If $$\sec{\theta}+\tan{\theta}=P$$, then the value of $$\sin{\theta}$$ is

$$\displaystyle\frac{{P}^{2}-1}{{P}^{2}+1}$$

$$\displaystyle\frac{{P}^{2}+1}{2P}$$

$$\displaystyle\frac{{P}^{2}-1}{2P}$$

$$\displaystyle\frac{{P}^{2}+1}{{P}^{2}-1}$$

MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 6

$\sin30^{\small\circ} + \cos60^{\small\circ}$ equals

0%
$\displaystyle\frac{1+\sqrt3}{2}$
0%
$\sqrt3$
0%
$1$
0%
None of these

Explanation

$\sin30^{0}+\cos60^{0}$

$=\sin30^{0}+\sin(90^{0}-60^{0})$

$=2\sin30^{0}$

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

$=1$

$\triangle ABC$ ,

$\angle B = 90^{\small\circ}$ . If

$AB = 14\space cm$ and

$AC = 50\space cm$ then

$\tan A$ equals:

0%
$\displaystyle\frac{24}{25}$
0%
$\displaystyle\frac{24}{7}$
0%
$\displaystyle\frac{7}{24}$
0%
$\displaystyle\frac{25}{24}$

Explanation

In triangle ABC

$B=90^{0}$
Hence

$AC=50cm$ will be the hypotenuse.
And

$AB=14cm$
Therefore

$BC=\sqrt{AC^{2}-AB^{2}}cm$

$=48cm$
Hence

$tanA$

$=\dfrac{perpendicular}{base}$

$=\dfrac{BC}{AB}$

$=\dfrac{48}{14}$

$=\dfrac{24}{7}$

The value of the expression

$[\text{cosec(}75^{\small\circ}+\theta) - \sec(15^{\small\circ}- \theta) - \tan(55^{\small\circ} + \theta) + \cot(35^{\small\circ} - \theta)]$ is

0%
$-1$
0%
$0$
0%
$1$
0%
$\displaystyle\frac{3}{2}$

Explanation

$cosec(75^{0}+\theta)-\sec(15^{0}-\theta)-\tan(55^{0}+\theta)+\cot(35^{0}-\theta)$

$=\sec(90^{0}-75^{0}-\theta)-\sec(15^{0}-\theta)-\cot(90^{0}-55^{0}-\theta)+\cot(35^{0}-\theta)$

$=\sec(15^{0}-\theta)-\sec(15^{0}-\theta)-\cot(35^{0}+\theta)+\cot(35^{0}+\theta)$

$=0$

$\sec\theta = \displaystyle\frac{\sqrt{p^2 + q^2}}{q}$ , then the value of

$\displaystyle\frac{p\sin\theta - q\cos\theta}{p\sin\theta + q\cos\theta}$

0%
$\displaystyle\frac{p}{q}$
0%
$\displaystyle\frac{p^2}{q^2}$
0%
$\displaystyle\frac{p^2 - q^2}{p^2+q^2}$
0%
$\displaystyle\frac{p^2 + q^2}{p^2-q^2}$

Explanation

$\sec\theta=\cfrac{hypotenuse}{base}=\cfrac{\sqrt{p^{2}+q^{2}}}{q}$
Hence,

$hypotenuse=\sqrt{p^{2}+q^{2}}$

$base=q$
Therefore by applying, Pythagoras theorem, we get the altitude as

$p$ .
Hence

$\sin\theta=\cfrac{p}{\sqrt{p^{2}+q^{2}}}$

$\cos\theta=\cfrac{q}{\sqrt{p^{2}+q^{2}}}$
Hence substituting, the values in the above question we get

$\cfrac{p\sin\theta-q\cos\theta}{p\sin\theta+q\cos\theta}$

$=\cfrac{p^{2}-q^{2}}{p^{2}+q^{2}}$

Is LHS=RHS?

$\left\{\displaystyle\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}\right\}^2 = \displaystyle\frac{1+\sin\theta}{1-\sin\theta},\quad 1-\cos\theta \ne 0$

0%
Yes
0%
No
0%
Ambiguos
0%
Data insufficient

Explanation

Given

LHS

$\left\{\dfrac{1+\cos \theta+ \sin \theta}{1+\cos \theta- \sin \theta} \right\}^2$

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

$\dfrac{2 +2\cos \theta +2 \sin \theta + 2\cos \theta \sin \theta }{2 +2\cos \theta -2 \sin \theta -2\cos \theta \sin \theta }$

$\dfrac{(1+\cos \theta) + \sin \theta (1 + \cos \theta) }{(1+\cos \theta) - \sin \theta (1 + \cos \theta)}$

$\dfrac{(1+\cos \theta )(1+\sin \theta)}{(1+\cos \theta )(1-\sin \theta)}$

$\implies \dfrac {1+\sin \theta }{1-\sin \theta }$

$\sec{\theta}+\tan{\theta}=P$ , then the value of

$\sin{\theta}$ is

0%
$\displaystyle\frac{{P}^{2}+1}{2P}$
0%
$\displaystyle\frac{{P}^{2}-1}{2P}$
0%
$\displaystyle\frac{{P}^{2}-1}{{P}^{2}+1}$
0%
$\displaystyle\frac{{P}^{2}+1}{{P}^{2}-1}$

Explanation

$\sec { \theta } +\tan { \theta } =P\\ \Rightarrow { \left( \sec { \theta } +\tan { \theta } \right) }^{ 2 }={ P }^{ 2 }\\ \Rightarrow \sec ^{ 2 }{ \theta } +\tan ^{ 2 }{ \theta } +2\tan { \theta } \sec { \theta } ={ P }^{ 2 }$

$\displaystyle \Rightarrow 2\tan ^{ 2 }{ \theta } +1+\frac { 2\sin { \theta } }{ \cos { \theta } } ={ P }^{ 2 }$

$\displaystyle \Rightarrow \frac { 2\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } }{ \cos ^{ 2 }{ \theta } } ={ P }^{ 2 }$

Applying componendo and dividendo

$\displaystyle \frac { 2\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } -\cos ^{ 2 }{ \theta } }{ 2\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } -\cos ^{ 2 }{ \theta } } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 }$

$\displaystyle \Rightarrow \frac { 2\sin { \theta } \left( \sin { \theta } +1 \right) }{ 2\left( \sin { \theta } +1 \right) } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } \Rightarrow \sin { \theta } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 }$

Is LHS=RHS?

$\sqrt{\displaystyle\frac{\csc\theta - \cot\theta}{\csc\theta + \cot\theta}} + \sqrt{\displaystyle\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta}} = 2\csc\theta$

Say true or false.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

$LHS =\sqrt{\displaystyle\frac{\csc\theta - \cot\theta}{\csc\theta + \cot\theta}} + \sqrt{\displaystyle\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta}}\\=\dfrac{\csc\theta - \cot\theta + \csc\theta + \cot\theta}{\sqrt{\csc^{2}\theta - \cot^{2}\theta}}$

$= \dfrac{2\csc\theta}{1}$

$= 2\csc\theta \\= RHS$

$\sin\theta+co\sec\theta=2$ , then the value of

$\sin^8\theta+co\sec^8\theta$ is equal to

0%
$2$
0%
$2^8$
0%
$2^4$
0%
none of these

Explanation

$\sin\theta +cosec\theta =2\\ \Rightarrow \sin\theta +\cfrac { 1 }{ \sin\theta } =2$
Squaring both sides

${ \left( \sin\theta +\cfrac { 1 }{ \sin\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta } +2\sin\theta \cfrac { 1 }{ \sin\theta } =4\\ \Rightarrow { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta } =2$
Again squaring bot sides

${ \left( { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { \sin }^{ 4 }\theta +\cfrac { 1 }{ { \sin }^{ 4 }\theta } =2$
Similarly

${ \sin }^{ 8 }\theta +\cfrac { 1 }{ { \sin }^{ 8 }\theta } =2$

$\Rightarrow { \sin }^{ 8 }\theta +{ cosec }^{ 8 }\theta =2$

$\cos 9\alpha = \sin \alpha$ , and

$9\alpha < 90^{\circ}$ , then

$\tan 5\alpha = .....$

0%
$\dfrac {1}{\sqrt {3}}$
0%
$\sqrt {3}$
0%
$1$
0%
$0$

Explanation

$\cos 9\alpha = \sin \alpha$ ,

So,

$\cos 9\alpha = \cos (90 - \alpha)\\\therefore 9\alpha = 90 - \alpha\Rightarrow 9\alpha + \alpha = 90^{\circ}\Rightarrow 10\alpha=90^o\Rightarrow \alpha=9^o$

