# The following question consists of two statements, one labeled as ‘Assertion (A)’ and the other labeled as ‘Reason (R)’. You have to examine these two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the reason (R) is a correct explanation for the given Assertion (A). Select your answer to these items using the codes given below and then select the correct option.

Codes:

(A) Both A and R are individually true and R is the correct explanation of A

(B) Both A and R are individually true but R is not the correct explanation of A

(C) A is true but R is false

(D) A is false but R is true

Assertion (A): Derivative of ${{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right)$ with respect to ${{\cos }^{-1}}\theta \left( \dfrac{1-\mathop{x}^{2}}{1+\mathop{x}^{2}} \right)$ is 1 for $0Reason (R): \[{{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right)={{\cos }^{-1}}\left( \dfrac{1-\mathop{x}^{2}}{1+\mathop{x}^{2}} \right)\] for $-1\le x\le 1$

a) A

b) B

c) C

d) D

a) A

b) B

c) C

d) D

Last updated date: 25th Mar 2023

•

Total views: 309.9k

•

Views today: 5.86k

Answer

Verified

309.9k+ views

Hint: To verify the assertion use the formula given below.

\[\dfrac{d}{dx}\left[ f(x), with \text{ } respect \text{ } to ,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

And to verify the Reason use the formulae of inverse trigonometric functions with considering their domains.

Firstly we will check the Assertion (a) is correct or not,

Assertion (A):$\begin{align}

& p={{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right) and q={{\cos }^{-1}}\theta \left( \dfrac{1-\mathop{x}^{2}}{1+\mathop{x}^{2}} \right) \\

& Domain(p)\equiv -1\le x\le 1 \\

\end{align}$

Consider,

$p={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ and, $q={{\cos }^{-1}}\theta \left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ For $ 0\le x\le 1 $

Substitute,$x=\tan \theta $ and $\therefore \theta ={{\tan }^{-1}}x$…………………………………………. (1)

$p={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$ and $q={{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)$

Now, to proceed further we should know the Half Angle formula of ‘sin’ and ‘cosine’ in the form of ‘tan’

Formulae:

$\sin 2\theta =\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$

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

By using above formulae we can write,

$\therefore p={{\sin }^{-1}}\left( \sin 2\theta \right)$ And $q={{\cos }^{-1}}\left( \cos 2\theta \right)$ ………………………………. (2)

As we know the domains of p and q,

$Domain(p)\equiv -1\le x\le 1$ and $Domain(q)=-1\le x\le 1$………………………… (A)

By observing both the domains carefully we can say that the required interval is the subset of given intervals i.e. $\left( 0Therefore now we can solve equation (2) with the use of formulae given below,

Formulae:

\[{{\sin }^{-1}}\left( \sin 2\theta \right)=2\theta \] $-1\le x\le 1$

\[{{\cos }^{-1}}\theta \left( \cos 2\theta \right)=2\theta \] $-1\le x\le 1$

Therefore equation (2) will become,

\[\therefore p=2\theta \] And \[q=2\theta \]

Now put the value of \[\theta ={{\tan }^{-1}}x\] as evaluated in (1),

\[\therefore p=2{{\tan }^{-1}}x\] And \[q=2{{\tan }^{-1}}x\]

Differentiating p and q with respect to x using we will get,

\[\therefore \dfrac{dp}{dx}=2\dfrac{d}{dx}{{\tan }^{-1}}x\] And \[\dfrac{dq}{dx}=2\dfrac{d}{dx}{{\tan }^{-1}}x\]

Formula:

$\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$

By using above formula we can write p and q as,

$\therefore \dfrac{dp}{dx}=2\left( \dfrac{1}{1+{{x}^{2}}} \right)$ And $\dfrac{dq}{dx}=2\left( \dfrac{1}{1+{{x}^{2}}} \right)$ …………………………… (3)

As we have to calculate derivative of p with respect to q we should know the formula given below,

Formula:

If f(x) and g(x) are two functions then,

\[\dfrac{d}{dx}\left[ f(x),with\text{ }respect\text{ }to,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

Therefore,

\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]={}^{\dfrac{dp}{dx}}/{}_{\dfrac{dq}{dx}}\]

From (3) we can write,

\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]=\dfrac{2\left( \dfrac{1}{1+{{x}^{2}}} \right)}{2\left( \dfrac{1}{1+{{x}^{2}}} \right)}\]

\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]=1\]

Therefore Assertion (a) is true.

