Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$${ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\dfrac { x+y }{ 1-xy } $$, $$xy<1$$
$$=\pi +{ tan }^{ -1 }\dfrac { x+y }{ 1-xy } $$, $$xy>1$$.

Evaluate: $${ tan }^{ -1 }\dfrac { 3sin2\alpha }{ 5+3cos2\alpha } +{ tan }^{ -1 }\left( \dfrac { tan\alpha }{ 4 } \right) $$
where $$-\dfrac { \pi }{ 2 } <\alpha <\dfrac { \pi }{ 2 } $$

A
$$\alpha$$

B
$$2\alpha$$

C
$$3\alpha$$

D
$$4\alpha$$

Solution

(a) $$sin2\alpha =\dfrac { 2t }{ 1+{ t }^{ 2 } } ,cos2\alpha =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } $$, where $$t=tan\alpha $$
L.H.S.$$={ tan }^{ -1 }\dfrac { 3t }{ 4+{ t }^{ 2 } } +{ tan }^{ -1 }\dfrac { t }{ 4 } $$
$$={ tan }^{ -1 }t\dfrac { (16+{ t }^{ 2 }) }{ (16+{ t }^{ 2 }) } ={ tan }^{ -1 }(tan\alpha )=\alpha $$

(b) Let us put $$tan\theta =t$$,
$$2\theta =\dfrac { 2t }{ 1+{ t }^{ 2 } } ,cos2\theta =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } $$.
The 2nd term is
$$\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 3.2t }{ 5(1+{ t }^{ 2 })+(1+{ t }^{ 2 }) } =\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 6t }{ 9+{ t }^{ 2 } } $$
$$=\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 2.t/3 }{ 1+{ \left( t/3 \right) }^{ 2 } } =\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 2T }{ 1+{ T }^{ 2 } } $$
$$=\dfrac { 1 }{ 2 } .2{ tan }^{ -1 }T={ tan }^{ -1 }(t/3)$$
$$\therefore $$ $$\theta ={ tan }^{ -1 }2{ t }^{ 2 }-{ tan }^{ -1 }\dfrac { t }{ 3 } ={ tan }^{ -1 }\dfrac { 2{ t }^{ 2 }-t/3 }{ 1+2{ t }^{ 2 }.t/3 } $$
or $$tan\theta =\dfrac { t(6t-1) }{ 3+2{ t }^{ 3 } } $$ or $$t(3+2{ t }^{ 3 })=t(6t-1)$$
$$\therefore $$ $$t=0$$ or $$2{ t }^{ 3 }-6t+4=0$$
or $${ t }^{ 3 }-3t+2=0$$
$$\therefore $$ $$t=0$$ or $$(t-1)({ t }^{ 2 }+t-2)=0$$
or $$t=0$$ or $$(t-1)(t-1)(t+2)=0$$
$$\therefore $$ $$t=0,1,-2$$
$$\therefore $$ $$\theta =n\pi ,n\pi +\pi /4,n\pi +\alpha $$ where $$tan\alpha =-2$$.

(c) Ans (i),(ii)
From given relation, if $$tan\theta =t$$
$$tan\theta =t=\dfrac { 2{ t }^{ 2 }-\dfrac { 1 }{ 3 } t }{ 1+\dfrac { 2 }{ 3 } { t }^{ 3 } } $$
$$\therefore $$ $$t=0$$ or $$1.\left( 1+\dfrac { 2 }{ 3 } { t }^{ 3 } \right) =2t-\dfrac { 1 }{ 3 } $$.
or $$2{ t }^{ 3 }-6t+4=0$$ or $${ t }^{ 3 }-3t+2=0$$
or $$(t+2)({t}^{2}-2t+1)=0$$
or $$(t+2){ \left( t-1 \right) }^{ 2 }=0$$
$$\therefore $$ $$t=tan\theta =-2,1\Rightarrow $$ (i),(ii).

