\frac{400 - 609}{580}$

$=$

$\frac{- 209}{580}$ $

$x=\csc^2\theta,\ y=\sec^2\theta,$ and

$z=\dfrac {1}{1-\sin^2\theta \cos^2\theta},$ then

$xyz$ is equal to

0%
$x+y+z$
0%
$xy+z$
0%
$x+y-z$
0%
$\displaystyle \frac {x+y}{z}$

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

$\sin^2 \theta \cos^2 \theta$

$=\dfrac{\dfrac{1}{\sin^2 \theta \cos^2 \theta}}{\dfrac{1}{\sin^2 \theta \cos^2 \theta}-1}$

$=\dfrac { { \csc }^{ 2 }\theta. { \sec }^{ 2 }\theta }{ { \csc }^{ 2 }\theta .{ \sec }^{ 2 }\theta -1 } =\dfrac { xy }{ xy-1 }$

$...(1)$

And

$xy={ \csc }^{ 2 }\theta \times { \sec }^{ 2 }\theta$

$=\dfrac { 1 }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta }$

$=\cfrac { { \sin }^{ 2 }\theta+ { \cos }^{ 2 }\theta }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta }$

$=\cfrac { 1 }{ { \sin }^{ 2 }\theta } +\cfrac { 1 }{ { \cos }^{ 2 }\theta }$

$={ \csc }^{ 2 }\theta +{ \sec }^{ 2 }\theta$

$...(2)$

From

$(1)$ and

$(2),$ we get

$xyz-z=xy$

$\Rightarrow xyz=xy+z$

$xyz=x+y+z$

$tan\, x^{\circ}\, =\, \dfrac{5}{12},\, tan\, y^{\circ}\, =\, \dfrac{3}{4}$ and

$AB = 48 m$ ; find the length of

$CD$ .

0%
$50m$
0%
$45m$
0%
$40m$
0%
$55m$

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

0%
$\displaystyle \frac{x^3}{y^3}$
0%
$\displaystyle \frac{x^4}{y^4}$
0%
$\displaystyle \frac{x^5}{y^5}$
0%
$\displaystyle \frac{y^2}{x^2}$

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.

$\tan48^{\small\circ}\tan23^{\small\circ}\tan42^{\small\circ}\tan67^{\small\circ} = 1$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

$\quad LHS = \tan48^{\small\circ}\tan23^{\small\circ}\tan42^{\small\circ}\tan67^{\small\circ}$

$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times\tan(90^{\small\circ} - 48^{\small\circ})\times\tan(90^{\small\circ} - 23^{\small\circ})$

$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times cot 48^{\small\circ}\times cot 23^{\small\circ}$

$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times\displaystyle\frac{1}{\tan48^{\small\circ}}\times\displaystyle\frac{1}{\tan23^{\small\circ}} = 1$

$\therefore \quad LHS = RHS$

$\displaystyle \left (\frac{\sin\, 50^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, 62^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$

0%
$0$
0%
$\dfrac{1}{2}$
0%
$1$
0%
$-2$

Explanation

$\displaystyle \left (\frac{\sin\, 50^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, 62^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$

$\displaystyle \left (\frac{\sin\, (90 - 40)^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\sin\, (90 - 28)^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$

$\displaystyle \left (\frac{\cos\, 40^{\circ}}{\cos\, 40^{\circ}} \right)^{2}\, +\, \left (\frac{\cos\, 28^{\circ}}{\cos\, 28^{\circ}} \right)^{2}\, -\, 2\, \tan^{2}\, 45^{\circ}$

$1 + 1 - 2 (1)^2$

$0$

$\displaystyle \cfrac {\cos\alpha}{\cos\beta}=a, \cfrac {\sin\alpha}{\sin\beta}=b$ , then

0%
$\sin^2\beta=\displaystyle \frac {a^2-1}{a^2-b^2}$
0%
$\cos^2\beta=\displaystyle \frac {a^2-1}{a^2-b^2}$
0%
$\cos^2\alpha=\displaystyle \frac {(b^2-1)a^2}{b^2-a^2}$
0%
$\sin^2\alpha=\displaystyle \frac {(b^2-1)a^2}{b^2-a^2}$

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

$\quad \displaystyle\frac{2\tan30^{\small\circ}}{1+(\tan30^{0})^{2}}$

0%
$\sin60^{\small\circ}$
0%
$\cos60^{\small\circ}$
0%
$\tan60^{\small\circ}$
0%
$\sin30^{\small\circ}$

