GSEB Class 12 Maths Solutions Chapter 2 ત્રિકોણમિતીય પ્રતિવિધેયો Exercise 2.2

GSEB Solutions Class 12 Maths Chapter 2 त्रिमितीय प्रतिવિધેયો Ex 2.2

સાબિત કરો :

Question 1. 3 sin⁻¹x = sin⁻¹(3x – 4x³), x ∈ \left[-\frac{1}{2}, \frac{1}{2}\right]
Answer:Let \( \sin^{-1}x = \theta \).
\( \implies x = \sin\theta \)
We are given that \( x \) is between \( -\frac{1}{2} \) and \( \frac{1}{2} \).
So, \( \sin^{-1}\left(-\frac{1}{2}\right) \le \sin^{-1}x \le \sin^{-1}\left(\frac{1}{2}\right) \)
\( \implies -\frac{\pi}{6} \le \theta \le \frac{\pi}{6} \)
Multiply everything by 3: \( -\frac{3\pi}{6} \le 3\theta \le \frac{3\pi}{6} \)
\( \implies -\frac{\pi}{2} \le 3\theta \le \frac{\pi}{2} \)
Now, let's look at the Right Hand Side (RHS):
RHS = \( \sin^{-1}(3x - 4x^3) \)
Substitute \( x = \sin\theta \):
RHS = \( \sin^{-1}(3\sin\theta - 4\sin^3\theta) \)
We know the identity: \( 3\sin\theta - 4\sin^3\theta = \sin(3\theta) \)
So, RHS = \( \sin^{-1}(\sin(3\theta)) \)
Since \( 3\theta \) is in the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we can write:
RHS = \( 3\theta \)
Substitute back \( \theta = \sin^{-1}x \):
RHS = \( 3\sin^{-1}x \)
This is the same as the Left Hand Side (LHS).
Thus, the equation is proven.
In simple words: We assume a variable for \( \sin^{-1}x \), then change \( x \) to \( \sin\theta \). We use a special math rule (identity) to simplify one side of the equation. We also check the range of the angles to make sure our steps are correct. Finally, both sides of the equation become the same.

🎯 Exam Tip: Remember to state the range of \( x \) and \( \theta \) clearly, as it is crucial for using the inverse trigonometric identity \( \sin^{-1}(\sin y) = y \). Showing the range check adds completeness and accuracy to your solution.

Question 2. 3 cos⁻¹x = cos⁻¹ (4x³ – 3x), x ∈ [\frac{1}{2}, 1]
Answer:Let \( \cos^{-1}x = \theta \).
\( \implies x = \cos\theta \)
We know that \( x \) is between \( \frac{1}{2} \) and \( 1 \).
Since \( \cos^{-1}x \) is a function that goes down (decreasing), we have:
\( \cos^{-1}(1) \le \cos^{-1}x \le \cos^{-1}\left(\frac{1}{2}\right) \)
\( \implies 0 \le \theta \le \frac{\pi}{3} \)
Multiply by 3: \( 0 \le 3\theta \le \pi \)
Now, let's look at the Right Hand Side (RHS):
RHS = \( \cos^{-1}(4x^3 - 3x) \)
Substitute \( x = \cos\theta \):
RHS = \( \cos^{-1}(4\cos^3\theta - 3\cos\theta) \)
We know the math rule: \( 4\cos^3\theta - 3\cos\theta = \cos(3\theta) \)
So, RHS = \( \cos^{-1}(\cos(3\theta)) \)
Since \( 3\theta \) is in the range \( [0, \pi] \), we can write:
RHS = \( 3\theta \)
Substitute back \( \theta = \cos^{-1}x \):
RHS = \( 3\cos^{-1}x \)
This is the same as the Left Hand Side (LHS).
Thus, the equation is proven.
In simple words: We set \( \cos^{-1}x \) to an angle, then change \( x \) to \( \cos\theta \). We use a special identity to simplify one side. We also check the range of the angle carefully because the inverse cosine function behaves differently. This helps us ensure the steps are valid, and both sides of the equation match.

🎯 Exam Tip: Remember that \( \cos^{-1}x \) is a decreasing function. This means the inequality signs flip when taking the inverse cosine of an inequality. Also, ensure the final angle \( 3\theta \) falls within the principal value branch of \( \cos^{-1}x \), which is \( [0, \pi] \).

