OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (G)

Exercise 8(G)

Differentiate the following w.r.t. x:

Question 1.
(i) \( \cos^{-1}(\cos x) \)
(ii) \( \tan^{-1}(\cot x) \)
Answer:
(i) Let \( y = \cos^{-1}(\cos x) \).
We know that \( \cos^{-1}(\cos \theta) = \theta \) when \( \theta \in [0, \pi] \). So, \( y = x \). The inverse cosine function effectively cancels out the cosine function within its principal range.
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(x) = 1 \).

(ii) Let \( y = \tan^{-1}(\cot x) \).
We can rewrite \( \cot x \) as \( \tan\left(\frac{\pi}{2}-x\right) \).
So, \( y = \tan^{-1}\left[\tan\left(\frac{\pi}{2}-x\right)\right] \). Using trigonometric identities to simplify the expression before differentiating often makes the process much easier.
This simplifies to \( y = \frac{\pi}{2} - x \).
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - x\right) = 0 - 1 = -1 \).
In simple words: For these problems, we first simplify the inverse trigonometric function by using its properties and basic trigonometric identities. This transforms the expression into a much simpler form that is easy to differentiate.

🎯 Exam Tip: Remember the principal value branches for inverse trigonometric functions to correctly simplify expressions before differentiation. Also, be familiar with co-function identities like \( \cot x = \tan(\frac{\pi}{2}-x) \).

Question 2.
(i) \( \tan^{-1} \frac{\sin x}{1+\cos x} \)
(ii) \( \tan^{-1} \frac{1-\cos x}{\sin x} \)
Answer:
(i) Let \( y = \tan^{-1} \frac{\sin x}{1+\cos x} \).
We use the half-angle formulas: \( \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} \) and \( 1+\cos x = 2\cos^2\frac{x}{2} \).
So, \( y = \tan^{-1} \left( \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} \right) \).
This simplifies to \( y = \tan^{-1}\left(\tan\frac{x}{2}\right) \). This method makes the differentiation straightforward by simplifying the expression first.
Therefore, \( y = \frac{x}{2} \).
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{2} \).

(ii) Let \( y = \tan^{-1} \frac{1-\cos x}{\sin x} \).
We use the half-angle formulas: \( 1-\cos x = 2\sin^2\frac{x}{2} \) and \( \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} \).
So, \( y = \tan^{-1} \left( \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \right) \).
This simplifies to \( y = \tan^{-1}\left(\tan\frac{x}{2}\right) \). Recognizing and applying the correct trigonometric identities is a key step in simplifying inverse trigonometric functions for easier differentiation.
Since \( \tan^{-1}(\tan \theta) = \theta \) for \( \theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we have \( y = \frac{x}{2} \).
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{2} \).
In simple words: We simplify the parts inside the inverse tangent using half-angle trigonometric formulas. This turns the complex fractions into a simple tangent form, making it easy to find the derivative.

🎯 Exam Tip: Half-angle formulas are incredibly useful for simplifying expressions involving \( 1 \pm \cos x \) and \( \sin x \) within inverse trigonometric functions.

Question 3. \( \tan^{-1} \frac{\cos x-\sin x}{\cos x+\sin x} \)
Answer:
Let \( y = \tan^{-1} \frac{\cos x-\sin x}{\cos x+\sin x} \).
To simplify, divide both the numerator and the denominator by \( \cos x \).
This gives us \( y = \tan^{-1} \left( \frac{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}} \right) = \tan^{-1} \left( \frac{1-\tan x}{1+\tan x} \right) \).
We know the identity \( \tan\left(\frac{\pi}{4}-x\right) = \frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x} = \frac{1-\tan x}{1+\tan x} \). The use of the \( \tan(A-B) \) formula helps transform the expression into a simpler form.
So, \( y = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right) \).
This simplifies to \( y = \frac{\pi}{4} - x \).
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} - x\right) = 0 - 1 = -1 \).
In simple words: We divide the top and bottom of the fraction by \( \cos x \) to turn it into a form of \( \tan(\frac{\pi}{4}-x) \). This simplifies the inverse tangent function, making it easy to differentiate.

🎯 Exam Tip: When you see expressions like \( \frac{\cos x \pm \sin x}{\cos x \mp \sin x} \), always try dividing by \( \cos x \) to convert it into a \( \tan(\frac{\pi}{4} \pm x) \) form.

Question 4. \( \cot^{-1}\left(\frac{\cos x}{1+\sin x}\right) \)
Answer:
Let \( y = \cot^{-1}\left(\frac{\cos x}{1+\sin x}\right) \).
We can transform the expression using \( \cos x = \sin\left(\frac{\pi}{2}-x\right) \) and \( \sin x = \cos\left(\frac{\pi}{2}-x\right) \).
So, \( y = \cot^{-1}\left(\frac{\sin\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)}\right) \).
Using half-angle formulas, \( \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} \) and \( 1+\cos A = 2\cos^2\frac{A}{2} \). Let \( A = \frac{\pi}{2}-x \).
So, \( y = \cot^{-1}\left(\frac{2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right) \).
This simplifies to \( y = \cot^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \). This approach transforms the expression using co-function identities before applying half-angle formulas.
We know \( \tan \theta = \cot\left(\frac{\pi}{2}-\theta\right) \). So, \( \tan\left(\frac{\pi}{4}-\frac{x}{2}\right) = \cot\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) = \cot\left(\frac{\pi}{4}+\frac{x}{2}\right) \).
Therefore, \( y = \cot^{-1}\left(\cot\left(\frac{\pi}{4}+\frac{x}{2}\right)\right) \).
This simplifies to \( y = \frac{\pi}{4}+\frac{x}{2} \).
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}+\frac{x}{2}\right) = 0 + \frac{1}{2} = \frac{1}{2} \).

Alternatively:
Let \( y = \cot^{-1}\left(\frac{\cos x}{1+\sin x}\right) \).
We use the identities \( \cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2} \) and \( 1+\sin x = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2 \).
So, \( y = \cot^{-1}\left(\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}\right) \).
This can be written as \( y = \cot^{-1}\left(\frac{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}\right) \).
After canceling common terms, \( y = \cot^{-1}\left(\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}\right) \).
Divide both the numerator and the denominator inside the bracket by \( \cos\frac{x}{2} \).
\( y = \cot^{-1}\left(\frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}}\right) \).
We know the identity \( \frac{1-\tan A}{1+\tan A} = \tan\left(\frac{\pi}{4}-A\right) \). Let \( A = \frac{x}{2} \).
So, \( y = \cot^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \).
Using \( \tan \theta = \cot\left(\frac{\pi}{2}-\theta\right) \), we get \( y = \cot^{-1}\left(\cot\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{x}{2}\right)\right)\right) \).
This simplifies to \( y = \cot^{-1}\left(\cot\left(\frac{\pi}{4}+\frac{x}{2}\right)\right) \).
Thus, \( y = \frac{\pi}{4}+\frac{x}{2} \).
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}+\frac{x}{2}\right) = 0 + \frac{1}{2} = \frac{1}{2} \).
In simple words: To differentiate this, we first simplify the expression inside the cot inverse function using trigonometric identities like half-angle formulas. Once simplified to a basic form, we can easily find its derivative. There are often multiple ways to simplify such expressions.