(d) If $$t=tan\theta $$, then $$sin2\theta =\dfrac { 2t }{ 1+{ t }^{ 2 } } ,cos2\theta =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } $$
$$\therefore $$ $$\dfrac { 3sin2\theta }{ 5+4cos2\theta } =\dfrac { 3.2t }{ 5(1+{ t }^{ 2 })+4(1+{ t }^{ 2 }) } =\dfrac { 6t }{ 9+{ t }^{ 2 } } =\dfrac { 2.t/2 }{ 1+{ \left( t/3 \right) }^{ 2 } } =\dfrac { 2tan\alpha }{ 1+{ tan }^{ 2 }\alpha } =sin2\alpha $$
where $$\dfrac { t }{ 3 } =tan\alpha $$
$$\dfrac { 1 }{ 2 } { sin }^{ -1 }\left( \dfrac { 3sin2\theta }{ 5+4cos2\theta } \right) =\alpha ={ tan }^{ -1 }\left( \dfrac { t }{ 3 } \right) ={ tan }^{ -1 }\left( \dfrac { 1 }{ 3 } tan\theta \right) $$
$$\Rightarrow $$ $${ tan }^{ -1 }x={ tan }^{ -1 }\left( \dfrac { 1 }{ 3 } tan\theta \right) $$ $$\therefore $$ $$x=\dfrac { 1 }{ 3 } tan\theta $$
Question 2
1 / -0

If sin $$^{-1}(x-\dfrac{x^2}{2}+\dfrac{x^3}{4}+..............\infty)+$$ cos $$^{-1}(x^2-\dfrac{x^4}{2}+\dfrac{x^6}{4}-...........\infty)$$$$=\dfrac {\pi}{2}$$ and $$0 < x < \sqrt{2}$$ then x =

A
$$\dfrac{1}{2}$$

B
$$1$$

C
$$-\dfrac{1}{2}$$

D
$$-1$$

Solution

Given $$\text{sin}^{-1}(x-\dfrac{x^2}{2}+\dfrac{x^3}{4}+..............\infty)+$$ $$\text{cos}^{-1}(x^2-\dfrac{x^4}{2}+\dfrac{x^6}{4}-...........\infty)$$$$=\dfrac {\pi}{2}$$
As we know that
$$\text{sin}^{-1} x+\text{cos}^{-1} x=\dfrac{\pi}{2}$$
So $$x-\dfrac{x^2}{2}+\dfrac{x^3}{4}+..............\infty=$$ $$x^2-\dfrac{x^4}{2}+\dfrac{x^6}{4}-...........\infty$$
As we know that
Sum of infinite $$GP$$ i.e $$a+a r+a r^2+\cdots\infty=\dfrac{a}{1-r}$$
$$\implies \dfrac{x}{1+\dfrac{x}{2}}=\dfrac{x^2}{1+\dfrac{x^2}{2}}$$
$$\implies \dfrac{1}{x}+\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{2}$$
$$\implies x=x^2\implies x=1$$
Question 3
1 / -0

if f(x)=$${\tan ^{ - 1}}$$(x)than f(x)+f(y) is equal to:

A
$${{\tan }^{-1}}x-{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$

B
$${{\tan }^{-1}}x+{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$$

C
$${{\tan }^{-1}}x+{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$

D
None of these

Solution

Given that ,$$f\left( x \right)={{\tan }^{-1}}x$$ ….(1)

Put x=y in equation (1) we get

$$f\left( y \right)={{\tan }^{-1}}y$$ ….(2)

Add equation (1) and (2)

$$f\left( x \right)+f\left( y \right)={{\tan }^{-1}}x+{{\tan }^{-1}}y$$

Using inverse trigonometry properties, we get

$${{\tan }^{-1}}x+{{\tan }^{-1}}y$$ $$={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$
This is the answer
Question 4
1 / -0