Question 3. tan⁻¹ \frac{2}{11} + tan⁻¹ \frac{7}{24} = tan⁻¹ \frac{1}{2}
Answer:Let's start with the Left Hand Side (LHS):
LHS = \( \tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} \)
We use the formula for adding two inverse tangents:
\( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \), assuming \( xy < 1 \).
Here, \( x = \frac{2}{11} \) and \( y = \frac{7}{24} \).
First, check if \( xy < 1 \): \( \frac{2}{11} \times \frac{7}{24} = \frac{14}{264} \), which is indeed less than 1.
Now, apply the formula:
LHS = \( \tan^{-1}\left(\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}\right) \)
Calculate the numerator:
\( \frac{2}{11} + \frac{7}{24} = \frac{(2 \times 24) + (7 \times 11)}{11 \times 24} = \frac{48 + 77}{264} = \frac{125}{264} \)
Calculate the denominator:
\( 1 - \frac{14}{264} = \frac{264 - 14}{264} = \frac{250}{264} \)
So, LHS = \( \tan^{-1}\left(\frac{125/264}{250/264}\right) \)
LHS = \( \tan^{-1}\left(\frac{125}{250}\right) \)
Simplify the fraction:
LHS = \( \tan^{-1}\left(\frac{1}{2}\right) \)
This matches the Right Hand Side (RHS).
Thus, the equation is proven.
In simple words: We started with the left side of the equation. We used a special formula to add two inverse tangent values. After doing the calculations and simplifying the fraction inside the inverse tangent, we found that the left side became exactly equal to the right side of the equation.

🎯 Exam Tip: Always verify the condition \( xy < 1 \) before applying the sum formula for \( \tan^{-1}x + \tan^{-1}y \). This ensures the correct formula is used and avoids potential errors related to the principal value branch.

નીચેનાં વિધેયોને સાદા સ્વરૂપમાં લખો 1

Question 4. 2tan⁻¹\frac{1}{2} + tan⁻¹ \frac{1}{7} = tan⁻¹\frac{31}{17}
Answer:Let's start with the Left Hand Side (LHS):
LHS = \( 2\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{7} \)
First, we use the formula for \( 2\tan^{-1}x \):
\( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \)
For \( 2\tan^{-1}\frac{1}{2} \), substitute \( x = \frac{1}{2} \):
\( 2\tan^{-1}\frac{1}{2} = \tan^{-1}\left(\frac{2 \times \frac{1}{2}}{1 - (\frac{1}{2})^2}\right) \)
\( = \tan^{-1}\left(\frac{1}{1 - \frac{1}{4}}\right) \)
\( = \tan^{-1}\left(\frac{1}{\frac{3}{4}}\right) \)
\( = \tan^{-1}\left(\frac{4}{3}\right) \)
Now, the LHS becomes:
LHS = \( \tan^{-1}\frac{4}{3} + \tan^{-1}\frac{1}{7} \)
Next, we use the formula for adding two inverse tangents:
\( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \), where \( x = \frac{4}{3} \) and \( y = \frac{1}{7} \).
Check if \( xy < 1 \): \( \frac{4}{3} \times \frac{1}{7} = \frac{4}{21} \), which is less than 1.
Apply the formula:
LHS = \( \tan^{-1}\left(\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}}\right) \)
Calculate the numerator:
\( \frac{4}{3} + \frac{1}{7} = \frac{(4 \times 7) + (1 \times 3)}{3 \times 7} = \frac{28 + 3}{21} = \frac{31}{21} \)
Calculate the denominator:
\( 1 - \frac{4}{21} = \frac{21 - 4}{21} = \frac{17}{21} \)
So, LHS = \( \tan^{-1}\left(\frac{31/21}{17/21}\right) \)
LHS = \( \tan^{-1}\left(\frac{31}{17}\right) \)
This matches the Right Hand Side (RHS).
Thus, the equation is proven.
In simple words: We first changed the \( 2\tan^{-1}(\frac{1}{2}) \) part into a single \( \tan^{-1} \) using a specific rule. Then, we added this new \( \tan^{-1} \) value with the other \( \tan^{-1}(\frac{1}{7}) \) using another rule for adding inverse tangents. After doing all the math steps, the final answer matched the right side of the problem.