🎯 Exam Tip: Always look for ways to simplify the expression inside inverse trigonometric functions using identities. This is generally much easier than differentiating the complex function directly.

Question 5. \( \sin^{-1} \sqrt{\frac{1-\cos 2 x}{2}} \)
Answer:
Let \( y = \sin^{-1} \sqrt{\frac{1-\cos 2 x}{2}} \).
We know the double-angle identity \( 1-\cos 2x = 2\sin^2 x \).
Substitute this into the expression:
\( y = \sin^{-1} \sqrt{\frac{2\sin^2 x}{2}} \).
This simplifies to \( y = \sin^{-1} \sqrt{\sin^2 x} \).
Therefore, \( y = \sin^{-1}(\sin x) \). This further simplifies to \( y = x \) (assuming \( x \) is within the principal range of \( \sin^{-1} \)).
Now, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(x) = 1 \).
In simple words: We simplify the inner part of the sine inverse function by using a known trigonometric rule about \( \cos 2x \). This makes the whole expression much simpler, allowing us to find its derivative easily.

🎯 Exam Tip: Always remember the fundamental trigonometric identities like \( 1-\cos 2x = 2\sin^2 x \) and \( 1+\cos 2x = 2\cos^2 x \), as they frequently appear in inverse trigonometric differentiation problems.

Question 6. \( \sin^{-1}\left(\sqrt{1-x^2}\right) \)
Answer:
Let \( y = \sin^{-1}\left(\sqrt{1-x^2}\right) \).
To simplify, we substitute \( x = \sin \theta \). This means \( \theta = \sin^{-1} x \).
Substitute \( x = \sin \theta \) into the expression:
\( y = \sin^{-1}\left(\sqrt{1-\sin^2 \theta}\right) \).
Using the identity \( 1-\sin^2 \theta = \cos^2 \theta \), we get:
\( y = \sin^{-1}\left(\sqrt{\cos^2 \theta}\right) \).
This simplifies to \( y = \sin^{-1}(\cos \theta) \).
We can rewrite \( \cos \theta \) as \( \sin\left(\frac{\pi}{2}-\theta\right) \).
So, \( y = \sin^{-1}\left(\sin\left(\frac{\pi}{2}-\theta\right)\right) \). This substitution helps transform the function into a more manageable form.
This simplifies to \( y = \frac{\pi}{2} - \theta \).
Now, substitute back \( \theta = \sin^{-1} x \).
\( y = \frac{\pi}{2} - \sin^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - \sin^{-1} x\right) = 0 - \frac{1}{\sqrt{1-x^2}} = -\frac{1}{\sqrt{1-x^2}} \).
In simple words: We make a substitution, changing \( x \) to \( \sin \theta \). This helps simplify the square root part using basic trigonometry. Then, we can use the rule for inverse sine to find the answer.

🎯 Exam Tip: When you see \( \sqrt{1-x^2} \) in an inverse trigonometric function, try substituting \( x = \sin \theta \) or \( x = \cos \theta \). This often simplifies the expression significantly.

Question 7. \( \sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \)
Answer:
Let \( y = \sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \).
To simplify this, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \sin^{-1}\left(\frac{1}{\sqrt{1+\tan^2 \theta}}\right) \).
Using the identity \( 1+\tan^2 \theta = \sec^2 \theta \), we get:
\( y = \sin^{-1}\left(\frac{1}{\sqrt{\sec^2 \theta}}\right) \).
This simplifies to \( y = \sin^{-1}\left(\frac{1}{\sec \theta}\right) \).
We know that \( \frac{1}{\sec \theta} = \cos \theta \).
So, \( y = \sin^{-1}(\cos \theta) \). This substitution greatly simplifies the expression inside the inverse sine function.
We can rewrite \( \cos \theta \) as \( \sin\left(\frac{\pi}{2}-\theta\right) \).
Therefore, \( y = \sin^{-1}\left(\sin\left(\frac{\pi}{2}-\theta\right)\right) \).
This simplifies to \( y = \frac{\pi}{2} - \theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = \frac{\pi}{2} - \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - \tan^{-1} x\right) = 0 - \frac{1}{1+x^2} = -\frac{1}{1+x^2} \).
In simple words: We change \( x \) to \( \tan \theta \) to simplify the fraction inside the sine inverse function. This transforms it into a simpler form that is easy to differentiate.

🎯 Exam Tip: When you see \( \sqrt{1+x^2} \) in an inverse trigonometric function, substituting \( x = \tan \theta \) or \( x = \cot \theta \) is often the best strategy to simplify the expression.

Question 8. \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \)
Answer:
Let \( y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \).
To simplify, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \).
We recognize the double-angle identity \( \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta \). This is a standard transformation for this type of inverse function.
So, \( y = \cos^{-1}(\cos 2\theta) \).
This simplifies to \( y = 2\theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = 2 \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \).
In simple words: By changing \( x \) to \( \tan \theta \), the expression inside the inverse cosine becomes a known formula for \( \cos 2\theta \). This makes the differentiation very quick and easy.

🎯 Exam Tip: Expressions like \( \frac{1-x^2}{1+x^2} \) (for \( \cos^{-1} \)), \( \frac{2x}{1+x^2} \) (for \( \sin^{-1} \)), and \( \frac{2x}{1-x^2} \) (for \( \tan^{-1} \)) often suggest a substitution of \( x = \tan \theta \) because they relate to double angle formulas for tangent.

Question 9. \( \tan^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right) \)
Answer:
Let \( y = \tan^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right) \).
To simplify, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \tan^{-1}\left(\frac{3 \tan \theta-\tan^3 \theta}{1-3 \tan^2 \theta}\right) \).
We recognize the triple-angle identity \( \frac{3 \tan \theta-\tan^3 \theta}{1-3 \tan^2 \theta} = \tan 3\theta \). This form makes the expression much simpler to handle.
So, \( y = \tan^{-1}(\tan 3\theta) \).
This simplifies to \( y = 3\theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = 3 \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = 3 \cdot \frac{1}{1+x^2} = \frac{3}{1+x^2} \).
In simple words: We replace \( x \) with \( \tan \theta \) to turn the complex fraction into a simple triple-angle tangent formula. This makes the inverse tangent function easy to simplify and differentiate.

🎯 Exam Tip: Recognize the forms \( \frac{3x-x^3}{1-3x^2} \) as \( \tan 3\theta \), \( 2x\sqrt{1-x^2} \) as \( \sin 2\theta \), and \( 4x^3-3x \) as \( \cos 3\theta \) for appropriate substitutions in inverse trigonometric differentiation.

Question 10. \( \sin^{-1}\left(\frac{2 x}{1+x^2}\right) \)
Answer:
Let \( y = \sin^{-1}\left(\frac{2 x}{1+x^2}\right) \).
To simplify, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) \).
We recognize the double-angle identity \( \frac{2 \tan \theta}{1+\tan^2 \theta} = \sin 2\theta \). This transformation is key to simplifying the problem.
So, \( y = \sin^{-1}(\sin 2\theta) \).
This simplifies to \( y = 2\theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = 2 \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \).
In simple words: We replace \( x \) with \( \tan \theta \) to transform the fraction into a simple double-angle sine formula. This allows us to easily find the derivative.