Now,

$\tan5\alpha=\tan(5*9^o)=\tan45^o=1$

Therefore, Answer is

$1$

$\sin{x}+\sin^{2}{x}=1$ , then the value of

$\cos^{2}{ x}+\cos^{4}{x}$ is

0%
$0$
0%
$2$
0%
$1$
0%
None of these

Explanation

$\textbf{Find the value of the given expression}$

$\text{We have given,}$

$\sin{x}+\sin^{2}{x}=1$

$\Rightarrow\sin{x}=1-\sin^{2}{x}$

$\Rightarrow \sin x=\cos^{2}{x}$

$\boldsymbol{[\because\sin^2 x+ \cos^2 x=1]}$

$\text{Now value of required eq. is obtained by}$

$\text{Substituting the value of}\cos^2 x$

$\text{in given equation}$

$\cos^{2}{x}+\cos^{4}{x}$

$=\sin{x}+\sin^{2}{x}$

$=1$

$\Rightarrow \cos^{2}{x}+\cos^{4}{x}=1$

$\textbf{Hence value of the given expression is 1}$

$2(\sin^{6}{\theta}+\cos^{6}{\theta})-3(\sin^{4}{\theta}+\cos^{4}{\theta})+1$ is equal to

0%
$2$
0%
$0$
0%
$4$
0%
$6$

Explanation

$2\left( \sin ^{ 6 }{ \theta } +\cos ^{ 6 }{ \theta } \right) -3\left( \sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } \right) +1$

Using

$(a^3+b^3=(a+b)^3-3\cdot a\cdot b(a+b)) \ and \ (a^2+b^2=(a+b)^2-2ab)$ , we get

$=2\left( { \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) }^{ 3 }-3\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) \right) -3\left( { \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) }^{ 2 }-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } \right) +1$

$=2-6\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } -3+6\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } +1=0$

$\sin \theta + \cos \theta = 1$ , then

$\sin \theta \cos \theta =$ .

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

Explanation

$\textbf{Step 1: Apply relevant identity of a trigonometric function and simplify}$

$\text{We have,}$

$\sin\theta + \cos\theta = 1$

$\text{Squaring both sides, we get,}$

$\left(\sin\theta + \cos\theta\right)^2 = \left(1\right)^2$

$\Rightarrow \sin^2\theta + \cos^2\theta + 2\sin\theta.\cos\theta = 1$

$\boldsymbol{\left[\because\left(a + b\right)^2 = a^2 + b^2 + 2ab\right]}$

$\Rightarrow 1 + 2\sin\theta\cos\theta = 1$

$\Rightarrow \sin\theta\cos\theta = 0$

$\textbf{Hence, the answer is 0}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

Given,

$\sin\theta+\dfrac{1}{\sin\theta}=2$

$(\sin\theta+\dfrac{1}{\sin\theta})^{2}=4$

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

$(\sin^{2}\theta+\dfrac{1}{\sin^{2}\theta})^{2}=4$

$\sin^{4}\theta+\dfrac{1}{\sin^{4}\theta}=2$

Therefore

$\sin^{2k}\theta+\dfrac{1}{\sin^{2k}\theta}=2$ where

$k=1,2,..N$

Hence assertion is incorrect.

$a+b=2$ and

$ab=1$

Then

$a=b=1$ satisfies the above equation.

Hence reason is correct.

Find

$\theta$ , if

$\displaystyle\frac{2\tan\displaystyle\frac{\theta}{2}}{1 + \tan^2\displaystyle\frac{\theta}{2}} = 1,\quad 0^{\small\circ} < \theta \le 90^{\small\circ}$

0%
$\theta = 60^{\small\circ}$
0%
$\theta = 40^{\small\circ}$
0%
$\theta = 90^{\small\circ}$
0%
$\theta = 70^{\small\circ}$

Explanation

Put

$\displaystyle\frac{\theta}{2} = A$ , then we must have

$\displaystyle\frac{2\tan A}{1+\tan^2A} = 1$

$\Rightarrow 1+\tan^2A - 2\tan A = 0$

$\Rightarrow(1-\tan A)^2 = 0$

$\Rightarrow\tan A = 1 = \tan45^{\small\circ}$

$A = 45^{\small\circ}$
We know

$\displaystyle\frac{\theta}{2} = A = 45^{\small\circ}$

$\Rightarrow \displaystyle\frac{\theta}{2} = 45^{\small\circ}$

$\Rightarrow \theta = 90^{\small\circ}$

$(\sec{A}-\tan{A})(\sec{B}-\tan{B})(\sec{C}-\tan{C})=(\sec{A}+\tan{A})(\sec{B}+\tan{B})(\sec{C}+\tan{C})$ represents each side of a equilateral triangle, then each side is equal to -

0%
$0$
0%
$1$
0%
$-1$
0%
$\pm 1$

Explanation

Let

$\left ( \sec A+\tan A \right )\left ( \sec B+\tan B \right )\left ( \sec C+\tan C \right )=x$ ............(i)

$\therefore \left ( \sec A-\tan A \right )\left ( \sec B-\tan B \right )\left ( \sec C-\tan C \right )=x$ .............(ii)

Multiplying Eqs. (i) and (ii), we get

$\left ( \sec ^{2}A-\tan ^{2}A \right )\left ( \sec ^{2}B-\tan ^{2}B \right )\left ( \sec ^{2}C-\tan ^{2}C \right )=x^{2}$ ...........

$\left ( \sec ^{2}A-\tan ^{2}A \right )=1$

$x^{2}=1$

$\therefore$

$x=\pm 1$

Hence, each side is equal to

$\pm 1$ .

$\displaystyle \sin \theta +\sin ^{2}\theta = 1,$ then

$\displaystyle \cos ^{2}\theta +\cos ^{4}\theta =$

0%
$1$
0%
$\displaystyle \sqrt{2}$
0%
$0$
0%
$2$

Explanation

Given:

$\displaystyle \sin\theta + \sin^{2}\theta=1$

$\therefore \displaystyle \sin\theta=1-\sin^{2}\theta$

$\therefore \sin \theta = \cos^{2}\theta$
Now,

$\cos^{2}\theta+\cos^{4}\theta$

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

$\displaystyle 4\sin \theta =3\cos \theta ,$ Then

$\displaystyle \frac{\sec ^{2}\theta }{4(1-\tan ^{2}\theta )}$ is

0%
$\displaystyle \frac{25}{16}$
0%
$\displaystyle \frac{25}{28}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{16}{25}$

Explanation

Given:

$4 \sin \theta = 3 \cos \theta$

$\implies \tan \theta = \dfrac{3}{4}$

Now,

$\displaystyle \dfrac{\sec ^{2}\theta }{4(1-\tan ^{2}\theta )}$

$=\displaystyle \dfrac{1 + \tan ^{2}\theta }{4(1-\tan ^{2}\theta )}$

$=\displaystyle \dfrac{1 + \left(\dfrac{3}{4}\right)^{2} }{4\left[1 - \left(\dfrac{3}{4}\right)^{2}\right]}$

$=\displaystyle \dfrac{16+ 9}{4(16 - 9)}$

$=\displaystyle \dfrac{25}{28}$

$\displaystyle \tan \theta =\frac{x}{y},$ then

$\displaystyle \frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$ is equal to

0%
$\displaystyle \frac{x^{2}+y^{2}}{x^{2}-y^{2}}$
0%
$\displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$
0%
$\displaystyle \frac{x}{\sqrt{x^{2}+y^{2}}}$
0%
$\displaystyle \frac{y}{\sqrt{x^{2}+y^{2}}}$

Explanation

$\displaystyle \cfrac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$

Multiply and divide by

$\cos \theta$

$=\cfrac{x\cfrac{\sin \theta }{\cos \theta }+y\cfrac{\cos \theta }{\cos \theta }}{x\cfrac{\sin \theta }{\cos \theta }-y\cfrac{\cos \theta}{\cos \theta }}$