$$\tan^{-1} \dfrac{1}{2} + \tan^{-1} \dfrac{1}{3} $$ equals

A
$$-\cfrac{\pi }{4}$$

B
$$\cfrac{\pi }{6}$$

C
$$\cfrac{\pi }{4}$$

D
None of these

Solution

Let $$\tan^{-1}\cfrac{1}{2}=x$$
$$\Rightarrow \tan x=\cfrac{1}{2}$$

Let $$\tan^{-1}\cfrac{1}{3}=y$$
$$\Rightarrow \tan y=\cfrac{1}{3}$$

$$\tan (x+y)=\cfrac{\tan x + \tan y}{1-\tan x.\tan y}$$

$$=\cfrac{\cfrac{1}{2}+\cfrac{1}{3}}{1-\cfrac{1}{2}\times \cfrac{1}{3}}$$

$$=1$$

$$x+y=\tan^{-1}1=\cfrac{\pi }{4}$$
$$\tan^{-1}\cfrac{1}{2}+\tan^{-1}\cfrac{1}{3}=\cfrac{\pi }{4}$$
Question 5
1 / -0

For $$x\in(0,\pi/2) $$
$${\sin ^{ - 1}}(\cos x)=?$$

A
$$\pi-x$$

B
$$\dfrac {\pi}{2}-x$$

C
$$\pi-\dfrac{x}{2}$$

D
$$\pi-2x$$

Solution

We have,
$$ {{\sin }^{-1}}\left( \cos x \right) $$

$$ ={{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{2}-x \right) \right) $$

$$ =\dfrac{\pi }{2}-x\,\,\,\,\,\,\,\,\,\,\,\, $$

Hence, this is the answer.
Question 6
1 / -0

The number of real values of x satisfying the equation $$\tan^{-1}\left(\dfrac{x}{1-x^2}\right)+\tan^{-1}\left(\dfrac{1}{x^3}\right)=\dfrac{3\pi}{4}$$, is?

A
$$0$$

B
$$1$$

C
$$2$$

D
Infinitely many

Solution

$$\tan ^{-1} \cfrac{x^4+1-x^2}{x^3-x^5-x} = \cfrac{3\pi}{4}$$
$$\cfrac{x^4+1-x^2}{x^3-x^5-x} = -1$$
$$x^4 + 1 -x^2 + x^3 - x^5-x = 0$$
Real roots = $$1$$
Question 7
1 / -0

If $$\sin^{-1}\dfrac{1}{3} + \sin^{-1}\dfrac{2}{3} = \sin^{-1}x$$, then $$x$$ is equal to-

A
$$0$$

B
$$\dfrac{\sqrt{5} - 4\sqrt{2}}{9}$$

C
$$\dfrac{\sqrt{5} + 4\sqrt{2}}{9}$$

D
$$\dfrac{\pi}{2}$$

Solution

Given $$ sin^{-1}\dfrac{1}{3}+sin^{-1}\dfrac{2}{3} = sin^{-1}x $$
To find x
Sol :- $$ sin^{-1}\dfrac{1}{3}+sin^{-1}\dfrac{2}{3} $$
$$ = sin^{-1}\left ( \dfrac{1}{3}\sqrt{1-\dfrac{4}{9}}+\dfrac{2}{3}\sqrt{1-\dfrac{1}{9}} \right ) $$ $$ \left [ sin^{-1}x+sin^{-1}y = sin^{-1}(x+\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}) \right ] $$
$$ = sin^{-1}\left ( \dfrac{1}{3}\dfrac{\sqrt{5}}{3}+\dfrac{2}{3}\dfrac{\sqrt{8}}{3} \right ) $$
$$ = sin^{-1}\left ( \dfrac{\sqrt{5}+4\sqrt{2}}{9} \right ) $$
$$ \because sin^{-1}\dfrac{1}{3}+sin^{-1}\dfrac{2}{3} = sin^{-1}x $$ (Given)
$$ \therefore x = \dfrac{\sqrt{5}+4\sqrt{2}}{9} $$
Question 8
1 / -0

$${\tan ^{ - 1}}2 + {\tan ^{ - 1}}3$$ is equal to

A
$$ - \dfrac{\pi }{4}$$

B
$${\tan ^{ - 1}}\left( { 1} \right)$$

C
$${\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$$

D
$$\dfrac{{3\pi }}{4}$$

Solution

A) We know that
tan$$^ { - 1 } \mathrm { x } + \tan ^ { - 1 } \mathrm { y } = \tan ^ { - 1 } \left( \dfrac { \mathrm { x } + \mathrm { y } } { 1 - \mathrm { xy } } \right)$$