🎯 Exam Tip: The forms \( \frac{2x}{1+x^2} \), \( \frac{1-x^2}{1+x^2} \), and \( \frac{2x}{1-x^2} \) are strong indicators to use the substitution \( x = \tan \theta \) because they are direct results of trigonometric double-angle formulas.

Question 11. \( \cos^{-1} (2x^2 - 1) \)
Answer:
Let \( y = \cos^{-1} (2x^2 - 1) \).
To simplify, we substitute \( x = \cos \theta \). This means \( \theta = \cos^{-1} x \).
Substitute \( x = \cos \theta \) into the expression:
\( y = \cos^{-1} (2\cos^2 \theta - 1) \).
We recognize the double-angle identity \( 2\cos^2 \theta - 1 = \cos 2\theta \). This is a direct application of a trigonometric identity.
So, \( y = \cos^{-1}(\cos 2\theta) \).
This simplifies to \( y = 2\theta \).
Now, substitute back \( \theta = \cos^{-1} x \).
\( y = 2 \cos^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = 2 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{2}{\sqrt{1-x^2}} \).
In simple words: We use the substitution \( x = \cos \theta \) to turn the expression into a simpler form using a double-angle cosine identity. This makes the inverse cosine function easy to differentiate.

🎯 Exam Tip: The expression \( 2x^2-1 \) within an inverse cosine function strongly suggests the substitution \( x = \cos \theta \) to utilize the identity \( \cos 2\theta = 2\cos^2 \theta - 1 \).

Question 12. \( \sin^{-1} (3x - 4x^3) \)
Answer:
Let \( y = \sin^{-1} (3x - 4x^3) \).
To simplify, we substitute \( x = \sin \theta \). This means \( \theta = \sin^{-1} x \).
Substitute \( x = \sin \theta \) into the expression:
\( y = \sin^{-1} (3\sin \theta - 4\sin^3 \theta) \).
We recognize the triple-angle identity \( 3\sin \theta - 4\sin^3 \theta = \sin 3\theta \). This identity is crucial for simplification.
So, \( y = \sin^{-1}(\sin 3\theta) \).
This simplifies to \( y = 3\theta \).
Now, substitute back \( \theta = \sin^{-1} x \).
\( y = 3 \sin^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{3}{\sqrt{1-x^2}} \).
In simple words: We change \( x \) to \( \sin \theta \) to use a triple-angle sine formula. This helps simplify the inverse sine function, making its derivative easy to find.

🎯 Exam Tip: The expression \( 3x-4x^3 \) inside an inverse sine function or \( 4x^3-3x \) inside an inverse cosine function suggests using \( x = \sin \theta \) or \( x = \cos \theta \) respectively to simplify with triple-angle identities.

Question 13. \( \cot^{-1}\left(\frac{1-x^2}{2 x}\right) \)
Answer:
Let \( y = \cot^{-1}\left(\frac{1-x^2}{2 x}\right) \).
To simplify, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \cot^{-1}\left(\frac{1-\tan^2 \theta}{2 \tan \theta}\right) \).
We recognize that \( \frac{2 \tan \theta}{1-\tan^2 \theta} = \tan 2\theta \). So, the inverse expression is \( \frac{1-\tan^2 \theta}{2 \tan \theta} = \frac{1}{\tan 2\theta} = \cot 2\theta \). This helps simplify the function for differentiation.
Thus, \( y = \cot^{-1}(\cot 2\theta) \).
This simplifies to \( y = 2\theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = 2 \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \).
In simple words: We replace \( x \) with \( \tan \theta \) to use a double-angle formula for cotangent. This simplifies the inverse cotangent function, making its derivative easy to find.

🎯 Exam Tip: The form \( \frac{1-x^2}{2x} \) is the reciprocal of the \( \tan 2\theta \) formula, so \( x = \tan \theta \) is the appropriate substitution for \( \cot^{-1} \) functions involving this expression.

Question 14. \( \text{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right) \)
Answer:
Let \( y = \text{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right) \).
To simplify, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \text{cosec}^{-1}\left(\frac{1+\tan^2 \theta}{2 \tan \theta}\right) \).
Using the identity \( 1+\tan^2 \theta = \sec^2 \theta \), we get:
\( y = \text{cosec}^{-1}\left(\frac{\sec^2 \theta}{2 \tan \theta}\right) \).
Rewrite in terms of sine and cosine: \( \frac{\sec^2 \theta}{2 \tan \theta} = \frac{1/\cos^2 \theta}{2\sin \theta/\cos \theta} = \frac{1}{2\sin \theta \cos \theta} \).
We know that \( 2\sin \theta \cos \theta = \sin 2\theta \). This substitution helps convert the complex fraction into a single trigonometric function.
So, \( y = \text{cosec}^{-1}\left(\frac{1}{\sin 2\theta}\right) \).
This simplifies to \( y = \text{cosec}^{-1}(\text{cosec } 2\theta) \).
Therefore, \( y = 2\theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = 2 \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \).
In simple words: We replace \( x \) with \( \tan \theta \) and use trigonometric identities to change the complex fraction into \( \text{cosec } 2\theta \). This allows for easy differentiation of the simplified inverse function.

🎯 Exam Tip: When dealing with \( \text{cosec}^{-1} \) or \( \sec^{-1} \) functions, converting them to \( \sin^{-1} \) or \( \cos^{-1} \) using \( \text{cosec}^{-1} A = \sin^{-1} (1/A) \) and then making substitutions can simplify the process.

Question 15. \( \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) \)
Answer:
Let \( y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) \).
To simplify, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \sin^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \).
We recognize the double-angle identity \( \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta \). This is a common pattern in simplifying inverse trigonometric functions.
So, \( y = \sin^{-1}(\cos 2\theta) \).
We can rewrite \( \cos 2\theta \) as \( \sin\left(\frac{\pi}{2}-2\theta\right) \).
Therefore, \( y = \sin^{-1}\left(\sin\left(\frac{\pi}{2}-2\theta\right)\right) \).
This simplifies to \( y = \frac{\pi}{2} - 2\theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = \frac{\pi}{2} - 2 \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2 \tan^{-1} x\right) = 0 - 2 \cdot \frac{1}{1+x^2} = -\frac{2}{1+x^2} \).
In simple words: We change \( x \) to \( \tan \theta \) to simplify the fraction inside the sine inverse function. This turns it into \( \cos 2\theta \), which we then convert to \( \sin(\frac{\pi}{2}-2\theta) \) for easy differentiation.

🎯 Exam Tip: Remember to convert \( \cos A \) to \( \sin(\frac{\pi}{2}-A) \) or vice versa if you have a \( \sin^{-1}(\cos A) \) or \( \cos^{-1}(\sin A) \) form after substitution.