🎯 Exam Tip: When simplifying expressions like \( 2\tan^{-1}x + \tan^{-1}y \), always simplify the \( 2\tan^{-1}x \) term first into a single \( \tan^{-1} \) term before combining it with other inverse tangents. This prevents errors and follows standard simplification procedures.

Question 5. tan⁻¹\left(\frac{\sqrt{1+x^2}-1}{x}\right), x ≠ 0.
Answer:Let's make a substitution to simplify the expression.
Let \( x = \tan\theta \).
This means \( \theta = \tan^{-1}x \). Since \( x \ne 0 \), it also means \( \theta \ne 0 \).
Substitute \( x = \tan\theta \) into the given expression:
\( \tan^{-1}\left(\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right) \)
We know that \( 1+\tan^2\theta = \sec^2\theta \). So, the expression becomes:
\( = \tan^{-1}\left(\frac{\sqrt{\sec^2\theta}-1}{\tan\theta}\right) \)
\( = \tan^{-1}\left(\frac{|\sec\theta|-1}{\tan\theta}\right) \)
Since \( x = \tan\theta \) and we are likely in the principal value range for \( \tan^{-1} \), \( \sec\theta \) is positive, so \( |\sec\theta| = \sec\theta \).
\( = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right) \)
Now, change \( \sec\theta \) to \( \frac{1}{\cos\theta} \) and \( \tan\theta \) to \( \frac{\sin\theta}{\cos\theta} \):
\( = \tan^{-1}\left(\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\right) \)
Simplify the fraction:
\( = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right) \)
Next, use the half-angle formulas: \( 1-\cos\theta = 2\sin^2\frac{\theta}{2} \) and \( \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \):
\( = \tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) \)
Cancel out common terms:
\( = \tan^{-1}\left(\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\right) \)
\( = \tan^{-1}\left(\tan\frac{\theta}{2}\right) \)
Since \( x = \tan\theta \) and \( x \ne 0 \), \( \theta \in (-\frac{\pi}{2}, \frac{\pi}{2}) \) and \( \theta \ne 0 \).
This means \( \frac{\theta}{2} \in (-\frac{\pi}{4}, \frac{\pi}{4}) \) and \( \frac{\theta}{2} \ne 0 \). This range is valid for \( \tan^{-1}(\tan y) = y \).
So, the expression simplifies to:
\( = \frac{\theta}{2} \)
Substitute back \( \theta = \tan^{-1}x \):
\( = \frac{1}{2}\tan^{-1}x \)
In simple words: To make this problem easier, we pretended that \( x \) was equal to \( \tan\theta \). Then we put \( \tan\theta \) everywhere we saw \( x \) in the math problem. After using some basic math rules and special angle formulas, we were able to simplify the whole expression down to a much smaller form, which is half of \( \tan^{-1}x \).

🎯 Exam Tip: When you see expressions involving \( \sqrt{1+x^2} \), a common and effective substitution is \( x = \tan\theta \) or \( x = \cot\theta \). This helps eliminate the square root and simplifies the expression using trigonometric identities. Always remember to state the substitution and its implications for the range of the angle.

Question 6. tan⁻¹\left(\frac{1}{\sqrt{x^2-1}}\right), |x| > 1
Answer:We are given that \( |x| > 1 \), which means \( x > 1 \) or \( x < -1 \).
Let's substitute \( x = \sec\theta \).
This means \( \theta = \sec^{-1}x \). The range for \( \theta \) is \( [0, \pi] \), but \( \theta \ne \frac{\pi}{2} \).
Since \( |x| > 1 \), for \( x = \sec\theta \), \( \theta \) will be in \( (0, \frac{\pi}{2}) \) if \( x > 1 \), or \( (\frac{\pi}{2}, \pi) \) if \( x < -1 \).
Substitute \( x = \sec\theta \) into the expression:
\( \tan^{-1}\left(\frac{1}{\sqrt{\sec^2\theta-1}}\right) \)
We know the identity \( \sec^2\theta - 1 = \tan^2\theta \). So, the expression becomes:
\( = \tan^{-1}\left(\frac{1}{\sqrt{\tan^2\theta}}\right) \)
\( = \tan^{-1}\left(\frac{1}{|\tan\theta|}\right) \)
For the inverse tangent function, we usually consider angles in \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Given \( |x|>1 \), if we assume \( x>1 \), then \( \theta \in (0, \frac{\pi}{2}) \). In this range, \( \tan\theta \) is positive, so \( |\tan\theta| = \tan\theta \).
\( = \tan^{-1}\left(\frac{1}{\tan\theta}\right) \)
\( = \tan^{-1}(\cot\theta) \)
We know that \( \cot\theta = \tan\left(\frac{\pi}{2}-\theta\right) \).
\( = \tan^{-1}\left(\tan\left(\frac{\pi}{2}-\theta\right)\right) \)
Since \( \theta \in (0, \frac{\pi}{2}) \), then \( -\theta \in (-\frac{\pi}{2}, 0) \).
So, \( \frac{\pi}{2}-\theta \in (0, \frac{\pi}{2}) \). This angle is within the valid range for \( \tan^{-1}(\tan y) = y \).
Therefore, the expression simplifies to:
\( = \frac{\pi}{2}-\theta \)
Substitute back \( \theta = \sec^{-1}x \):
\( = \frac{\pi}{2}-\sec^{-1}x \)
In simple words: We changed \( x \) to \( \sec\theta \) to help simplify the square root part of the problem. Using a known math rule, \( \sec^2\theta - 1 = \tan^2\theta \), we got rid of the square root. Then, we changed \( \cot\theta \) into \( \tan(\frac{\pi}{2}-\theta) \) and simplified the expression to its final, simpler form. We also made sure the angles were in the correct range for the inverse functions to work properly.

🎯 Exam Tip: For expressions with \( \sqrt{x^2-1} \), choose a substitution like \( x=\sec\theta \) or \( x=\csc\theta \). Pay careful attention to the domain \( |x|>1 \) and the principal value branches of the inverse trigonometric functions to correctly resolve \( \sqrt{\tan^2\theta} \) to \( |\tan\theta| \) and the final \( \tan^{-1}(\tan y) \) simplification.

Question 7. tan⁻¹\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0 < x < π
Answer:Let's use some trigonometric half-angle formulas to simplify the expression inside the square root.
We know that:
\( 1-\cos x = 2\sin^2\frac{x}{2} \)
\( 1+\cos x = 2\cos^2\frac{x}{2} \)
Substitute these into the expression:
\( \tan^{-1}\left(\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}\right) \)
Cancel out the 2s:
\( = \tan^{-1}\left(\sqrt{\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}\right) \)
\( = \tan^{-1}\left(\sqrt{\tan^2\frac{x}{2}}\right) \)
This simplifies to:
\( = \tan^{-1}\left(|\tan\frac{x}{2}|\right) \)
We are given that \( 0 < x < \pi \).
This means that \( 0 < \frac{x}{2} < \frac{\pi}{2} \).
In this range (first quadrant), \( \tan\frac{x}{2} \) is always positive.
So, \( |\tan\frac{x}{2}| = \tan\frac{x}{2} \).
The expression becomes:
\( = \tan^{-1}\left(\tan\frac{x}{2}\right) \)
Since \( \frac{x}{2} \) is in the interval \( (0, \frac{\pi}{2}) \), which is part of the principal value range for \( \tan^{-1} \) (i.e., \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)), we can simply write:
\( = \frac{x}{2} \)
In simple words: We used special rules for cosine to change the terms inside the square root. This helped us simplify the fraction to \( \tan^2(\frac{x}{2}) \). Since the angle \( x \) is between 0 and \( \pi \), the half-angle \( \frac{x}{2} \) is between 0 and \( \frac{\pi}{2} \), making \( \tan(\frac{x}{2}) \) a positive value. This allowed us to remove the square root and finally get the simple answer \( \frac{x}{2} \).

🎯 Exam Tip: Always look for half-angle identities when you see expressions like \( 1-\cos x \) or \( 1+\cos x \). Remember to simplify \( \sqrt{\text{function}^2} \) as \( |\text{function}| \) and then determine the sign based on the given domain of \( x \).