$=\cfrac{x\times\cfrac{x}{y}+y}{x\times\cfrac{x}{y}-y}$

$=\cfrac{\cfrac{x^{2}+y^{2}}{y}}{\cfrac{x^{2}-y^{2}}{y}}$

$=\cfrac{x^{2}+y^{2}}{x^{2}-y^{2}}$

$x = a \sec$

$\displaystyle \theta$ + b tan

$\displaystyle \theta$ and

$y = b \sec$

$\displaystyle \theta$ + a tan

$\displaystyle \theta$ , then

$\displaystyle x^{2}-y^{2}$ is equal to

0%
$\displaystyle 4ab \sec \theta \tan \theta$
0%
$\displaystyle a^{2}-b^{2}$
0%
$\displaystyle b^{2}-a^{2}$
0%
$\displaystyle a^{2}+b^{2}$

Explanation

Given,

$x=a\sec \theta+b\tan \theta, y=b\sec \theta +a \tan \theta$

$\displaystyle x^{2}-y^{2}$

$=\left ( a\sec\theta+b\tan\theta \right )^{2}-\left ( b\sec\theta+a\tan\theta \right )^{2}$

$=a^{2}\sec^{2}\theta+2ab\sec\theta\tan\theta+b^{2}tan^2 \theta-\left ( b^{2}\sec^{2}\theta+2ab\sec\theta\tan\theta+a^{2}\tan^{2}\theta \right )$

$=\left ( a^{2}-b^{2} \right )\sec^{2}\theta+\left ( b^{2}-a^{2} \right )\tan^{2}\theta$

$=\left ( a^{2} - b^{2} \right )\left ( \sec^{2}\theta-\tan^{2}\theta \right )\left ......[ \because a^{2}-b^{2} =-\left ( b^{2}-a^{2} \right ) \right ]$

$=a^{2}-b^{2}$

$......\left ( \because\sec^{2}\theta-\tan^{2}\theta=1 \right )$

Hence option

$'B'$ is the answer.

Consider the following:
1.

$\displaystyle \tan ^{2}\theta -\sin ^{2}\theta =\tan ^{2}\theta \sin ^{2}\theta$
2.

$\displaystyle (1+\cot ^{2}\theta )(1-\cos \theta )(1+\cos \theta )=1$
Which of the statements given below is correct ?

0%
1 only is the identity
0%
2 only is the identity
0%
Both 1 and 2 are identities
0%
Neither 1 nor 2 is the identity

Explanation

$\displaystyle 1.\tan^{2}\theta-\sin^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}-\sin^{2}\theta$

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

$=\dfrac{\sin^{2}\theta\left ( 1-\cos^{2}\theta \right )}{\cos^{2}\theta}$

$=\dfrac{\sin^{2}\theta}{\cos^{2}\theta}\times\sin^{2}\theta=\tan^{2}\theta\sin^{2}\theta\:$

$2.\left ( 1+\cot^{2}\theta \right )\left ( 1-\cos\theta \right )\left ( 1+cos\theta \right )$

$=\left ( 1+\cot^{2}\theta \right )\left ( 1-\cos^{2}\theta \right )$

$=\left ( 1+\frac{\cos^{2}\theta}{\sin^{2}\theta} \right )\left ( 1-\cos^{2}\theta \right )$

$= \left ( \dfrac{\sin^{2}\theta+\cos^{2}\theta}{\sin^{2}\theta} \right )\times\sin^{2}\theta$ .......[Since,

$(1-\cos^{2}\theta=\sin^{2}\theta )$ ]

$=1$ .......[Since,

$\left ( \cos^{2}\theta+\sin^{2}\theta=1 \right )$ ]

Therefore, both are Identities.

$\sin{x}+\sin^{2}{x}=1$ , then

$\cos^{12}{x}+3\cos^{10}{x}+3\cos^{8}{x}+\cos^{6}{x}-2$ is equals to

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

Explanation

Given

$\sin{x}+\sin^{2}{x}=1$

$\Rightarrow \sin x=\cos^2 x$

Now,

$\cos^{12}{x}+3\cos^{10}{x}+3\cos^{8}{x}+\cos^{6}{x}-2$

$= \sin^{6}{x}+3\sin^{5}{x}+3\sin^{4}{x}+\sin^{3}{x}-2$

$={ (\sin ^{ 2 }{ x } ) }^{ 3 }+3{ (\sin ^{ 2 }{ x } ) }^{ 2 }\sin { x } +3{ (\sin ^{ 2 }{ x } ) }^{ 2 }+\sin ^{ 2 }{ x } \sin { x } -2$