Question 16. \( \sec^{-1}\left(\frac{x^2+1}{x^2-1}\right) \)
Answer:
Let \( y = \sec^{-1}\left(\frac{x^2+1}{x^2-1}\right) \).
To simplify, we substitute \( x = \tan \theta \). This means \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the expression:
\( y = \sec^{-1}\left(\frac{\tan^2 \theta+1}{\tan^2 \theta-1}\right) \).
Using the identity \( 1+\tan^2 \theta = \sec^2 \theta \), and recognizing \( \tan^2 \theta-1 = - (1-\tan^2 \theta) \), we get:
\( y = \sec^{-1}\left(\frac{\sec^2 \theta}{- (1-\tan^2 \theta)}\right) = \sec^{-1}\left(-\frac{1}{\cos^2 \theta \cdot \frac{1-\tan^2 \theta}{1}}\right) \).
This expression can be rewritten as: \( y = \sec^{-1}\left(\frac{1+ \tan^2 \theta}{-(1-\tan^2 \theta)}\right) = \sec^{-1}\left(-\frac{1}{\cos 2\theta}\right) \).
So, \( y = \sec^{-1}(-\sec 2\theta) \).
We use the property \( \sec^{-1}(-A) = \pi - \sec^{-1}(A) \). This identity is essential for handling negative arguments in inverse secant functions.
\( \implies y = \pi - \sec^{-1}(\sec 2\theta) \).
This simplifies to \( y = \pi - 2\theta \).
Now, substitute back \( \theta = \tan^{-1} x \).
\( y = \pi - 2 \tan^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\pi - 2 \tan^{-1} x\right) = 0 - 2 \cdot \frac{1}{1+x^2} = -\frac{2}{1+x^2} \).
In simple words: We substitute \( x = \tan \theta \) to transform the fraction. This results in \( -\sec 2\theta \). Then, we use a special rule for inverse secant with negative values to simplify and differentiate it.

🎯 Exam Tip: Be very careful with the properties of inverse trigonometric functions for negative arguments (e.g., \( \sec^{-1}(-x) = \pi - \sec^{-1}x \)). Misapplying these can lead to incorrect signs in the final derivative.

Question 17.
(i) \( \tan^{-1} \sqrt{\frac{1+x}{1-x}} \)
(ii) \( \cot^{-1} \sqrt{\frac{1+x}{1-x}} \)
Answer:
(i) Let \( y = \tan^{-1} \sqrt{\frac{1+x}{1-x}} \).
To simplify, we substitute \( x = \cos \theta \). This means \( \theta = \cos^{-1} x \).
Substitute \( x = \cos \theta \) into the expression:
\( y = \tan^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} \).
Using the half-angle identities \( 1+\cos \theta = 2\cos^2\frac{\theta}{2} \) and \( 1-\cos \theta = 2\sin^2\frac{\theta}{2} \):
\( y = \tan^{-1} \sqrt{\frac{2\cos^2\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}} \).
This simplifies to \( y = \tan^{-1} \sqrt{\cot^2\frac{\theta}{2}} \). This particular substitution is very effective here.
So, \( y = \tan^{-1}\left(\cot\frac{\theta}{2}\right) \).
We can rewrite \( \cot\frac{\theta}{2} \) as \( \tan\left(\frac{\pi}{2}-\frac{\theta}{2}\right) \).
Therefore, \( y = \tan^{-1}\left(\tan\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \).
This simplifies to \( y = \frac{\pi}{2} - \frac{\theta}{2} \).
Now, substitute back \( \theta = \cos^{-1} x \).
\( y = \frac{\pi}{2} - \frac{1}{2} \cos^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - \frac{1}{2} \cos^{-1} x\right) = 0 - \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) = \frac{1}{2\sqrt{1-x^2}} \).

(ii) Let \( y = \cot^{-1} \sqrt{\frac{1+x}{1-x}} \).
To simplify, we substitute \( x = \cos \theta \). This means \( \theta = \cos^{-1} x \).
Substitute \( x = \cos \theta \) into the expression:
\( y = \cot^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} \).
Using the half-angle identities \( 1+\cos \theta = 2\cos^2\frac{\theta}{2} \) and \( 1-\cos \theta = 2\sin^2\frac{\theta}{2} \):
\( y = \cot^{-1} \sqrt{\frac{2\cos^2\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}} \).
This simplifies to \( y = \cot^{-1} \sqrt{\cot^2\frac{\theta}{2}} \). This makes the function easy to work with.
So, \( y = \cot^{-1}\left(\cot\frac{\theta}{2}\right) \).
This simplifies to \( y = \frac{\theta}{2} \).
Now, substitute back \( \theta = \cos^{-1} x \).
\( y = \frac{1}{2} \cos^{-1} x \).
Finally, differentiate both sides with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2} \cos^{-1} x\right) = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{1}{2\sqrt{1-x^2}} \).
In simple words: For both parts, we replace \( x \) with \( \cos \theta \) and use half-angle trigonometric formulas to simplify the square root part into a simple cotangent expression. Then, we can easily find the derivatives.

🎯 Exam Tip: When you see expressions like \( \sqrt{\frac{1+x}{1-x}} \) or \( \sqrt{\frac{1-x}{1+x}} \), the substitution \( x = \cos \theta \) is almost always the most effective way to simplify using half-angle identities.

Question 17.
(i) \( \tan ^{-1} \sqrt{\frac{1+x}{1-x}} \)
(ii) \( \cot ^{-1} \sqrt{\frac{1+x}{1-x}} \)
Answer:
(i) Let \( y = \tan^{-1} \sqrt{\frac{1+x}{1-x}} \)
Let \( x = \cos \theta \). This means \( \theta = \cos^{-1}x \).
So, \( y = \tan^{-1} \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} \)
\( \implies y = \tan^{-1} \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} \)
\( \implies y = \tan^{-1} \left( \cot \frac{\theta}{2} \right) \). We use trigonometric identities to simplify.
\( \implies y = \tan^{-1} \left( \tan \left( \frac{\pi}{2} - \frac{\theta}{2} \right) \right) \)
\( \implies y = \frac{\pi}{2} - \frac{\theta}{2} \)
\( \implies y = \frac{\pi}{2} - \frac{1}{2} \cos^{-1}x \). This simplifies the expression for y.
Now, we differentiate with respect to x.
\( \frac{dy}{dx} = 0 - \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} \right) \)
\( \implies \frac{dy}{dx} = \frac{1}{2\sqrt{1-x^2}} \).
(ii) Let \( y = \cot^{-1} \sqrt{\frac{1+x}{1-x}} \)
Let \( x = \cos \theta \). This means \( \theta = \cos^{-1}x \).
So, \( y = \cot^{-1} \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} \)
\( \implies y = \cot^{-1} \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} \)
\( \implies y = \cot^{-1} \left( \cot \frac{\theta}{2} \right) \). The cotangent functions cancel each other out.
\( \implies y = \frac{\theta}{2} \)
\( \implies y = \frac{1}{2} \cos^{-1}x \). This simplifies the expression for y.
Now, we differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} \right) \)
\( \implies \frac{dy}{dx} = \frac{-1}{2\sqrt{1-x^2}} \).
In simple words: For both parts, we replaced 'x' with 'cos θ' to simplify the expressions inside the inverse trigonometric functions. After simplifying them using basic trigonometry, we found 'y' in terms of 'cos inverse x'. Finally, we differentiated these simplified forms to get the answers.

🎯 Exam Tip: When you see expressions like \( \sqrt{\frac{1+x}{1-x}} \) inside inverse trigonometric functions, try substituting \( x = \cos \theta \) or \( x = \cos 2\theta \) to simplify the expression significantly before differentiation.

Question 18.
(i) \( \cos ^{-1} \sqrt{\frac{1+x}{2}} \)
(ii) \( \cos ^{-1} \sqrt{\frac{1-x^2}{2}} \)
(iii) \( \sin ^{-1} \sqrt{\frac{1-x^2}{2}} \)
(iv) \( \cos ^{-1} \sqrt{\frac{1+x^2}{2}} \)
Answer:
(i) Let \( y = \cos^{-1} \sqrt{\frac{1+x}{2}} \)
Let \( x = \cos \theta \). This means \( \theta = \cos^{-1}x \).
So, \( y = \cos^{-1} \sqrt{\frac{1+\cos\theta}{2}} \)
\( \implies y = \cos^{-1} \sqrt{\frac{2\cos^2(\theta/2)}{2}} \)
\( \implies y = \cos^{-1} \left( \cos \frac{\theta}{2} \right) \). The inverse cosine and cosine functions cancel out.
\( \implies y = \frac{\theta}{2} \)
\( \implies y = \frac{1}{2} \cos^{-1}x \). This gives a much simpler form for y.
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} \right) \)
\( \implies \frac{dy}{dx} = \frac{-1}{2\sqrt{1-x^2}} \).
(ii) Let \( y = \cos^{-1} \sqrt{\frac{1-x^2}{2}} \)
Let \( x^2 = \cos \theta \). This means \( \theta = \cos^{-1}(x^2) \).
So, \( y = \cos^{-1} \sqrt{\frac{1-\cos\theta}{2}} \)
\( \implies y = \cos^{-1} \sqrt{\frac{2\sin^2(\theta/2)}{2}} \)
\( \implies y = \cos^{-1} \left( \sin \frac{\theta}{2} \right) \). We convert sin to cos using a complementary angle.
\( \implies y = \cos^{-1} \left( \cos \left( \frac{\pi}{2} - \frac{\theta}{2} \right) \right) \)
\( \implies y = \frac{\pi}{2} - \frac{\theta}{2} \)
\( \implies y = \frac{\pi}{2} - \frac{1}{2} \cos^{-1}(x^2) \). This simplifies the expression for y.
Now, differentiate with respect to x.
\( \frac{dy}{dx} = 0 - \frac{1}{2} \left( \frac{-1}{\sqrt{1-(x^2)^2}} \right) (2x) \). Remember to apply the chain rule.
\( \implies \frac{dy}{dx} = \frac{x}{\sqrt{1-x^4}} \).
(iii) Let \( y = \sin^{-1} \sqrt{\frac{1-x^2}{2}} \)
Let \( x^2 = \cos \theta \). This means \( \theta = \cos^{-1}(x^2) \).
So, \( y = \sin^{-1} \sqrt{\frac{1-\cos\theta}{2}} \)
\( \implies y = \sin^{-1} \sqrt{\frac{2\sin^2(\theta/2)}{2}} \)
\( \implies y = \sin^{-1} \left( \sin \frac{\theta}{2} \right) \). The inverse sine and sine functions cancel out.
\( \implies y = \frac{\theta}{2} \)
\( \implies y = \frac{1}{2} \cos^{-1}(x^2) \). This makes the function easier to differentiate.
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-(x^2)^2}} \right) (2x) \)
\( \implies \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^4}} \).
(iv) Let \( y = \cos^{-1} \sqrt{\frac{1+x^2}{2}} \)
Let \( x^2 = \cos \theta \). This means \( \theta = \cos^{-1}(x^2) \).
So, \( y = \cos^{-1} \sqrt{\frac{1+\cos\theta}{2}} \)
\( \implies y = \cos^{-1} \sqrt{\frac{2\cos^2(\theta/2)}{2}} \)
\( \implies y = \cos^{-1} \left( \cos \frac{\theta}{2} \right) \). This simplification is key to solving the problem.
\( \implies y = \frac{\theta}{2} \)
\( \implies y = \frac{1}{2} \cos^{-1}(x^2) \). This gives a simple form for y.
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-(x^2)^2}} \right) (2x) \)
\( \implies \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^4}} \).
In simple words: For all these problems, we use a trick by replacing 'x' or 'x squared' with 'cos θ'. This helps us use common trigonometric formulas to make the expression much simpler. Once the expression is simple, we can easily find its derivative. Remember to use the chain rule when differentiating if you replaced 'x squared' instead of just 'x'.

🎯 Exam Tip: Always look for substitutions like \( x=\cos\theta \) or \( x=\sin\theta \) that simplify the expression inside inverse trigonometric functions, especially when dealing with \( 1+x \), \( 1-x \), \( 1+x^2 \), or \( 1-x^2 \) terms.

Question 19. \( \tan ^{-1} \frac{x}{\sqrt{\left(a^2-x^2\right)}} \)
Answer:
Let \( y = \tan^{-1} \frac{x}{\sqrt{a^2-x^2}} \)
To simplify, let \( x = a \sin \theta \). This means \( \sin \theta = \frac{x}{a} \), so \( \theta = \sin^{-1} \left( \frac{x}{a} \right) \).
Substitute \( x = a \sin \theta \) into the expression for y:
\( y = \tan^{-1} \frac{a \sin \theta}{\sqrt{a^2-(a \sin \theta)^2}} \)
\( \implies y = \tan^{-1} \frac{a \sin \theta}{\sqrt{a^2-a^2 \sin^2 \theta}} \)
\( \implies y = \tan^{-1} \frac{a \sin \theta}{\sqrt{a^2(1-\sin^2 \theta)}} \)
\( \implies y = \tan^{-1} \frac{a \sin \theta}{\sqrt{a^2 \cos^2 \theta}} \)
\( \implies y = \tan^{-1} \frac{a \sin \theta}{a \cos \theta} \)
\( \implies y = \tan^{-1} (\tan \theta) \). The inverse tangent and tangent functions cancel out.
\( \implies y = \theta \). This is a great simplification.
Since \( \theta = \sin^{-1} \left( \frac{x}{a} \right) \), we have \( y = \sin^{-1} \left( \frac{x}{a} \right) \).
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^2}} \cdot \frac{d}{dx} \left( \frac{x}{a} \right) \). Remember to use the chain rule for differentiation.
\( \implies \frac{dy}{dx} = \frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \cdot \frac{1}{a} \)
\( \implies \frac{dy}{dx} = \frac{1}{\sqrt{\frac{a^2-x^2}{a^2}}} \cdot \frac{1}{a} \)
\( \implies \frac{dy}{dx} = \frac{a}{\sqrt{a^2-x^2}} \cdot \frac{1}{a} \)
\( \implies \frac{dy}{dx} = \frac{1}{\sqrt{a^2-x^2}} \).
In simple words: We used a special substitution, \( x = a \sin \theta \), to make the complex fraction much simpler. After simplification, 'y' became equal to 'θ', which is \( \sin^{-1}(x/a) \). Then, we simply differentiated this easy expression to find the answer.

🎯 Exam Tip: For expressions involving \( \sqrt{a^2-x^2} \), a common and effective substitution is \( x = a \sin \theta \) or \( x = a \cos \theta \) to simplify the square root term into a single trigonometric function.

Question 20. \( \tan ^{-1} \frac{x}{\sqrt{\left(a^2-x^2\right)}} \)
Answer:
Let \( y = \tan^{-1} \frac{1-x}{1+x} \). This type of function is often simplified using trigonometric identities.
Let \( x = \cos \theta \). This means \( \theta = \cos^{-1}x \).
So, \( y = \tan^{-1} \left( \frac{1-\cos\theta}{1+\cos\theta} \right) \)
\( \implies y = \tan^{-1} \left( \frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)} \right) \). We use the half-angle formulas for sine and cosine.
\( \implies y = \tan^{-1} \left( \tan^2 \frac{\theta}{2} \right) \). This simplifies the expression further.
The solution in the source has \( y = \tan^{-1} (\tan (\theta/2)) \) leading to \( y = \theta/2 \). Assuming the square root was implied or a typo in the intermediate step of the source:
Assuming the form \( \tan^{-1} \sqrt{\frac{1-x}{1+x}} \), then \( y = \tan^{-1} \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \tan^{-1} \sqrt{\tan^2(\theta/2)} = \tan^{-1}(\tan(\theta/2)) = \frac{\theta}{2} \).
So, \( y = \frac{1}{2} \cos^{-1}x \). This is a common simplification for this structure.
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} \right) \)
\( \implies \frac{dy}{dx} = \frac{-1}{2\sqrt{1-x^2}} \).
In simple words: We changed 'x' to 'cos θ' to simplify the expression using known formulas. After making it simpler, we found 'y' in terms of 'cos inverse x'. Then, we just took the derivative of that simpler expression.

🎯 Exam Tip: When faced with inverse trigonometric functions involving \( \frac{1-x}{1+x} \) (or similar forms), a useful substitution is \( x = \cos \theta \) to leverage half-angle trigonometric identities for simplification before differentiation.

Question 21. \( \tan ^{-1} \frac{x}{1+\sqrt{1-x^2}} \)
Answer:
Let \( y = \tan^{-1} \frac{x}{1+\sqrt{1-x^2}} \). To simplify this, we use a trigonometric substitution.
Let \( x = \sin \theta \). This means \( \theta = \sin^{-1}x \).
So, \( y = \tan^{-1} \left( \frac{\sin \theta}{1+\sqrt{1-\sin^2 \theta}} \right) \)
\( \implies y = \tan^{-1} \left( \frac{\sin \theta}{1+\sqrt{\cos^2 \theta}} \right) \)
\( \implies y = \tan^{-1} \left( \frac{\sin \theta}{1+\cos \theta} \right) \). Now we use half-angle identities.
\( \implies y = \tan^{-1} \left( \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} \right) \)
\( \implies y = \tan^{-1} \left( \frac{\sin(\theta/2)}{\cos(\theta/2)} \right) \)
\( \implies y = \tan^{-1} \left( \tan \frac{\theta}{2} \right) \). The inverse tangent and tangent cancel each other out.
\( \implies y = \frac{\theta}{2} \). This is a significant simplification.
Since \( \theta = \sin^{-1}x \), we have \( y = \frac{1}{2} \sin^{-1}x \).
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{2} \left( \frac{1}{\sqrt{1-x^2}} \right) \)
\( \implies \frac{dy}{dx} = \frac{1}{2\sqrt{1-x^2}} \).
In simple words: We replaced 'x' with 'sin θ' to make the expression simpler using trigonometry. This turned the whole problem into a simple derivative of \( \frac{1}{2} \sin^{-1}x \).

🎯 Exam Tip: For inverse tangent expressions involving \( \sqrt{1-x^2} \), substituting \( x=\sin\theta \) or \( x=\cos\theta \) often leads to straightforward simplification using half-angle formulas.

Question 22. \( \tan ^{-1}\left(\frac{\sqrt{a}-\sqrt{x}}{1+\sqrt{a x}}\right) \)
Answer:
Let \( y = \tan^{-1}\left(\frac{\sqrt{a}-\sqrt{x}}{1+\sqrt{a x}}\right) \). This form matches a known identity.
We know the identity: \( \tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \).
Comparing our expression with the identity, we can see that \( A = \sqrt{a} \) and \( B = \sqrt{x} \).
So, \( y = \tan^{-1}\sqrt{a} - \tan^{-1}\sqrt{x} \). This decomposition simplifies the function greatly.
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}\sqrt{a}) - \frac{d}{dx}(\tan^{-1}\sqrt{x}) \).
Since \( \tan^{-1}\sqrt{a} \) is a constant (it does not contain x), its derivative is 0.
For \( \tan^{-1}\sqrt{x} \), we use the chain rule: \( \frac{d}{dx}(\tan^{-1}u) = \frac{1}{1+u^2} \frac{du}{dx} \). Here \( u = \sqrt{x} = x^{1/2} \).
So, \( \frac{d}{dx}(\tan^{-1}\sqrt{x}) = \frac{1}{1+(\sqrt{x})^2} \cdot \frac{d}{dx}(x^{1/2}) \)
\( = \frac{1}{1+x} \cdot \frac{1}{2}x^{(1/2)-1} \)
\( = \frac{1}{1+x} \cdot \frac{1}{2}x^{-1/2} \)
\( = \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} \).
Therefore, \( \frac{dy}{dx} = 0 - \frac{1}{2\sqrt{x}(1+x)} \).
\( \implies \frac{dy}{dx} = \frac{-1}{2\sqrt{x}(1+x)} \).
In simple words: This problem uses a special formula for inverse tangent subtraction. By splitting the given expression into two simpler inverse tangents, one of which is a constant, differentiation becomes very easy. We just need to differentiate \( \tan^{-1}\sqrt{x} \) using the chain rule.

🎯 Exam Tip: Recognize the inverse tangent addition/subtraction formula \( \tan^{-1}A \pm \tan^{-1}B = \tan^{-1}\left(\frac{A \pm B}{1 \mp AB}\right) \). This identity is frequently used to simplify complex inverse trigonometric functions before differentiating.

Question 23. \( \tan ^{-1} \sqrt{\frac{2-x}{2+x}} \)
Answer:
Let \( y = \tan^{-1} \sqrt{\frac{2-x}{2+x}} \). To simplify this, we use a substitution involving cosine.
Let \( x = 2 \cos \theta \). This means \( \cos \theta = \frac{x}{2} \), so \( \theta = \cos^{-1}\left(\frac{x}{2}\right) \).
Substitute \( x = 2 \cos \theta \) into the expression for y:
\( y = \tan^{-1} \sqrt{\frac{2-2\cos \theta}{2+2\cos \theta}} \)
\( \implies y = \tan^{-1} \sqrt{\frac{2(1-\cos \theta)}{2(1+\cos \theta)}} \)
\( \implies y = \tan^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \). Now, use half-angle trigonometric identities.
\( \implies y = \tan^{-1} \sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} \)
\( \implies y = \tan^{-1} \sqrt{\tan^2(\theta/2)} \)
\( \implies y = \tan^{-1} \left( \tan \frac{\theta}{2} \right) \). The inverse tangent and tangent cancel out.
\( \implies y = \frac{\theta}{2} \). This is a very simple form.
Since \( \theta = \cos^{-1}\left(\frac{x}{2}\right) \), we have \( y = \frac{1}{2} \cos^{-1}\left(\frac{x}{2}\right) \).
Now, differentiate with respect to x.
\( \frac{dy}{dx} = \frac{1}{2} \cdot \frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^2}} \cdot \frac{d}{dx}\left(\frac{x}{2}\right) \). Remember to apply the chain rule.
\( \implies \frac{dy}{dx} = \frac{1}{2} \cdot \frac{-1}{\sqrt{1-\frac{x^2}{4}}} \cdot \frac{1}{2} \)
\( \implies \frac{dy}{dx} = \frac{-1}{4\sqrt{\frac{4-x^2}{4}}} \)
\( \implies \frac{dy}{dx} = \frac{-1}{4 \cdot \frac{\sqrt{4-x^2}}{2}} \)
\( \implies \frac{dy}{dx} = \frac{-1}{2\sqrt{4-x^2}} \).
In simple words: We used the substitution \( x = 2 \cos \theta \) to make the fraction inside the square root simple. After using trigonometric rules, the whole expression became just \( \frac{\theta}{2} \). Then, we found the derivative of this simple term.

🎯 Exam Tip: For expressions involving \( \sqrt{\frac{a-x}{a+x}} \) or \( \sqrt{\frac{a+x}{a-x}} \), try the substitution \( x = a \cos \theta \) to simplify the fraction using half-angle identities before proceeding with differentiation.

Question 24. \( \sin ^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \)
Answer:
Let \( y = \sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \). We can rewrite the terms to make a substitution easier.
\( \implies y = \sin^{-1}\left(\frac{2^x \cdot 2^1}{1+(2^2)^x}\right) \)
\( \implies y = \sin^{-1}\left(\frac{2 \cdot 2^x}{1+(2^x)^2}\right) \). This form resembles a trigonometric identity.
Let \( 2^x = \tan \theta \). This means \( \theta = \tan^{-1}(2^x) \).
Substitute \( 2^x = \tan \theta \) into the expression for y:
\( y = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) \). This is a well-known double angle identity.
\( \implies y = \sin^{-1}(\sin 2\theta) \). The inverse sine and sine functions cancel out.
\( \implies y = 2\theta \). This simplifies the expression for y considerably.
Since \( \theta = \tan^{-1}(2^x) \), we have \( y = 2 \tan^{-1}(2^x) \).
Now, differentiate with respect to x.
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+(2^x)^2} \cdot \frac{d}{dx}(2^x) \). Remember to apply the chain rule.
\( \implies \frac{dy}{dx} = 2 \cdot \frac{1}{1+4^x} \cdot (2^x \log 2) \). The derivative of \( a^x \) is \( a^x \log a \).
\( \implies \frac{dy}{dx} = \frac{2 \cdot 2^x \log 2}{1+4^x} \).
\( \implies \frac{dy}{dx} = \frac{2^{x+1} \log 2}{1+4^x} \).
In simple words: We first rewrote the expression to highlight \( 2^x \) and then substituted \( 2^x \) with \( \tan \theta \). This turned the complex expression into a simple \( \sin^{-1}(\sin 2\theta) \), which simplifies to \( 2\theta \). Finally, we differentiated this simple form using the chain rule.

🎯 Exam Tip: For expressions involving \( 2^x \), \( 4^x \), or other powers, always look for substitutions like \( a^x = \tan \theta \) or \( a^x = \sin \theta \) to utilize double angle formulas like \( \sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} \) or \( \cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} \).

Question 25.
(i) \( \tan ^{-1}\left(\sqrt{1+x^2}+x\right) \)
(ii) \( \cot ^{-1}\left(\sqrt{1+x^2}-x\right) \)
Answer:
(i) Let \( y = \tan^{-1}(\sqrt{1+x^2}+x) \). To simplify this, we use a trigonometric substitution.
Let \( x = \tan \theta \). This means \( \theta = \tan^{-1}x \).
Substitute \( x = \tan \theta \) into the expression for y:
\( y = \tan^{-1}(\sqrt{1+\tan^2 \theta}+\tan \theta) \)
\( \implies y = \tan^{-1}(\sqrt{\sec^2 \theta}+\tan \theta) \)
\( \implies y = \tan^{-1}(\sec \theta+\tan \theta) \). We express these in terms of sine and cosine.
\( \implies y = \tan^{-1}\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) \)
\( \implies y = \tan^{-1}\left(\frac{1+\sin \theta}{\cos \theta}\right) \). Now, use half-angle identities.
\( \implies y = \tan^{-1}\left(\frac{1+\cos(\frac{\pi}{2}-\theta)}{\sin(\frac{\pi}{2}-\theta)}\right) \). (Using \( \sin\theta = \cos(\pi/2-\theta) \) and \( \cos\theta = \sin(\pi/2-\theta) \)).
\( \implies y = \tan^{-1}\left(\frac{2\cos^2(\frac{\pi}{4}-\frac{\theta}{2})}{2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})}\right) \)
\( \implies y = \tan^{-1}\left(\cot\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right) \). We convert cot to tan.
\( \implies y = \tan^{-1}\left(\tan\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)\right) \)
\( \implies y = \tan^{-1}\left(\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) \)
\( \implies y = \frac{\pi}{4}+\frac{\theta}{2} \). This is a great simplification.
Since \( \theta = \tan^{-1}x \), we have \( y = \frac{\pi}{4}+\frac{1}{2}\tan^{-1}x \).
Now, differentiate with respect to x.
\( \frac{dy}{dx} = 0 + \frac{1}{2} \cdot \frac{1}{1+x^2} \).
\( \implies \frac{dy}{dx} = \frac{1}{2(1+x^2)} \).
(ii) Let \( y = \cot^{-1}(\sqrt{1+x^2}-x) \). To simplify this, we use a trigonometric substitution.
Let \( x = \tan \theta \). This means \( \theta = \tan^{-1}x \).
Substitute \( x = \tan \theta \) into the expression for y:
\( y = \cot^{-1}(\sqrt{1+\tan^2 \theta}-\tan \theta) \)
\( \implies y = \cot^{-1}(\sqrt{\sec^2 \theta}-\tan \theta) \)
\( \implies y = \cot^{-1}(\sec \theta-\tan \theta) \). We express these in terms of sine and cosine.
\( \implies y = \cot^{-1}\left(\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\right) \)
\( \implies y = \cot^{-1}\left(\frac{1-\sin \theta}{\cos \theta}\right) \). Now, use half-angle identities.
\( \implies y = \cot^{-1}\left(\frac{1-\cos(\frac{\pi}{2}-\theta)}{\sin(\frac{\pi}{2}-\theta)}\right) \). (Using \( \sin\theta = \cos(\pi/2-\theta) \) and \( \cos\theta = \sin(\pi/2-\theta) \)).
\( \implies y = \cot^{-1}\left(\frac{2\sin^2(\frac{\pi}{4}-\frac{\theta}{2})}{2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})}\right) \)
\( \implies y = \cot^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right) \). We convert tan to cot.
\( \implies y = \cot^{-1}\left(\cot\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)\right) \)
\( \implies y = \cot^{-1}\left(\cot\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) \)
\( \implies y = \frac{\pi}{4}+\frac{\theta}{2} \). This simplifies the expression for y considerably.
Since \( \theta = \tan^{-1}x \), we have \( y = \frac{\pi}{4}+\frac{1}{2}\tan^{-1}x \).
Now, differentiate with respect to x.
\( \frac{dy}{dx} = 0 + \frac{1}{2} \cdot \frac{1}{1+x^2} \).
\( \implies \frac{dy}{dx} = \frac{1}{2(1+x^2)} \).
In simple words: For both parts, we replaced 'x' with 'tan θ'. This helped simplify the expression inside the inverse function into a simple trigonometric ratio. After using identities, the function 'y' became a simple expression of \( \frac{\pi}{4} + \frac{1}{2} \tan^{-1}x \), which is then easy to differentiate.

🎯 Exam Tip: When you see \( \sqrt{1+x^2} \) inside an inverse trigonometric function, almost always try the substitution \( x=\tan\theta \) or \( x=\cot\theta \). This helps simplify \( \sqrt{1+\tan^2\theta} \) to \( \sec\theta \) and makes further trigonometric transformations possible.

Question 26.
(i) \( \tan ^{-1}\left(\sqrt{1+x^2}+x\right) \)
(ii) \( \cot ^{-1}\left(\sqrt{1+x^2}-x\right) \)
Answer:
(i) Let \( y = \tan^{-1}(\sqrt{1+x^2}+x) \). We substitute \( x = \tan \theta \), which means \( \theta = \tan^{-1}x \).
So, \( y = \tan^{-1}(\sqrt{1+\tan^2\theta} + \tan \theta) \)
\( y = \tan^{-1}(\sec \theta + \tan \theta) \)
Next, we rewrite in terms of sine and cosine:
\( y = \tan^{-1}\left(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}\right) \)
\( y = \tan^{-1}\left(\frac{1+\sin \theta}{\cos \theta}\right) \)
Using trigonometric identities, \( 1+\sin \theta = 1+\cos\left(\frac{\pi}{2}-\theta\right) = 2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \) and \( \cos \theta = \sin\left(\frac{\pi}{2}-\theta\right) = 2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \).
Therefore, the expression becomes:
\( y = \tan^{-1}\left(\frac{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\right) \)
\( y = \tan^{-1}\left(\cot\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right) \)
We know that \( \cot A = \tan\left(\frac{\pi}{2}-A\right) \), so:
\( y = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)\right) \)
\( y = \tan^{-1}\left(\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) \)
\( y = \frac{\pi}{4}+\frac{\theta}{2} \)
Substitute back \( \theta = \tan^{-1}x \):
\( y = \frac{\pi}{4}+\frac{1}{2}\tan^{-1}x \)
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{d}{dx}\left(\frac{1}{2}\tan^{-1}x\right) \)
\( \frac{dy}{dx} = 0 + \frac{1}{2} \cdot \frac{1}{1+x^2} \)
\( \implies \frac{dy}{dx} = \frac{1}{2(1+x^2)} \)
(ii) Let \( y = \cot^{-1}(\sqrt{1+x^2}-x) \). We put \( x = \tan \theta \), so \( \theta = \tan^{-1}x \).
Then, \( y = \cot^{-1}(\sqrt{1+\tan^2\theta} - \tan \theta) \)
\( y = \cot^{-1}(\sec \theta - \tan \theta) \)
Now, express this in terms of sine and cosine:
\( y = \cot^{-1}\left(\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}\right) \)
\( y = \cot^{-1}\left(\frac{1-\sin \theta}{\cos \theta}\right) \)
Using known trigonometric identities, \( 1-\sin \theta = 1-\cos\left(\frac{\pi}{2}-\theta\right) = 2\sin^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \) and \( \cos \theta = \sin\left(\frac{\pi}{2}-\theta\right) = 2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \).
Substituting these into the expression:
\( y = \cot^{-1}\left(\frac{2\sin^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\right) \)
\( y = \cot^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right) \)
Since \( \tan A = \cot\left(\frac{\pi}{2}-A\right) \), we write:
\( y = \cot^{-1}\left(\cot\left(\frac{\pi}{2} - \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)\right) \)
\( y = \cot^{-1}\left(\cot\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) \)
\( y = \frac{\pi}{4}+\frac{\theta}{2} \)
Substitute back \( \theta = \tan^{-1}x \):
\( y = \frac{\pi}{4}+\frac{1}{2}\tan^{-1}x \)
Finally, differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{d}{dx}\left(\frac{1}{2}\tan^{-1}x\right) \)
\( \frac{dy}{dx} = 0 + \frac{1}{2} \cdot \frac{1}{1+x^2} \)
\( \implies \frac{dy}{dx} = \frac{1}{2(1+x^2)} \)
In simple words: For both problems, we replace x with \( \tan \theta \) to simplify the expression inside the inverse trigonometric function. After using identity formulas, the expression becomes a simple linear function of \( \theta \). Then we replace \( \theta \) with \( \tan^{-1}x \) and finally differentiate to find the answer.

🎯 Exam Tip: When differentiating inverse trigonometric functions with square roots like \( \sqrt{1+x^2} \), remember to use the substitution \( x = \tan \theta \) to simplify the expression and eliminate the square root, making the differentiation much easier.

Question 27. \( \sin ^{2}\left[\cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right]. \)
Answer: Let \( y = \sin^2\left[\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right] \).
To simplify the expression inside the inverse cotangent, we use the substitution \( x = \cos \theta \). This means \( \theta = \cos^{-1}x \).
Substitute this into the equation:
\( y = \sin^2\left[\cot^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\right] \)
We use the half-angle trigonometric identities: \( 1-\cos\theta = 2\sin^2(\theta/2) \) and \( 1+\cos\theta = 2\cos^2(\theta/2) \).
Replacing these, we get:
\( y = \sin^2\left[\cot^{-1}\sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}}\right] \)
\( y = \sin^2\left[\cot^{-1}\sqrt{\tan^2(\theta/2)}\right] \)
\( y = \sin^2\left[\cot^{-1}\left(\tan\frac{\theta}{2}\right)\right] \)
We know that \( \tan A = \cot\left(\frac{\pi}{2}-A\right) \), so we can write:
\( y = \sin^2\left[\cot^{-1}\left(\cot\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right)\right] \)
\( y = \sin^2\left[\frac{\pi}{2}-\frac{\theta}{2}\right] \)
Using the identity \( \sin\left(\frac{\pi}{2}-A\right) = \cos A \), the equation becomes:
\( y = \cos^2\left(\frac{\theta}{2}\right) \)
Another half-angle identity states \( \cos^2(\theta/2) = \frac{1+\cos\theta}{2} \).
So, \( y = \frac{1+\cos\theta}{2} \)
Now, substitute back \( \cos\theta = x \):
\( y = \frac{1+x}{2} \)
Finally, we differentiate this simple expression with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1+x}{2}\right) \)
\( \frac{dy}{dx} = \frac{1}{2} \cdot \frac{d}{dx}(1+x) \)
\( \frac{dy}{dx} = \frac{1}{2} \cdot (0+1) \)
\( \implies \frac{dy}{dx} = \frac{1}{2} \)
In simple words: First, we use a substitution \( x = \cos \theta \) to make the expression inside the brackets much simpler. We apply some trigonometry rules to change \( \cot^{-1}(\tan(...)) \) into a simpler angle. Then, we use more trigonometric rules to change \( \sin^2(...) \) into an expression with \( \cos \theta \). After replacing \( \cos \theta \) with x, we get a very simple function that is easy to differentiate.

🎯 Exam Tip: For expressions involving \( \sqrt{\frac{1-x}{1+x}} \), substituting \( x = \cos \theta \) is a common and effective strategy. This simplifies the fraction to \( \tan(\theta/2) \) or \( \cot(\theta/2) \), making the problem much easier to solve.