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

$\displaystyle \left( \frac{3 \cos 43^{\circ}}{\sin 47^{\circ}} \right)^2 -\frac{\cos 37 ^{\circ}. \text{cosec} 53^{\circ}}{\tan 5^{\circ}. \tan 25^{\circ}. \tan 45^{\circ}. \tan 65^{\circ} \tan 85^{\circ}}$

0%
$7$
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

0%
$x=\cos A \cos B \cos C$
0%
$y=\sin A \sin B \sin C$
0%
$y=-\cos A \cos B \cos C$
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)

$\Rightarrow { P }^{ 2 }=\left( 1+\sin A \right) \left( 1+\sin B \right) \left( 1+\sin C \right) \times \left( 1-\sin A \right) \left( 1-\sin B \right) \left( 1-\sin C \right)$

$\Rightarrow { P }^{ 2 }=\left( 1-\sin ^{ 2 }{ A } \right) \left( 1-\sin ^{ 2 }{ B } \right) \left( 1-\sin ^{ 2 }{ C } \right) =\cos ^{ 2 }{ A } \cos ^{ 2 }{ B } \cos ^{ 2 }{ C }$

$\therefore P=\pm \cos { A } \cos { B } \cos { C }$

$\cos A + \cos^2A = 1$ then

$\sin^2A + \sin^4A =1$ .Is it true or false?

0%
True
0%
False
0%
Ambiguous
0%
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?

$\displaystyle\frac{cosec\theta + \cot\theta}{cosec\theta - \cot\theta} = 1+ 2\cot^2\theta + 2cosec\theta\cot\theta$

0%
Yes
0%
No
0%
Can't say
0%
Ambiguous

$\tan A = \displaystyle\dfrac{3}{4}$ and

$A+B = 90^{\small\circ}$ , then what is the value of

$\cot B$ ?

0%
$\displaystyle\dfrac{1}{2}$
0%
$-\displaystyle\dfrac{2}{5}$
0%
$\displaystyle\dfrac{3}{4}$
0%
$-\displaystyle\dfrac{7}{5}$

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.

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

$(i)\quad \displaystyle\frac{1+\tan^230^{\small\circ}}{1-\tan^230^{\small\circ}}+cosec^260^{\small\circ} - \cos^245^{\small\circ} + \sin^245^{\small\circ} + \displaystyle\frac{1+\cot^260^{\small\circ}}{1-\cot^260^{\small\circ}}$

$(ii)\quad 4(\sin^430^{\small\circ} + \cos^460^{\small\circ}) - 3(\cos^245^{\small\circ} - \sin^290^{\small\circ})$

0%
$(i)\quad \displaystyle\frac{16}{3} \\ (ii)\quad 4$
0%
$(i)\quad \displaystyle\frac{19}{3} \\ (ii)\quad 7$
0%
$(i)\quad \displaystyle\frac{16}{7} \\ (ii)\quad 11$
0%
$(i)\quad \displaystyle\frac{17}{5} \\ (ii)\quad 5$

Explanation

$\dfrac{1+tan^{2}30^{0}}{1-tan^{2}30^{0}}+cosec^{2}60^{0}-cos^{2}45^{0}+sin^{2}45^{0}+\dfrac{1+cot^{2}60^{0}}{1-cot^{2}60^{0}}$

$=\frac{3+1}{3-1}+\frac{4}{3}-\frac{1}{2}+\frac{1}{2}+\frac{3+1}{3-1}$

$=2(2)+\frac{4}{3}$

$=\frac{16}{3}$

$4(sin^{4}30^{0}+cos^{4}30^{0})-3(cos^{2}45^{0}-sin^{2}90^{0})$

$=4(\frac{1+9}{16})-3(\frac{1}{2}-1)$

$=\frac{10}{4}+\frac{3}{2}$

$=\frac{8}{2}$

$=4$

$\sin\theta - \cos\theta = 0$ , then the value of

$(\sin^4\theta + \cos^4\theta)$ is

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

$\sin^{4}\theta+\cos^{4}\theta$

$=\sin^{4}45^{0}+\cos^{4}45^{0}$

$=2\times(\dfrac{1}{\sqrt{2}})^{4}$

$=2\times\dfrac{1}{4}$

$=\dfrac{1}{2}$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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