# 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

Answer

Verified

365.7k+ 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.

Last updated date: 27th Sep 2023

•

Total views: 365.7k

•

Views today: 5.65k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

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

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE