Samacheer Kalvi Class 12 Maths Solutions Chapter 4 Inverse Trigonometric Functions Exercise 4.6

Choose the Most Suitable Answer from the Given Four Alternatives:

Question 1. The value of \( \sin^{-1}(\cos x) \), \( 0 \le x \le \pi \) is
(a) \( \pi - x \)
(b) \( x - \frac{\pi}{2} \)
(c) \( \frac{\pi}{2} - x \)
Answer: (c) \( \frac{\pi}{2} - x \)
To find the value of `\( \sin^{-1}(\cos x) \)`, we use the trigonometric identity that relates sine and cosine through complementary angles. We know that `\( \cos x = \sin(\frac{\pi}{2} - x) \)`.
Substitute this identity into the expression:
`\( \sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right) \)`
For the given range `\( 0 \le x \le \pi \)`, the argument `\( \left(\frac{\pi}{2} - x\right) \)` will lie in the interval `\( [-\frac{\pi}{2}, \frac{\pi}{2}] \)`. This range is the principal value branch for `\( \sin^{-1} \)` when its argument is `\( \sin \theta \)`.
Therefore, `\( \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x \)`.
The problem demonstrates the simplification of inverse trigonometric functions using fundamental identities.
In simple words: We change the cosine part to sine using a common trigonometric rule: `\( \cos x = \sin(\frac{\pi}{2} - x) \)`. Then, `\( \sin^{-1} \)` and `\( \sin \)` cancel each other out, giving us the remaining expression `\( \frac{\pi}{2} - x \)`. This works because `x` is in a specific range.

🎯 Exam Tip: Remember to use the identity `\( \cos \theta = \sin(\frac{\pi}{2} - \theta) \)` to simplify inverse trigonometric expressions. Also, pay attention to the given range of `x` for correct principal values, ensuring that `\( \left(\frac{\pi}{2} - x\right) \)` falls within `\( [-\frac{\pi}{2}, \frac{\pi}{2}] \)`.

Question 2. If \( \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \); then \( \cos^{-1} x + \cos^{-1} y \) is equal to
(a) \( \frac{2\pi}{3} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{6} \)
(d) \( \pi \)
Answer: (b) \( \frac{\pi}{3} \)
We know the important identity for inverse trigonometric functions:
`\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)` for `\( A \in [-1, 1] \)`.
Applying this identity for both `x` and `y`:
`\( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \)`
`\( \sin^{-1} y + \cos^{-1} y = \frac{\pi}{2} \)`
Now, let's add these two equations together:
`\( (\sin^{-1} x + \cos^{-1} x) + (\sin^{-1} y + \cos^{-1} y) = \frac{\pi}{2} + \frac{\pi}{2} \)`
Rearrange the terms:
`\( (\sin^{-1} x + \sin^{-1} y) + (\cos^{-1} x + \cos^{-1} y) = \pi \)`
We are given that `\( \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \)`. Substitute this value into the equation:
`\( \frac{2\pi}{3} + (\cos^{-1} x + \cos^{-1} y) = \pi \)`
Now, solve for `\( \cos^{-1} x + \cos^{-1} y \)`:
`\( \cos^{-1} x + \cos^{-1} y = \pi - \frac{2\pi}{3} \)`
`\( \cos^{-1} x + \cos^{-1} y = \frac{3\pi - 2\pi}{3} \)`
`\( \cos^{-1} x + \cos^{-1} y = \frac{\pi}{3} \)`
This problem beautifully illustrates the application of fundamental inverse trigonometric identities.
In simple words: We use the rule that `\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)`. We apply this rule for both `x` and `y` and add them together. Then, we put in the given value for the sum of `\( \sin^{-1} \)` terms. After doing the subtraction, we find that the sum of `\( \cos^{-1} \)` terms is `\( \frac{\pi}{3} \)`.

🎯 Exam Tip: Mastering the fundamental identity `\( \sin^{-1} z + \cos^{-1} z = \frac{\pi}{2} \)` is crucial. This identity often simplifies problems involving sums of inverse trigonometric functions by allowing you to transform one type of sum into another or to find a specific value for `z`.

Question 3. \( \sin^{-1}\left(\frac{3}{5}\right) - \cos^{-1}\left(\frac{12}{13}\right) + \sec^{-1}\left(\frac{5}{3}\right) - \operatorname{cosec}^{-1}\left(\frac{13}{12}\right) \) is equal to
(a) \( 2\pi \)
(b) \( \pi \)
(c) \( 0 \)
(d) \( \tan^{-1}\left(\frac{12}{65}\right) \)
Answer: (c) \( 0 \)
The given expression is:
`\( \sin^{-1}\left(\frac{3}{5}\right) - \cos^{-1}\left(\frac{12}{13}\right) + \sec^{-1}\left(\frac{5}{3}\right) - \operatorname{cosec}^{-1}\left(\frac{13}{12}\right) \)`
We use the reciprocal identities for inverse trigonometric functions:
`\( \sec^{-1} x = \cos^{-1}\left(\frac{1}{x}\right) \)`
`\( \operatorname{cosec}^{-1} x = \sin^{-1}\left(\frac{1}{x}\right) \)`
Applying these identities to the terms in the expression:
`\( \sec^{-1}\left(\frac{5}{3}\right) = \cos^{-1}\left(\frac{3}{5}\right) \)`
`\( \operatorname{cosec}^{-1}\left(\frac{13}{12}\right) = \sin^{-1}\left(\frac{12}{13}\right) \)`
Substitute these transformed terms back into the original expression:
`\( = \sin^{-1}\left(\frac{3}{5}\right) - \cos^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) - \sin^{-1}\left(\frac{12}{13}\right) \)`
Now, rearrange the terms to group `\( \sin^{-1} A + \cos^{-1} A \)` pairs:
`\( = \left(\sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{3}{5}\right)\right) - \left(\cos^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac{12}{13}\right)\right) \)`
Using the identity `\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)` for each group:
`\( = \frac{\pi}{2} - \frac{\pi}{2} \)`
`\( = 0 \)`
The problem demonstrates the simplification of complex inverse trigonometric expressions through identity recognition.
In simple words: First, we use rules to change `\( \sec^{-1} \)` to `\( \cos^{-1} \)` and `\( \operatorname{cosec}^{-1} \)` to `\( \sin^{-1} \)`. After changing and rearranging the terms, we see two pairs. Each pair follows the rule `\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)`. Since one `\( \frac{\pi}{2} \)` is subtracted from the other, the answer becomes zero.

🎯 Exam Tip: It's important to be familiar with reciprocal identities of inverse trigonometric functions: `\( \sec^{-1} x = \cos^{-1} \frac{1}{x} \)` and `\( \operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x} \)`. This allows you to group terms and apply the fundamental sum identity `\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)`, simplifying complex expressions efficiently.

Question 4. If \( \sin^{-1} x = 2\sin^{-1} \alpha \) has a solution, then
(a) \( |\alpha| \le \frac{1}{\sqrt{2}} \)
(b) \( |\alpha| \ge \frac{1}{\sqrt{2}} \)
(c) \( \alpha < \frac{1}{\sqrt{2}} \)
(d) \( |\alpha| > \frac{1}{\sqrt{2}} \)
Answer: (a) \( |\alpha| \le \frac{1}{\sqrt{2}} \)
For the function `\( \sin^{-1} x \)` to be defined, `\( x \)` must lie in the interval `\( [-1, 1] \)`. Additionally, the range of `\( \sin^{-1} x \)` is `\( [-\frac{\pi}{2}, \frac{\pi}{2}] \)`.
We are given the equation `\( \sin^{-1} x = 2\sin^{-1} \alpha \)`.
For this equation to have a solution, the value `\( 2\sin^{-1} \alpha \)` must be within the range of `\( \sin^{-1} x \)`.
So, we must have:
`\( -\frac{\pi}{2} \le 2\sin^{-1} \alpha \le \frac{\pi}{2} \)`
Divide all parts of the inequality by 2:
`\( -\frac{\pi}{4} \le \sin^{-1} \alpha \le \frac{\pi}{4} \)`
Now, apply the sine function to all parts of the inequality. Since `\( \sin \theta \)` is an increasing function in the interval `\( [-\frac{\pi}{4}, \frac{\pi}{4}] \)`, the inequality signs remain the same:
`\( \sin\left(-\frac{\pi}{4}\right) \le \alpha \le \sin\left(\frac{\pi}{4}\right) \)`
`\( -\frac{1}{\sqrt{2}} \le \alpha \le \frac{1}{\sqrt{2}} \)`
This inequality can be written in terms of absolute value as `\( |\alpha| \le \frac{1}{\sqrt{2}} \)`. This condition ensures a valid solution exists.
In simple words: For the equation to work, the value `\( 2\sin^{-1} \alpha \)` must be between `\( -\frac{\pi}{2} \)` and `\( \frac{\pi}{2} \)`. We divide by 2, and then take the sine of each part. This shows that `\( \alpha \)` must be between `\( -\frac{1}{\sqrt{2}} \)` and `\( \frac{1}{\sqrt{2}} \)`, which can also be written as `\( |\alpha| \le \frac{1}{\sqrt{2}} \)`.

🎯 Exam Tip: Always consider the domain and range of inverse trigonometric functions when solving equations or inequalities. The range of `\( \sin^{-1} x \)` is `\( [-\frac{\pi}{2}, \frac{\pi}{2}] \)`, while for `\( \cos^{-1} x \)` it is `\( [0, \pi] \)`. The argument of the inverse function must also be within `[-1, 1]`.

Question 5. \( \sin^{-1}(\cos x) = \frac{\pi}{2} - x \) is valid for
(b) \( 0 \le x \le \pi \)
(c) \( -\frac{\pi}{2} \le x \le \frac{\pi}{2} \)
(d) \( -\frac{\pi}{4} \le x \le \frac{3\pi}{4} \)
Answer: (b) \( 0 \le x \le \pi \)
The given identity is `\( \sin^{-1}(\cos x) = \frac{\pi}{2} - x \)`.
For this identity to be valid, the expression on the right-hand side, `\( \frac{\pi}{2} - x \)`, must fall within the principal value range of the `\( \sin^{-1} \)` function. The principal value range of `\( \sin^{-1} \)` is `\( [-\frac{\pi}{2}, \frac{\pi}{2}] \)`.
So, we set up the inequality:
`\( -\frac{\pi}{2} \le \frac{\pi}{2} - x \le \frac{\pi}{2} \)`
To isolate `\( -x \)`, subtract `\( \frac{\pi}{2} \)` from all parts of the inequality:
`\( -\frac{\pi}{2} - \frac{\pi}{2} \le -x \le \frac{\pi}{2} - \frac{\pi}{2} \)`
`\( -\pi \le -x \le 0 \)`
Now, multiply all parts of the inequality by -1. Remember to reverse the direction of the inequality signs when multiplying by a negative number:
`\( 0 \le x \le \pi \)`
This means the identity is valid for `\( x \)` in the interval `\( [0, \pi] \)`. This range ensures that `\( \cos x \)` covers all its possible values, and `\( \frac{\pi}{2} - x \)` maps correctly to the principal values of `\( \sin^{-1} \)`.
In simple words: For the equation `\( \sin^{-1}(\cos x) = \frac{\pi}{2} - x \)` to be correct, the right side (`\( \frac{\pi}{2} - x \)` ) must be within the normal output range of `\( \sin^{-1} \)` (which is from `\( -\frac{\pi}{2} \)` to `\( \frac{\pi}{2} \)`). We solve this range for `x` and find that `x` must be between `0` and `\( \pi \)`.

🎯 Exam Tip: When an inverse trigonometric function equals an expression (e.g., `\( \sin^{-1}(A) = B \)`), always ensure that the value of `B` lies within the principal value branch of that inverse function. For `\( \sin^{-1} \)`, the output `\( y \)` must satisfy `\( -\frac{\pi}{2} \le y \le \frac{\pi}{2} \)`. For `\( \cos^{-1} \)`, it must satisfy `\( 0 \le y \le \pi \)`.

Question 6. If \( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2} \), the value of \( x^{2017} + y^{2018} + z^{2019} - \frac{9}{x^{101}+y^{101}+z^{101}} \) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a) 0
The maximum possible value for `\( \sin^{-1} A \)` is `\( \frac{\pi}{2} \)`. This occurs when `\( A = 1 \)`.
We are given the equation: `\( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2} \)`.
Since each of the `\( \sin^{-1} \)` terms can be at most `\( \frac{\pi}{2} \)`, for their sum to be exactly `\( \frac{3\pi}{2} \)`, each individual term must attain its maximum value:
`\( \sin^{-1} x = \frac{\pi}{2} \implies x = 1 \)`
`\( \sin^{-1} y = \frac{\pi}{2} \implies y = 1 \)`
`\( \sin^{-1} z = \frac{\pi}{2} \implies z = 1 \)`
So, we have `\( x=1, y=1, z=1 \)`.
Now, we substitute these values into the given expression we need to evaluate:
`\( x^{2017} + y^{2018} + z^{2019} - \frac{9}{x^{101}+y^{101}+z^{101}} \)`
`\( = 1^{2017} + 1^{2018} + 1^{2019} - \frac{9}{1^{101}+1^{101}+1^{101}} \)`
Since any positive integer power of 1 is simply 1:
`\( = 1 + 1 + 1 - \frac{9}{1+1+1} \)`
`\( = 3 - \frac{9}{3} \)`
`\( = 3 - 3 \)`
`\( = 0 \)`
This demonstrates how understanding the range of inverse trigonometric functions can simplify complex algebraic expressions.
In simple words: The largest value for `\( \sin^{-1} \)` of any number is `\( \frac{\pi}{2} \)`. Since the sum of three `\( \sin^{-1} \)` terms is `\( \frac{3\pi}{2} \)`, it means each one must be `\( \frac{\pi}{2} \)`. This only happens if `x`, `y`, and `z` are all 1. We then put `1` in place of `x`, `y`, and `z` in the given math problem. When we calculate it, the answer turns out to be 0.

🎯 Exam Tip: Recognize that `\( \sin^{-1} x \)` has a maximum value of `\( \frac{\pi}{2} \)` (when `\( x=1 \)`). If the sum of multiple `\( \sin^{-1} \)` terms equals the sum of their maximum possible values, then each term must be at its maximum. This often simplifies problems by giving specific values for variables, which can then be substituted into other expressions.

Question 7. If \( \cot^{-1} x = \frac{2\pi}{5} \) for some \( x \in \mathbb{R} \), the value of \( \tan^{-1} x \) is
(a) \( -\frac{\pi}{10} \)
(b) \( \frac{\pi}{5} \)
(c) \( \frac{\pi}{10} \)
(d) \( -\frac{\pi}{5} \)
Answer: (c) \( \frac{\pi}{10} \)
We know the important identity that connects `\( \tan^{-1} x \)` and `\( \cot^{-1} x \)`:
`\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)`
We are given that `\( \cot^{-1} x = \frac{2\pi}{5} \)`.
Substitute this value into the identity:
`\( \tan^{-1} x + \frac{2\pi}{5} = \frac{\pi}{2} \)`
Now, we solve for `\( \tan^{-1} x \)` by subtracting `\( \frac{2\pi}{5} \)` from `\( \frac{\pi}{2} \)`:
`\( \tan^{-1} x = \frac{\pi}{2} - \frac{2\pi}{5} \)`
To perform the subtraction of fractions, find a common denominator for 2 and 5, which is 10:
`\( \tan^{-1} x = \frac{5\pi}{10} - \frac{4\pi}{10} \)`
`\( \tan^{-1} x = \frac{5\pi - 4\pi}{10} \)`
`\( \tan^{-1} x = \frac{\pi}{10} \)`
Thus, the value of `\( \tan^{-1} x \)` is `\( \frac{\pi}{10} \)`. This problem relies on the direct application of a fundamental identity.
In simple words: We use the math rule that `\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)`. Since we know `\( \cot^{-1} x \)` is `\( \frac{2\pi}{5} \)`, we just subtract that from `\( \frac{\pi}{2} \)` to find `\( \tan^{-1} x \)`. After calculating the fractions, the answer is `\( \frac{\pi}{10} \)`.

🎯 Exam Tip: Familiarize yourself with all the complementary inverse trigonometric identities. They are very useful for quickly simplifying expressions and solving equations where one inverse function value is given, and you need to find its complementary function's value. These include `\( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \)`, `\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)`, and `\( \sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2} \) `.

Question 8. The domain of the function defined by \( f(x) = \sin^{-1} \sqrt{x-1} \) is
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Answer: (a) [1, 2]
For the function `\( f(x) = \sin^{-1} \sqrt{x-1} \)` to be defined, two conditions must be satisfied:
1. The expression inside the square root must be non-negative:
`\( x-1 \ge 0 \)`
`\( x \ge 1 \)`
2. The argument of the `\( \sin^{-1} \)` function, which is `\( \sqrt{x-1} \)`, must lie within its domain `\( [-1, 1] \)`.
So, `\( -1 \le \sqrt{x-1} \le 1 \)`
From condition 1, we already know that `\( \sqrt{x-1} \ge 0 \)`. This means the `\( -1 \le \sqrt{x-1} \)` part of the inequality in condition 2 is automatically satisfied.
Therefore, we only need to consider the upper bound from condition 2:
`\( \sqrt{x-1} \le 1 \)`
Square both sides of the inequality (since both sides are non-negative):
`\( (\sqrt{x-1})^2 \le 1^2 \)`
`\( x-1 \le 1 \)`
Add 1 to both sides:
`\( x \le 2 \)`
Combining both conditions, `\( x \ge 1 \)` and `\( x \le 2 \)`, the domain of the function `f(x)` is the closed interval `\( [1, 2] \)`. This ensures that both the square root and inverse sine are well-defined.
In simple words: For the function `\( f(x) \)` to work, two things must be true. First, the number under the square root (`\( x-1 \)` ) must be 0 or more, meaning `\( x \)` must be `\( 1 \)` or higher. Second, the entire square root part (`\( \sqrt{x-1} \)` ) must be between -1 and 1 for `\( \sin^{-1} \)` to work. Since a square root is never negative, we only need `\( \sqrt{x-1} \le 1 \)`. Squaring both sides gives `\( x-1 \le 1 \)`, which simplifies to `\( x \le 2 \)`. Putting these two conditions together, `x` must be in the range from `1` to `2`.

🎯 Exam Tip: When finding the domain of composite functions, always consider the domain restrictions of all inner and outer functions. For `\( \sqrt{f(x)} \)`, ensure `\( f(x) \ge 0 \)`. For `\( \sin^{-1}(g(x)) \)`, ensure `\( -1 \le g(x) \le 1 \)`. Combine all these conditions systematically to find the overall domain.

Question 9. If \( x = \frac{1}{5} \), the value of \( \cos(\cos^{-1}x + 2\sin^{-1}x) \) is
(a) \( -\sqrt{\frac{24}{25}} \)
(b) \( \sqrt{\frac{24}{25}} \)
(c) \( \frac{1}{5} \)
(d) \( -\frac{1}{5} \)
Answer: (d) \( -\frac{1}{5} \)
We need to find the value of `\( \cos(\cos^{-1}x + 2\sin^{-1}x) \)` when `\( x = \frac{1}{5} \)`.
First, let's simplify the expression inside the cosine function:
`\( \cos^{-1}x + 2\sin^{-1}x \)` can be split as `\( \cos^{-1}x + \sin^{-1}x + \sin^{-1}x \)`
We know the fundamental inverse trigonometric identity:
`\( \cos^{-1}x + \sin^{-1}x = \frac{\pi}{2} \)`
Substitute this identity into our expression:
`\( = \frac{\pi}{2} + \sin^{-1}x \)`
Now, the original expression becomes:
`\( \cos\left(\frac{\pi}{2} + \sin^{-1}x\right) \)`
We use the trigonometric identity `\( \cos(\frac{\pi}{2} + \theta) = -\sin\theta \)`. Here, `\( \theta = \sin^{-1}x \)`.
`\( = -\sin(\sin^{-1}x) \)`
Since `\( \sin(\sin^{-1}x) = x \)`, the expression simplifies further to:
`\( = -x \)`
Given that `\( x = \frac{1}{5} \)`, substitute this value:
`\( = -\frac{1}{5} \)`
This problem demonstrates combining inverse trigonometric identities with general trigonometric identities.
In simple words: We can break `\( 2\sin^{-1}x \)` into `\( \sin^{-1}x + \sin^{-1}x \)`. Then we use the rule `\( \cos^{-1}x + \sin^{-1}x = \frac{\pi}{2} \)`. So, the inside of the cosine becomes `\( \frac{\pi}{2} + \sin^{-1}x \)`. Using another rule, `\( \cos(90^\circ + \text{angle}) = -\sin(\text{angle}) \)`, this becomes `\( -\sin(\sin^{-1}x) \)`. Since `\( \sin(\sin^{-1}x) = x \)`, the whole thing simplifies to `\( -x \)`. Because `\( x = \frac{1}{5} \)`, the answer is `\( -\frac{1}{5} \)`.

🎯 Exam Tip: This question requires combining two types of identities: the inverse trigonometric identity `\( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \)` and the trigonometric identity for complementary angles `\( \cos(\frac{\pi}{2} + \theta) = -\sin\theta \)`. Break down the problem into smaller, manageable steps, applying one identity at a time for accurate simplification.

Question 10. \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \) is equal to
(a) \( \frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right) \)
(b) \( \frac{1}{2}\sin^{-1}\left(\frac{3}{5}\right) \)
(c) \( \frac{1}{2}\tan^{-1}\left(\frac{3}{5}\right) \)
(d) \( \tan^{-1}\left(\frac{1}{2}\right) \)
Answer: (d) \( \tan^{-1}\left(\frac{1}{2}\right) \)
To solve this, we use the sum formula for inverse tangent functions:
`\( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \)`, provided that `\( xy < 1 \)`.
Here, `\( x = \frac{1}{4} \)` and `\( y = \frac{2}{9} \)`.
First, let's check the condition `\( xy < 1 \)`:
`\( xy = \frac{1}{4} \times \frac{2}{9} = \frac{2}{36} = \frac{1}{18} \)`
Since `\( \frac{1}{18} < 1 \)`, we can use the formula.
Now, substitute the values of `x` and `y` into the formula:
`\( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \times \frac{2}{9}}\right) \)`
Calculate the numerator:
`\( \frac{1}{4} + \frac{2}{9} = \frac{1 \times 9 + 2 \times 4}{4 \times 9} = \frac{9 + 8}{36} = \frac{17}{36} \)`
Calculate the denominator:
`\( 1 - \frac{1}{4} \times \frac{2}{9} = 1 - \frac{2}{36} = 1 - \frac{1}{18} = \frac{18 - 1}{18} = \frac{17}{18} \)`
Substitute these calculated values back into the `\( \tan^{-1} \)` expression:
`\( = \tan^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) \)`
To divide fractions, multiply the numerator by the reciprocal of the denominator:
`\( = \tan^{-1}\left(\frac{17}{36} \times \frac{18}{17}\right) \)`
Cancel out the common factor 17 and simplify the remaining fraction:
`\( = \tan^{-1}\left(\frac{18}{36}\right) \)`
`\( = \tan^{-1}\left(\frac{1}{2}\right) \)`
This calculation provides the simplified sum of the two inverse tangent functions.
In simple words: We use a special rule for adding `\( \tan^{-1} \)` values, which is `\( \tan^{-1} \left(\frac{x+y}{1-xy}\right) \)`. We put `\( \frac{1}{4} \)` for `x` and `\( \frac{2}{9} \)` for `y`. After calculating the top and bottom parts of the fraction separately, the result is `\( \frac{17}{36} \)` divided by `\( \frac{17}{18} \)`. When we simplify this division, we get `\( \frac{1}{2} \)`. So, the final answer is `\( \tan^{-1}\left(\frac{1}{2}\right) \)`.

🎯 Exam Tip: Always remember the formula for `\( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \)` when `\( xy < 1 \)`. If `\( xy > 1 \)`, the formula includes an additional `\( \pi \)` term. Precise fraction arithmetic is also essential to avoid errors in intermediate steps and arrive at the correct final value.

Question 11. If the function \( f(x) = \sin^{-1}(x^2 - 3) \), then \( x \) belongs to
(a) [-1, 1]
(b) [\( \sqrt{2} \), 2]
(c) [-2, \( -\sqrt{2} \)]U[ \( \sqrt{2} \), 2]
(d) [-2, \( -\sqrt{2} \)]
Answer: (c) [-2, \( -\sqrt{2} \)]U[ \( \sqrt{2} \), 2]
For the function `\( f(x) = \sin^{-1}(x^2 - 3) \)` to be defined, its argument, `\( x^2 - 3 \)`, must lie within the domain of the `\( \sin^{-1} \)` function. The domain of `\( \sin^{-1} \)` is `\( [-1, 1] \)`.
So, we must set up the inequality:
`\( -1 \le x^2 - 3 \le 1 \)`
To isolate `\( x^2 \)`, add 3 to all parts of the inequality:
`\( -1 + 3 \le x^2 - 3 + 3 \le 1 + 3 \)`
`\( 2 \le x^2 \le 4 \)`
Now, we need to find the values of `x` that satisfy this compound inequality. Taking the square root of all parts, we must consider both positive and negative roots, leading to absolute value:
`\( \sqrt{2} \le |x| \le \sqrt{4} \)`
`\( \sqrt{2} \le |x| \le 2 \)`
This absolute value inequality can be split into two separate conditions for `x`:
1. For positive values of `x`: `\( \sqrt{2} \le x \le 2 \)` (This forms the interval `\( [\sqrt{2}, 2] \)`).
2. For negative values of `x`: `\( \sqrt{2} \le -x \le 2 \)`
Multiplying by -1 reverses the inequality signs: `\( -2 \le x \le -\sqrt{2} \)` (This forms the interval `\( [-2, -\sqrt{2}] \)`).
Combining these two intervals, the domain of the function `f(x)` is `\( [-2, -\sqrt{2}] \cup [\sqrt{2}, 2] \)`. This ensures the argument `\( x^2-3 \)` remains within `[-1, 1]`.
In simple words: For the `\( \sin^{-1} \)` function to work, the value inside it (`\( x^2 - 3 \)` ) must be between -1 and 1. By doing some algebra (adding 3 to all parts), we find that `\( x^2 \)` must be between 2 and 4. This means `x` itself must be between `\( \sqrt{2} \)` and 2, but it also includes the negative values from `\( -2 \)` to `\( -\sqrt{2} \)`. So, `x` can be in either of these two separate ranges.

🎯 Exam Tip: When solving inequalities involving `\( x^2 \)` (e.g., `\( a \le x^2 \le b \)`), remember that `\( \sqrt{a} \le |x| \le \sqrt{b} \)`. This typically results in two separate intervals for `x`: `\( [-\sqrt{b}, -\sqrt{a}] \)` and `\( [\sqrt{a}, \sqrt{b}] \)`. Always include both positive and negative roots and use the union symbol `\( \cup \)` to combine such disjoint intervals.

Question 12. If \( \cot^{-1} 2 \) and \( \cot^{-1} 3 \) are two angles of a triangle, then the third angle is
(a) \( \frac{\pi}{4} \)
(b) \( \frac{3\pi}{4} \)
(c) \( \frac{\pi}{6} \)
(d) \( \frac{\pi}{3} \)
Answer: (b) \( \frac{3\pi}{4} \)
Let the two given angles of the triangle be `\( A = \cot^{-1} 2 \)` and `\( B = \cot^{-1} 3 \)`. Let `C` be the third angle.
We know that the sum of the interior angles of a triangle is `\( \pi \)` radians (or `\( 180^\circ \)`):
`\( A + B + C = \pi \)`
Substitute the given angles into this equation:
`\( \cot^{-1} 2 + \cot^{-1} 3 + C = \pi \)`
Now, we use the sum formula for inverse cotangent functions: `\( \cot^{-1} x + \cot^{-1} y = \cot^{-1}\left(\frac{xy-1}{x+y}\right) \)`
Apply this formula to `\( \cot^{-1} 2 + \cot^{-1} 3 \)`:
`\( \cot^{-1}\left(\frac{2 \times 3 - 1}{2 + 3}\right) + C = \pi \)`
`\( \cot^{-1}\left(\frac{6 - 1}{5}\right) + C = \pi \)`
`\( \cot^{-1}\left(\frac{5}{5}\right) + C = \pi \)`
`\( \cot^{-1}(1) + C = \pi \)`
We know that the principal value of `\( \cot^{-1}(1) \)` is `\( \frac{\pi}{4} \)`.
`\( \frac{\pi}{4} + C = \pi \)`
Now, solve for the third angle `C`:
`\( C = \pi - \frac{\pi}{4} \)`
`\( C = \frac{4\pi - \pi}{4} \)`
`\( C = \frac{3\pi}{4} \)`
Thus, the third angle of the triangle is `\( \frac{3\pi}{4} \)`. This demonstrates how inverse trigonometric functions can represent geometrical angles.
In simple words: The three angles inside any triangle always add up to `\( \pi \)`. We add the two given angles, `\( \cot^{-1} 2 \)` and `\( \cot^{-1} 3 \)`, using a special `\( \cot^{-1} \)` addition rule. This sum simplifies to `\( \cot^{-1}(1) \)`, which is `\( \frac{\pi}{4} \)`. So, `\( \frac{\pi}{4} \)` plus the third angle equals `\( \pi \)`. By subtracting `\( \frac{\pi}{4} \)` from `\( \pi \)`, we find the third angle is `\( \frac{3\pi}{4} \)`.

🎯 Exam Tip: Remember the angle sum property of a triangle (`\( A+B+C = \pi \)`). Be proficient with the addition formulas for inverse trigonometric functions like `\( \cot^{-1} x + \cot^{-1} y = \cot^{-1}\left(\frac{xy-1}{x+y}\right) \)` and `\( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \)`. Always use radians when working with trigonometric formulas unless degrees are explicitly requested.

Question 13. \( \sin^{-1}(\tan\frac{\pi}{4}) - \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{6} \). Then \( x \) is a root of the equation
(a) \( x^2 - x - 6 = 0 \)
(b) \( x^2 - x - 12 = 0 \)
(c) \( x^2 + x - 12 = 0 \)
(d) \( x^2 + x - 6 = 0 \)
Answer: (b) \( x^2 - x - 12 = 0 \)
We start with the given equation: `\( \sin^{-1}(\tan\frac{\pi}{4}) - \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{6} \)`
First, we evaluate `\( \tan\frac{\pi}{4} \)`. We know that `\( \tan\frac{\pi}{4} = 1 \)`. Substitute this into the equation:
`\( \sin^{-1}(1) - \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{6} \)`
We also know that `\( \sin^{-1}(1) = \frac{\pi}{2} \)`. So, the equation becomes:
`\( \frac{\pi}{2} - \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{6} \)`
Now, we isolate the `\( \sin^{-1} \)` term:
`\( \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{2} - \frac{\pi}{6} \)`
Combine the terms on the right side by finding a common denominator (6):
`\( \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{3\pi}{6} - \frac{\pi}{6} \)`
`\( \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{2\pi}{6} \)`
`\( \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{3} \)`
Next, take the sine of both sides to remove the inverse sine function:
`\( \sqrt{\frac{3}{x}} = \sin\left(\frac{\pi}{3}\right) \)`
We know that `\( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)`. So:
`\( \sqrt{\frac{3}{x}} = \frac{\sqrt{3}}{2} \)`
To eliminate the square roots, square both sides of the equation:
`\( \left(\sqrt{\frac{3}{x}}\right)^2 = \left(\frac{\sqrt{3}}{2}\right)^2 \)`
`\( \frac{3}{x} = \frac{3}{4} \)`
Now, cross-multiply to solve for `x`:
`\( 3 \times 4 = 3 \times x \)`
`\( 12 = 3x \)`
`\( x = \frac{12}{3} \)`
`\( x = 4 \)`
Finally, we check which of the given quadratic equations has `\( x = 4 \)` as a root by substituting `\( x = 4 \)` into each option:
(a) `\( x^2 - x - 6 = 0 \implies 4^2 - 4 - 6 = 16 - 4 - 6 = 6 \ne 0 \)`
(b) `\( x^2 - x - 12 = 0 \implies 4^2 - 4 - 12 = 16 - 4 - 12 = 0 \)` (This is the correct option.)
(c) `\( x^2 + x - 12 = 0 \implies 4^2 + 4 - 12 = 16 + 4 - 12 = 8 \ne 0 \)`
(d) `\( x^2 + x - 6 = 0 \implies 4^2 + 4 - 6 = 16 + 4 - 6 = 14 \ne 0 \)`
Therefore, `\( x=4 \)` is a root of the equation `\( x^2 - x - 12 = 0 \)`. This problem requires careful step-by-step simplification.
In simple words: First, we simplify `\( \tan\frac{\pi}{4} \)` to 1 and `\( \sin^{-1}(1) \)` to `\( \frac{\pi}{2} \)`. The equation becomes `\( \frac{\pi}{2} - \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{6} \)`. By moving terms around, we find `\( \sin^{-1}\left(\sqrt{\frac{3}{x}}\right) = \frac{\pi}{3} \)`. Taking sine of both sides gives `\( \sqrt{\frac{3}{x}} = \frac{\sqrt{3}}{2} \)`. Squaring both sides results in `\( \frac{3}{x} = \frac{3}{4} \)`, so `\( x=4 \)`. Then, we check which of the given equations is true when `x=4`, and option (b) is the correct one.

🎯 Exam Tip: When solving equations involving inverse trigonometric functions, always simplify known trigonometric values first. Isolate the inverse trigonometric term, then apply the direct trigonometric function to both sides. Remember to square both sides carefully when dealing with square roots. Finally, verify your solution with the given options to ensure accuracy.

Question 14. \( \sin^{-1}(2\cos^2x - 1) + \cos^{-1}(1 - 2\sin^2x) = \)
(a) \( \frac{\pi}{2} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{4} \)
(d) \( \frac{\pi}{6} \)
Answer: (a) \( \frac{\pi}{2} \)
We need to simplify the expression: `\( \sin^{-1}(2\cos^2x - 1) + \cos^{-1}(1 - 2\sin^2x) \)`
We use the double angle identities for cosine:
1. `\( \cos(2x) = 2\cos^2x - 1 \)`
2. `\( \cos(2x) = 1 - 2\sin^2x \)`
Substitute these trigonometric identities into the inverse trigonometric expression:
`\( = \sin^{-1}(\cos(2x)) + \cos^{-1}(\cos(2x)) \)`
Now, let `\( A = \cos(2x) \)`. The expression becomes:
`\( = \sin^{-1}(A) + \cos^{-1}(A) \)`
We know the fundamental identity for inverse trigonometric functions:
`\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)`
This identity holds true for all `\( A \in [-1, 1] \)`. Since `\( A = \cos(2x) \)` and the range of `\( \cos(2x) \)` is indeed `\( [-1, 1] \)`, the identity can be applied directly.
Therefore, the value of the given expression is `\( \frac{\pi}{2} \)`. This problem elegantly utilizes double-angle formulas to simplify complex inverse trigonometric forms.
In simple words: We use the double angle rules for cosine to change `\( 2\cos^2x - 1 \)` into `\( \cos(2x) \)` and `\( 1 - 2\sin^2x \)` also into `\( \cos(2x) \)`. So, the problem becomes `\( \sin^{-1}(\cos(2x)) + \cos^{-1}(\cos(2x)) \)`. Since `\( \sin^{-1} A + \cos^{-1} A \)` always equals `\( \frac{\pi}{2} \)`, and `\( \cos(2x) \)` acts as `A`, the final answer is `\( \frac{\pi}{2} \)`.

🎯 Exam Tip: It is crucial to remember the double angle formulas for cosine: `\( \cos(2x) = \cos^2x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x \)`. These identities simplify expressions within inverse trigonometric functions, allowing you to use the core identity `\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)`. Always ensure the argument `A` is within its valid domain `[-1, 1]`.

Question 15. If \( \cot^{-1}(\sqrt{\sin\alpha}) + \tan^{-1}(\sqrt{\sin\alpha}) = u \), then \( \cos 2u \) is equal to
(a) \( \tan^2\alpha \)
(b) 0
(c) -1
(d) \( \tan 2\alpha \)
Answer: (c) -1
We are given the equation: `\( \cot^{-1}(\sqrt{\sin\alpha}) + \tan^{-1}(\sqrt{\sin\alpha}) = u \)`
We recall the fundamental identity for inverse trigonometric functions:
`\( \tan^{-1} A + \cot^{-1} A = \frac{\pi}{2} \)`
This identity holds true for any real number `A`. In our case, `\( A = \sqrt{\sin\alpha} \)`. For this term to be defined and real, we must have `\( \sin\alpha \ge 0 \)`. Assuming this condition is met, the sum of `\( \cot^{-1}(\sqrt{\sin\alpha}) \)` and `\( \tan^{-1}(\sqrt{\sin\alpha}) \)` simplifies directly to `\( \frac{\pi}{2} \)`.
Therefore, we have:
`\( u = \frac{\pi}{2} \)`
Now, the question asks for the value of `\( \cos 2u \)`. Substitute the value of `u`:
`\( \cos 2u = \cos\left(2 \times \frac{\pi}{2}\right) \)`
`\( = \cos(\pi) \)`
The value of `\( \cos(\pi) \)` (cosine of 180 degrees) is `\( -1 \)`.
So, `\( \cos 2u = -1 \)`. This problem highlights the direct application of a fundamental inverse trigonometric identity and basic cosine values.
In simple words: The sum of `\( \cot^{-1} \)` and `\( \tan^{-1} \)` of the same number (which is `\( \sqrt{\sin\alpha} \)` here) is always `\( \frac{\pi}{2} \)`. So, the value of `u` is `\( \frac{\pi}{2} \)`. We need to find `\( \cos 2u \)`, which means `\( \cos(2 \times \frac{\pi}{2}) \)`. This simplifies to `\( \cos(\pi) \)`, and the value of `\( \cos(\pi) \)` is -1.

🎯 Exam Tip: Immediately recognize and apply the fundamental inverse trigonometric identities like `\( \tan^{-1} A + \cot^{-1} A = \frac{\pi}{2} \)`. This allows for quick calculation of `u` and subsequent evaluation of `\( \cos 2u \)`. Always ensure that the argument `A` is valid for both functions involved (e.g., `\( A \ge 0 \)` for `\( \sqrt{A} \)` and `\( A \in \mathbb{R} \)` for `\( \tan^{-1} A \) ` and `\( \cot^{-1} A \) `).

Question 16. If \( |x| \le 1 \), then \( 2\tan^{-1} x - \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) is equal to
(a) \( \tan^{-1}x \)
(b) \( \sin^{-1}x \)
(c) 0
(d) \( \pi \)
Answer: (c) 0
We need to simplify the expression: `\( 2\tan^{-1} x - \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)`
A key identity related to `\( 2\tan^{-1} x \)` is:
`\( 2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)`
This identity is valid for `\( |x| \le 1 \)`, which is precisely the condition given in the problem statement.
Substitute this identity directly into the given expression:
`\( = \sin^{-1}\left(\frac{2x}{1+x^2}\right) - \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)`
Since we are subtracting an expression from itself, the result is:
`\( = 0 \)`
This problem simplifies directly due to the application of a fundamental inverse trigonometric identity.
In simple words: We know a special math rule that `\( 2\tan^{-1} x \)` is the same as `\( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)` when `\( x \)` is between -1 and 1. Since the problem tells us this condition for `x`, we can replace the first part of the expression with the second part. Then, we are subtracting the same thing from itself, so the answer is 0.

🎯 Exam Tip: It is vital to remember the various equivalent forms of `\( 2\tan^{-1} x \)` in terms of `\( \sin^{-1} \)`, `\( \cos^{-1} \)`, and `\( \tan^{-1} \)` functions. The choice of which identity to use depends on the problem's specific context and the given range of `x`. For `\( |x| \le 1 \)`, the `\( \sin^{-1} \)` form is appropriate here.

Question 17. The equation \( \tan^{-1} x - \cot^{-1} x = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \) has
(a) no solution
(b) unique solution
(c) two solutions
(d) infinite number of solutions
Answer: (b) unique solution
We are given the equation:
`\( \tan^{-1} x - \cot^{-1} x = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \)` (Equation 1)
First, evaluate the constant term on the right side:
`\( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \)`
So, Equation 1 becomes:
`\( \tan^{-1} x - \cot^{-1} x = \frac{\pi}{6} \)`
We also know a fundamental inverse trigonometric identity:
`\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)` (Equation 2)
Now we have a system of two linear equations in terms of `\( \tan^{-1} x \)` and `\( \cot^{-1} x \)`:
1. `\( \tan^{-1} x - \cot^{-1} x = \frac{\pi}{6} \)`
2. `\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)`
To solve for `\( \tan^{-1} x \)`, add Equation 1 and Equation 2:
`\( (\tan^{-1} x - \cot^{-1} x) + (\tan^{-1} x + \cot^{-1} x) = \frac{\pi}{6} + \frac{\pi}{2} \)`
`\( 2\tan^{-1} x = \frac{\pi}{6} + \frac{3\pi}{6} \)`
`\( 2\tan^{-1} x = \frac{4\pi}{6} \)`
`\( 2\tan^{-1} x = \frac{2\pi}{3} \)`
Divide both sides by 2:
`\( \tan^{-1} x = \frac{\pi}{3} \)`
Finally, take the tangent of both sides to find `x`:
`\( x = \tan\left(\frac{\pi}{3}\right) \)`
We know that `\( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \)`.
So, `\( x = \sqrt{3} \)`.
Since we found a single, specific value for `x`, the equation has a unique solution. This problem demonstrates the power of using simultaneous equations with inverse trigonometric identities.
In simple words: We use the given equation and another known math rule `\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)`. We have two equations. Adding them together cancels out `\( \cot^{-1} x \)`, leaving `\( 2\tan^{-1} x = \frac{\pi}{6} + \frac{\pi}{2} \)`. This simplifies to `\( \tan^{-1} x = \frac{\pi}{3} \)`. Taking the tangent of both sides gives `\( x = \tan(\frac{\pi}{3}) \)`, which is `\( \sqrt{3} \)`. Since there's only one value for `x`, it's a unique solution.

🎯 Exam Tip: When an equation involves both `\( \tan^{-1} x \)` and `\( \cot^{-1} x \)`, it's usually best to use the identity `\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)`. Set up a system of equations, and solve for `\( \tan^{-1} x \)` (or `\( \cot^{-1} x \)`), then find `x`. This method consistently leads to the correct number of solutions.

Question 18. If \( \sin^{-1} x + \cot^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} \), then \( x \) is equal to
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{\sqrt{5}} \)
(c) \( \frac{2}{\sqrt{5}} \)
(d) \( \frac{\sqrt{3}}{2} \)
Answer: (b) \( \frac{1}{\sqrt{5}} \)
We are given the equation: `\( \sin^{-1} x + \cot^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} \)`
We know the fundamental identity for inverse trigonometric functions:
`\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)`
Comparing the given equation with this identity, for the equation to hold, `\( \cot^{-1}\left(\frac{1}{2}\right) \)` must be equal to `\( \cos^{-1} x \)`.
So, we have: `\( \cos^{-1} x = \cot^{-1}\left(\frac{1}{2}\right) \)`
Now, we need to convert `\( \cot^{-1}\left(\frac{1}{2}\right) \)` into an equivalent `\( \cos^{-1} \)` expression. Let `\( \theta = \cot^{-1}\left(\frac{1}{2}\right) \)`.
This implies `\( \cot \theta = \frac{1}{2} \)`.
In a right-angled triangle, `\( \cot \theta = \frac{\text{adjacent side}}{\text{opposite side}} \)`. So, we can consider `adjacent = 1` and `opposite = 2`.
Using the Pythagorean theorem, the hypotenuse `h` is:
`\( h = \sqrt{\text{adjacent}^2 + \text{opposite}^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \)`
Now, we can find `\( \cos \theta \)` from this triangle:
`\( \cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}} \)`
Therefore, `\( \theta = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \)`.
Substituting this back into `\( \cos^{-1} x = \cot^{-1}\left(\frac{1}{2}\right) \)`:
`\( \cos^{-1} x = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right) \)`
This directly implies that `\( x = \frac{1}{\sqrt{5}} \)`. The conversion between different inverse trigonometric functions is a valuable skill.
In simple words: We are given `\( \sin^{-1} x + \cot^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} \)`. Since we know `\( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \)`, it means `\( \cot^{-1}\left(\frac{1}{2}\right) \)` must be equal to `\( \cos^{-1} x \)`. We draw a right triangle where `\( \cot \theta = \frac{1}{2} \)`. This means the adjacent side is 1, the opposite side is 2, and the longest side (hypotenuse) is `\( \sqrt{5} \)`. From this, `\( \cos \theta = \frac{1}{\sqrt{5}} \)`. So, `\( x = \frac{1}{\sqrt{5}} \)`.

🎯 Exam Tip: When an equation involves different inverse trigonometric functions, try to convert them to a common type (e.g., all to `\( \sin^{-1} \)` or `\( \cos^{-1} \)`). This can be done using right-angled triangles to find equivalent expressions. The identity `\( \sin^{-1} A + \cos^{-1} A = \frac{\pi}{2} \)` is very useful for comparing terms and simplifying equations.

Question 19. If \( \sin^{-1}\left(\frac{x}{5}\right) + \operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2} \), then the value of \( x \) is
(a) 4
(b) 5
(c) 2
(d) 3
Answer: (d) 3
We are given the equation: `\( \sin^{-1}\left(\frac{x}{5}\right) + \operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2} \)`
First, let's use the reciprocal identity `\( \operatorname{cosec}^{-1} A = \sin^{-1}\left(\frac{1}{A}\right) \)`.
So, `\( \operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \sin^{-1}\left(\frac{4}{5}\right) \)`
Substitute this back into the original equation:
`\( \sin^{-1}\left(\frac{x}{5}\right) + \sin^{-1}\left(\frac{4}{5}\right) = \frac{\pi}{2} \)`
Rearrange the equation to isolate one `\( \sin^{-1} \)` term:
`\( \sin^{-1}\left(\frac{x}{5}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{4}{5}\right) \)`
Now, use the complementary identity `\( \frac{\pi}{2} - \sin^{-1} A = \cos^{-1} A \)`.
So, `\( \sin^{-1}\left(\frac{x}{5}\right) = \cos^{-1}\left(\frac{4}{5}\right) \)`
To compare the arguments, we need to express `\( \cos^{-1}\left(\frac{4}{5}\right) \)` as a `\( \sin^{-1} \)` function. Let `\( \phi = \cos^{-1}\left(\frac{4}{5}\right) \)`. This means `\( \cos \phi = \frac{4}{5} \)`.
Consider a right-angled triangle where `adjacent = 4` and `hypotenuse = 5`.
Using the Pythagorean theorem, the opposite side is `\( \sqrt{\text{hypotenuse}^2 - \text{adjacent}^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \)`.
From this triangle, `\( \sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5} \)`.
Therefore, `\( \cos^{-1}\left(\frac{4}{5}\right) = \sin^{-1}\left(\frac{3}{5}\right) \)`.
Substitute this back into our equation:
`\( \sin^{-1}\left(\frac{x}{5}\right) = \sin^{-1}\left(\frac{3}{5}\right) \)`
This implies that the arguments must be equal:
`\( \frac{x}{5} = \frac{3}{5} \)`
Multiplying both sides by 5 gives:
`\( x = 3 \)`
This problem demonstrates a complete sequence of inverse trigonometric transformations to find the variable `x`.
In simple words: We start with the equation `\( \sin^{-1}\left(\frac{x}{5}\right) + \operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2} \)`. We change `\( \operatorname{cosec}^{-1}\left(\frac{5}{4}\right) \)` to `\( \sin^{-1}\left(\frac{4}{5}\right) \)`. Then, using `\( \frac{\pi}{2} - \sin^{-1} A = \cos^{-1} A \)`, we get `\( \sin^{-1}\left(\frac{x}{5}\right) = \cos^{-1}\left(\frac{4}{5}\right) \)`. Next, we convert `\( \cos^{-1}\left(\frac{4}{5}\right) \)` into `\( \sin^{-1}\left(\frac{3}{5}\right) \)` by drawing a right triangle. This means `\( \sin^{-1}\left(\frac{x}{5}\right) = \sin^{-1}\left(\frac{3}{5}\right) \)`, so `\( \frac{x}{5} = \frac{3}{5} \)`, which simplifies to `\( x = 3 \)`.

🎯 Exam Tip: When solving equations with multiple inverse trigonometric functions, a common strategy is to convert all functions to a single type (e.g., all to `\( \sin^{-1} \)` or `\( \tan^{-1} \)`). Use reciprocal identities (`\( \operatorname{cosec}^{-1}x = \sin^{-1}(1/x) \)`), complementary identities (`\( \frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x \)`), and right-triangle constructions for inter-conversion. Work step-by-step to simplify the equation effectively.

Question 20. \( \sin(\tan^{-1} x) \), `\( |x| < 1 \)` is equal to
(a) \( \frac{x}{\sqrt{1-x^2}} \)
(b) \( \frac{1}{\sqrt{1-x^2}} \)
(c) \( \frac{1}{\sqrt{1+x^2}} \)
(d) \( \frac{x}{\sqrt{1+x^2}} \)
Answer: (d) \( \frac{x}{\sqrt{1+x^2}} \)
We need to find the value of `\( \sin(\tan^{-1} x) \)`.
Let `\( \theta = \tan^{-1} x \)`.
This definition implies that `\( \tan \theta = x \)`. We can write `\( x \)` as `\( \frac{x}{1} \)`.
We can visualize this relationship by constructing a right-angled triangle where `\( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)`.
So, we can label the side opposite to `\( \theta \)` as `x` and the adjacent side as `1`.
Using the Pythagorean theorem, we can find the hypotenuse `h`:
`\( h = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} \)`
`\( h = \sqrt{x^2 + 1^2} \)`
`\( h = \sqrt{1+x^2} \)`
Now, we need to find `\( \sin \theta \)` from this triangle:
`\( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)`
`\( \sin \theta = \frac{x}{\sqrt{1+x^2}} \)`
Since `\( \theta = \tan^{-1} x \)`, we substitute it back to get the final expression:
`\( \sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}} \)`
The condition `\( |x| < 1 \)` ensures that `\( \tan^{-1} x \)` gives an angle in `\( (-\frac{\pi}{2}, \frac{\pi}{2}) \)`, where sine has the same sign as `x`. This construction correctly reflects that behavior.
In simple words: To find `\( \sin(\tan^{-1} x) \)`, we can think of `\( \tan^{-1} x \)` as an angle, let's call it `\( \theta \)`. This means `\( \tan \theta = x \)`. We draw a right triangle where the side opposite `\( \theta \)` is `x` and the side next to it (adjacent) is 1. Using Pythagoras' theorem, the longest side (hypotenuse) is `\( \sqrt{1+x^2} \)`. Now, `\( \sin \theta \)` is the opposite side divided by the hypotenuse, which means `\( \frac{x}{\sqrt{1+x^2}} \)`.

🎯 Exam Tip: For expressions of the form `\( \text{TrigonometricFunction}(\text{InverseTrigonometricFunction } x) \)`, drawing a right-angled triangle is the most reliable method. Let the inverse function (e.g., `\( \tan^{-1} x \)` ) be `\( \theta \)`. Then, `\( \tan \theta = x \)` (or `\( \sin \theta = x \)`). Fill in the sides of the triangle using `x` and 1, use the Pythagorean theorem to find the third side, and then read off the value of the outer trigonometric function.