$={ (\sin ^{ 2 }{ x } ) }^{ 3 }+3{ (\sin ^{ 2 }{ x } ) }^{ 2 }\sin { x } +3\sin ^{ 2 }{ x } { (\sin { x } ) }^{ 2 }+\sin ^{ 3 }{ x } -2$

$={(\sin^{2}{x}+\sin{x})}^{3}-2$

$=1-2=-1$

$\displaystyle \tan 2A= \cot (A-60^{\circ}),$ where 2A is an acute angle, then the value of A is

0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 50^{\circ}$
0%
$\displaystyle 24^{\circ}$

Explanation

$\displaystyle \tan 2A= \cot (A-60^{\circ})$

$\cot (90 - 2A) = \cot (A - 60)$

$90 - 2 A = A - 60$

$3A = 150$

$A = 50^{\circ}$

$\displaystyle x = a\cos ^{3}\theta$ and y = b

$\displaystyle \sin ^{3}\theta ,$ then the value of

$\displaystyle \left ( \frac{x}{a} \right )^{2/3}+\left ( \frac{y}{b} \right )^{2/3}$ is

0%
1
0%
-2
0%
2
0%
-1

Explanation

Given:

$x = a \cos^3 \theta$ and

$y = b \sin^3 \theta$

$\therefore \ \dfrac xa = \cos^3 \theta$ and

$\dfrac yb = \sin^3 \theta$

Hence

$\displaystyle \left ( \dfrac{x}{a} \right )^{\dfrac{2}{3}}+\left ( \dfrac{y}{b} \right )^{\dfrac{2}{3}}$

$=\left ( {{\cos^{3}\theta}} \right )^{\dfrac{2}{3}}+\left ( {{\sin^{3}\theta}} \right )^{\dfrac{2}{3}}$

$=\left ( \cos\theta \right )^{2}+\left ( \sin\theta \right )^{2}$

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

$=1$

The value of

$\displaystyle \cot 15^{\circ}\cot 16^{\circ}\cot 17^{\circ}.....\cot 73^{\circ}\cot 74^{\circ}\cot 75^{\circ}$ is

0%
$\displaystyle \frac{1}{2}$
0%
$0$
0%
$1$
0%
$-1$

Explanation

$\cot { 15° } \cot { 16° } \cot { 17° } .........\cot { 73° } \cot { 74° } \cot { 75° }$

$=\cot { \left( 90°-75° \right) } \cot { \left( 90°-74° \right) } \cot { \left( 90°-73° \right) } ......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° }$

$=\tan { 75° } \tan { 74° } \tan { 73° } .......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° }$

$=(\tan { 75° }.\cot75^o) (\tan { 74° }.\cot74^o)( \tan { 73° }.\cot73^o) .......\cot45^o$

$=1\times1\times1\times.........\times1$

$=1$

Hence, the answer is

$1.$

$\displaystyle \dfrac{1+\tan ^{2}A}{1+\cot ^{2}A}$ equals

0%
$\displaystyle \sec ^{2}A$
0%
-1
0%
$\displaystyle \cot ^{2}A$
0%
$\displaystyle \tan ^{2}A$

Explanation

$\displaystyle \frac{1+\tan ^{2}A}{1+\cot ^{2}A}$

$\displaystyle \dfrac{1+\dfrac{sin ^{2}A}{cos ^{2}A}}{1+\dfrac{cos ^{2}A}{sin ^{2}A}}$

$\displaystyle \dfrac{\dfrac{cos^2 A + sin ^{2}A}{cos ^{2}A}}{\dfrac{ sin^2 A + cos ^{2}A}{sin ^{2}A}}$

$\displaystyle \dfrac{\sin^2 A}{\cos^2 A}$

$\displaystyle \tan^2 A$

Using trigonometric identities

$\displaystyle 5 \text{ cosec} ^{2}\theta -5\text{ cot} ^{2}\theta -3$ expressed as an integer is

0%
5
0%
3
0%
2
0%
0

Explanation

The given equation is

$\displaystyle 5 \text{ cosec} ^{2}\theta -5\cot ^{2}\theta -3$
=

$5 ( \text{cosec} ^{2}\theta - \cot ^{2}\theta) -3$
=

$5 - 3$ (since,

$\text{cosec}^2 \theta - \text{cot}^2 \theta = 1$ )
=

$2$

$\displaystyle \frac{4}{3}\cot ^{2}30^{\circ}+3\sin ^{2}60^{\circ}-2\text{cosec} ^{2}60^{\circ}-\frac{3}{4}\tan ^{2}30^{\circ}$ is