Now, we will check whether the Reason is true or false,

Reason (R):

We will rewrite the equation given in the Reason,

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

Consider, Left Hand Side(L.H.S.)

$L.H.S.={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ $-1\le x\le 1$

Put,$x=\tan \theta $ in above equation, $\therefore \theta ={{\tan }^{-1}}x$

$\therefore L.H.S.={{\sin }^{-1}}\left( \dfrac{2\tan x}{1+{{\tan }^{2}}x} \right)$

By using Half Angle formula we can write,

$\therefore L.H.S.={{\sin }^{-1}}\left( \sin 2\theta \right)$ $-1\le x\le 1$

To solve further we should know the formula given below,

Formula:

\[{{\sin }^{-1}}\left( \sin 2\theta \right)=2\theta \] $-1\le x\le 1$

By using this formula we can write,

\[\therefore L.H.S.=2\theta \]

Put, \[\theta ={{\tan }^{-1}}x\] as evaluated earlier,

\[\therefore L.H.S.=2{{\tan }^{-1}}x\]…………………………………………….(4)

Also Consider,

$R.H.S.={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ $-1\le x\le 1$

Put,$x=\tan \theta $ in above equation, $\therefore \theta ={{\tan }^{-1}}x$

$\therefore R.H.S.={{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)$

By using Half Angle formula we can write,

\[\therefore R.H.S.={{\cos }^{-1}}\left( \cos 2\theta \right)\] $-1\le x\le 1$

To solve further we should know the formula given below,

Formula:

\[{{\cos }^{-1}}\left( \cos 2\theta \right)=2\theta \] $-1\le x\le 1$

By using this formula we can write,

\[\therefore R.H.S.=2\theta \]

Put, \[\theta ={{\tan }^{-1}}x\] as evaluated earlier,

\[\therefore R.H.S.=2{{\tan }^{-1}}x\]…………………………………………………. (5)

From (4) and (5) we can write,

\[\therefore L.H.S.=R.H.S.\]

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

Therefore, we can say that Reason (R) is true.

If we observe our solution carefully then we can say that the value obtained while solving could have covered nearly half the answer of Assertion and hence Reason is very much helpful in solving this example.

Therefore, both A and R is individually true and R is the correct explanation of A. Option (a) is the correct answer.

Note:

The formula given below is very much important to solve the problem,

\[\dfrac{d}{dx}\left[ f(x),with\text{ }respect\text{ }to,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

In this problem Answer is very much dependent on Limits and you should be very much careful during considering them while solving.

\[\dfrac{d}{dx}\left[ f(x), with \text{ } respect \text{ } to ,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

And to verify the Reason use the formulae of inverse trigonometric functions with considering their domains.

Firstly we will check the Assertion (a) is correct or not,

Assertion (A):$\begin{align}

& p={{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right) and q={{\cos }^{-1}}\theta \left( \dfrac{1-\mathop{x}^{2}}{1+\mathop{x}^{2}} \right) \\

& Domain(p)\equiv -1\le x\le 1 \\

\end{align}$

Consider,

$p={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ and, $q={{\cos }^{-1}}\theta \left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ For $ 0\le x\le 1 $

Substitute,$x=\tan \theta $ and $\therefore \theta ={{\tan }^{-1}}x$…………………………………………. (1)

$p={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$ and $q={{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)$

Now, to proceed further we should know the Half Angle formula of ‘sin’ and ‘cosine’ in the form of ‘tan’

Formulae:

$\sin 2\theta =\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$

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

By using above formulae we can write,

$\therefore p={{\sin }^{-1}}\left( \sin 2\theta \right)$ And $q={{\cos }^{-1}}\left( \cos 2\theta \right)$ ………………………………. (2)

As we know the domains of p and q,

$Domain(p)\equiv -1\le x\le 1$ and $Domain(q)=-1\le x\le 1$………………………… (A)

By observing both the domains carefully we can say that the required interval is the subset of given intervals i.e. $\left( 0

Formulae:

\[{{\sin }^{-1}}\left( \sin 2\theta \right)=2\theta \] $-1\le x\le 1$

\[{{\cos }^{-1}}\theta \left( \cos 2\theta \right)=2\theta \] $-1\le x\le 1$

Therefore equation (2) will become,

\[\therefore p=2\theta \] And \[q=2\theta \]

Now put the value of \[\theta ={{\tan }^{-1}}x\] as evaluated in (1),

\[\therefore p=2{{\tan }^{-1}}x\] And \[q=2{{\tan }^{-1}}x\]

Differentiating p and q with respect to x using we will get,

\[\therefore \dfrac{dp}{dx}=2\dfrac{d}{dx}{{\tan }^{-1}}x\] And \[\dfrac{dq}{dx}=2\dfrac{d}{dx}{{\tan }^{-1}}x\]

Formula:

$\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$

By using above formula we can write p and q as,

$\therefore \dfrac{dp}{dx}=2\left( \dfrac{1}{1+{{x}^{2}}} \right)$ And $\dfrac{dq}{dx}=2\left( \dfrac{1}{1+{{x}^{2}}} \right)$ …………………………… (3)

As we have to calculate derivative of p with respect to q we should know the formula given below,

Formula:

If f(x) and g(x) are two functions then,

\[\dfrac{d}{dx}\left[ f(x),with\text{ }respect\text{ }to,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

Therefore,

\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]={}^{\dfrac{dp}{dx}}/{}_{\dfrac{dq}{dx}}\]

From (3) we can write,

\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]=\dfrac{2\left( \dfrac{1}{1+{{x}^{2}}} \right)}{2\left( \dfrac{1}{1+{{x}^{2}}} \right)}\]

\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]=1\]

Therefore Assertion (a) is true.

Now, we will check whether the Reason is true or false,

Reason (R):

We will rewrite the equation given in the Reason,

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

Consider, Left Hand Side(L.H.S.)

$L.H.S.={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ $-1\le x\le 1$

Put,$x=\tan \theta $ in above equation, $\therefore \theta ={{\tan }^{-1}}x$

$\therefore L.H.S.={{\sin }^{-1}}\left( \dfrac{2\tan x}{1+{{\tan }^{2}}x} \right)$

By using Half Angle formula we can write,

$\therefore L.H.S.={{\sin }^{-1}}\left( \sin 2\theta \right)$ $-1\le x\le 1$

To solve further we should know the formula given below,

Formula:

\[{{\sin }^{-1}}\left( \sin 2\theta \right)=2\theta \] $-1\le x\le 1$

By using this formula we can write,

\[\therefore L.H.S.=2\theta \]

Put, \[\theta ={{\tan }^{-1}}x\] as evaluated earlier,

\[\therefore L.H.S.=2{{\tan }^{-1}}x\]…………………………………………….(4)

Also Consider,

$R.H.S.={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ $-1\le x\le 1$

Put,$x=\tan \theta $ in above equation, $\therefore \theta ={{\tan }^{-1}}x$

$\therefore R.H.S.={{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)$

By using Half Angle formula we can write,

\[\therefore R.H.S.={{\cos }^{-1}}\left( \cos 2\theta \right)\] $-1\le x\le 1$

To solve further we should know the formula given below,

Formula:

\[{{\cos }^{-1}}\left( \cos 2\theta \right)=2\theta \] $-1\le x\le 1$

By using this formula we can write,

\[\therefore R.H.S.=2\theta \]

Put, \[\theta ={{\tan }^{-1}}x\] as evaluated earlier,

\[\therefore R.H.S.=2{{\tan }^{-1}}x\]…………………………………………………. (5)

From (4) and (5) we can write,

\[\therefore L.H.S.=R.H.S.\]

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

Therefore, we can say that Reason (R) is true.

If we observe our solution carefully then we can say that the value obtained while solving could have covered nearly half the answer of Assertion and hence Reason is very much helpful in solving this example.

Therefore, both A and R is individually true and R is the correct explanation of A. Option (a) is the correct answer.

Note:

The formula given below is very much important to solve the problem,

\[\dfrac{d}{dx}\left[ f(x),with\text{ }respect\text{ }to,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

In this problem Answer is very much dependent on Limits and you should be very much careful during considering them while solving.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

Alfred Wallace worked in A Galapagos Island B Australian class 12 biology CBSE

Imagine an atom made up of a proton and a hypothetical class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Why is the cell called the structural and functional class 12 biology CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main