Thus
$$= \tan$$$$^ { - 1 } \mathrm { 2 } + \tan ^ { - 1 } \mathrm { 3 } = \tan ^ { - 1 } \left( \dfrac { \mathrm { 2 } + \mathrm { 3} } { 1 - \mathrm { 6 } } \right)$$

$$=\tan^{-1} {(-1)}$$

$$=- \dfrac { \pi } { 4 }$$
Question 9
1 / -0

The value of $$\cos^{-1} (\cos 12) - \sin^{-1} (\sin 12)$$ is

A
$$0$$

B
$$\pi$$

C
$$8 \pi - 24$$

D
none of these

Solution

$$12\,rad$$ lies in $$4th$$ quadrant
$$\dfrac{7\pi}{2}<12<4\pi$$
Let $$\theta $$ be an acute angle such that
$$12+\theta=4\pi$$
$$\therefore 12=4\pi-\theta$$ or $$\theta=4\pi-12$$

$$\cos^{-1}(\cos12)-\sin^{-1}(\sin12)$$
$$=\cos^{-1}(\cos(4\pi-\theta))-\sin^{-1}(\sin(4\pi-\theta))$$
$$=\cos^{-1}(\cos\theta)-\sin^{-1}(-\sin\theta)$$
$$=\cos^{-1}(\cos\theta)-\sin^{-1}(\sin(-\theta))$$
$$=\theta-(-\theta)$$
$$=2\theta$$
$$=2(4\pi-12)$$
$$=8\pi-24$$
$$\boxed{\therefore\,\cos^{-1}(\cos12)-\sin^{-1}(\sin12)=8\pi-24}$$
Question 10
1 / -0

$$\tan(\cot^{-1}x)$$ is equal to :

A
$$\dfrac{\pi}{2}-x$$

B
$$\cot(\tan^{-1}x)$$

C
$$\tan x$$

D
$$\dfrac1x$$

Solution

$$\tan(\cot^{-1}x)$$

$$=\tan \left(\tan^{-1}\left({\dfrac1x}\right)\right)$$

$$=\dfrac1x$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 21

Result

Inverse Trigonometric Functions Test - 21

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\alpha$$

$$2\alpha$$

$$3\alpha$$

$$4\alpha$$

Solution

Question 2

$$\dfrac{1}{2}$$

$$1$$

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

$$-1$$

Solution

Question 3

$${{\tan }^{-1}}x-{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$

$${{\tan }^{-1}}x+{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$$

$${{\tan }^{-1}}x+{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$

None of these

Solution

Question 4

$$-\cfrac{\pi }{4}$$

$$\cfrac{\pi }{6}$$

$$\cfrac{\pi }{4}$$

None of these

Solution

Question 5

$$\pi-x$$

$$\dfrac {\pi}{2}-x$$

$$\pi-\dfrac{x}{2}$$

$$\pi-2x$$

Solution

Question 6

$$0$$

$$1$$

$$2$$

Infinitely many

Solution

Question 7

$$0$$

$$\dfrac{\sqrt{5} - 4\sqrt{2}}{9}$$

$$\dfrac{\sqrt{5} + 4\sqrt{2}}{9}$$

$$\dfrac{\pi}{2}$$

Solution

Question 8

$$ - \dfrac{\pi }{4}$$

$${\tan ^{ - 1}}\left( { 1} \right)$$

$${\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$$

$$\dfrac{{3\pi }}{4}$$

Solution

Question 9

$$0$$

$$\pi$$

$$8 \pi - 24$$

none of these

Solution

Question 10

$$\dfrac{\pi}{2}-x$$

$$\cot(\tan^{-1}x)$$

$$\tan x$$

$$\dfrac1x$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.