0%
$1$
0%
$\displaystyle -\frac{20}{3}$
0%
$\displaystyle \frac{10}{3}$
0%
$5$

Explanation

$\displaystyle \frac{4}{3}\cot ^{2}30^{\circ}+3\sin ^{2}60^{\circ}-2\text{cosec}^{2}60^{\circ}-\frac{3}{4}\tan ^{2}30^{\circ}$

$= \displaystyle \frac{4}{3}(\sqrt 3)^2+3(\frac {\sqrt 3}{2})^2-2(\frac {2}{\sqrt 3})^2-\frac{3}{4}(\frac {1}{\sqrt 3})^2$

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

$=\displaystyle 4+ \frac {9}{4}- \frac {8}{3}-\frac{1}{4}$

$=\displaystyle \frac {10}{3}$
Option C is correct.

In the given figure,

$tan x^o = \dfrac{4}{3}$ and "T" is the mid-point of PR, calculate the length of PQ.

0%
$\sqrt 8$ m
0%
$9$ m
0%
$\sqrt{59}$ m
0%
$10$ m

Explanation

$\triangle RST$

$\Rightarrow \tan { x=\dfrac { RT }{ SR } }$

$\Rightarrow \dfrac { 4 }{ 3 } =\dfrac { RT }{ 3 }$

$\Rightarrow RT=4m$

$\therefore PR=2RT=8m$

Now,

$\triangle PSQ$ is a right angled triangle, such that

$\Rightarrow PQ{ PR }^{ 2 }+{ PQ }^{ 2 }={ PQ }^{ 2 }$

${ \Rightarrow PQ }^{ 2 }+{ PR }^{ 2 }={ RQ }^{ 2 }={ \left( 8 \right) }^{ 2 }+{ \left( 6 \right) }^{ 2 }=64+364$

$\Rightarrow PQ=\sqrt { 100 } =10m$

Hence, the answer is

$10m.$

Evaluate:

$\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$

0%
$\displaystyle \sqrt{3}$
0%
$2$
0%
$1$
0%
$\displaystyle \sqrt{2}$

Explanation

$\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$

$\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\tan (90^{\circ}-23^{\circ}).\tan (90^{\circ}-7^{\circ})$

$\displaystyle \left [ \because \tan (90^{\circ}-\theta )=\cot \theta \right ]$
=

$\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\cot 23^{\circ}.\cot 7^{\circ}$

$\displaystyle \tan 7^{\circ}.\cot 7^{\circ}\tan 23^{\circ}.\cot 23^{\circ}.\sqrt{3}$

$\displaystyle (\because \tan \theta .\cot \theta =1)$

$\displaystyle =1\times 1\times \sqrt{3}=\sqrt{3}$

Evaluate:

$\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$

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

Explanation

$\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$

$=\sin { \left[ { 90 }^{ o }-{ 40 }^{ o }+\theta \right] } -\cos { \left( { 40 }^{ o }-\theta \right) } +(\tan { { 1 }^{ o } } \tan {{89}^{o}})(\tan{{10}^{o}}\tan{{80}^{o}})(\tan {{20}^{o}}\tan {{70}^{o}})$

$=\sin {\left[{90}^{o}-({40}^{o}-\theta) \right]} -\cos {\left({40}^{o}-\theta \right)} +\tan {({90}^{o}} -{ 89 }^{ o })\tan { { 89 }^{ o } } \tan { { (90 }^{ o } } -{ 80 }^{ o })\tan { { 80}^{ o } } \tan { { (90 }^{ o } } -{ 70 }^{ o })\tan { { 70 }^{ o } }$

$=\cos { \left( { 40 }^{ o }-\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\cot { { 89 }^{ o } } \tan { { 89 }^{ o } } \cot { { 80 }^{ o } } \tan { { 80 }^{ o } } \cot { { 70 }^{ o } } \tan { { 70 }^{ o } }$

$=0+1$

$\because \cot {\theta} \tan {\theta}=1$ ]

$=1$

Hence,

$\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}=1$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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