Question 8. tan⁻¹\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right), 0 < x < π
Answer:To simplify this expression, we will divide the numerator and the denominator by \( \cos x \).
This is allowed because \( \cos x \) is not zero for all values in \( 0 < x < \pi \), except at \( x=\frac{\pi}{2} \).
\( \tan^{-1}\left(\frac{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}\right) \)
\( = \tan^{-1}\left(\frac{1-\tan x}{1+\tan x}\right) \)
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \). Let's use this:
\( = \tan^{-1}\left(\frac{\tan\left(\frac{\pi}{4}\right)-\tan x}{1+\tan\left(\frac{\pi}{4}\right)\tan x}\right) \)
This form matches the trigonometric identity for \( \tan(A-B) \), which is \( \frac{\tan A-\tan B}{1+\tan A \tan B} \).
Here, \( A = \frac{\pi}{4} \) and \( B = x \).
So, the expression becomes:
\( = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right) \)
We are given that \( 0 < x < \pi \).
Let's find the range for \( \frac{\pi}{4}-x \):
From \( 0 < x < \pi \), we get \( -\pi < -x < 0 \).
Adding \( \frac{\pi}{4} \) to all parts:
\( \frac{\pi}{4}-\pi < \frac{\pi}{4}-x < \frac{\pi}{4}-0 \)
\( -\frac{3\pi}{4} < \frac{\pi}{4}-x < \frac{\pi}{4} \)
For \( \tan^{-1}(\tan y) = y \) to hold simply, \( y \) must be in the principal value range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
In this case, the simplest form is:
\( = \frac{\pi}{4}-x \)
In simple words: We took the top and bottom parts of the fraction and divided both by \( \cos x \) to make it simpler. Then, we used a special math rule that helps combine tangent terms, turning the whole expression into \( \tan(\frac{\pi}{4}-x) \). Because of how inverse tangent works, the final simple answer is just \( \frac{\pi}{4}-x \), keeping in mind the angle has to be in a certain range.

🎯 Exam Tip: When you see expressions involving \( (\cos x - \sin x) / (\cos x + \sin x) \), immediately think of dividing by \( \cos x \) to get terms in \( \tan x \). This allows the use of the \( \tan(A \pm B) \) identities. Always verify the range of the resulting angle for \( \tan^{-1}(\tan y) = y \) to simplify correctly.

Question 9. tan⁻¹\frac{x}{\sqrt{a^2-x^2}}, |x| < a
Answer:We are given that \( |x| < a \), which means \( -a < x < a \).
Let's make a substitution: \( x = a\sin\theta \).
From this, we can say \( \sin\theta = \frac{x}{a} \).
Therefore, \( \theta = \sin^{-1}\left(\frac{x}{a}\right) \).
Since \( -a < x < a \), it means \( -1 < \frac{x}{a} < 1 \).
For \( \sin^{-1}\left(\frac{x}{a}\right) \), the angle \( \theta \) will be in the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
Now, substitute \( x = a\sin\theta \) into the given expression:
\( \tan^{-1}\left(\frac{a\sin\theta}{\sqrt{a^2-(a\sin\theta)^2}}\right) \)
\( = \tan^{-1}\left(\frac{a\sin\theta}{\sqrt{a^2-a^2\sin^2\theta}}\right) \)
Factor out \( a^2 \) from under the square root:
\( = \tan^{-1}\left(\frac{a\sin\theta}{\sqrt{a^2(1-\sin^2\theta)}}\right) \)
Use the identity \( 1-\sin^2\theta = \cos^2\theta \):
\( = \tan^{-1}\left(\frac{a\sin\theta}{\sqrt{a^2\cos^2\theta}}\right) \)
Take the square root: \( \sqrt{a^2\cos^2\theta} = |a\cos\theta| \). Since \( a > 0 \), this is \( a|\cos\theta| \).
So, \( = \tan^{-1}\left(\frac{a\sin\theta}{a|\cos\theta|}\right) \)
Cancel out \( a \):
\( = \tan^{-1}\left(\frac{\sin\theta}{|\cos\theta|}\right) \)
Since \( \theta \) is in \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), \( \cos\theta \) is always positive. So, \( |\cos\theta| = \cos\theta \).
\( = \tan^{-1}\left(\frac{\sin\theta}{\cos\theta}\right) \)
\( = \tan^{-1}(\tan\theta) \)
Because \( \theta \) is in \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), which is the valid range for \( \tan^{-1}(\tan y) = y \), the expression simplifies to:
\( = \theta \)
Substitute back \( \theta = \sin^{-1}\left(\frac{x}{a}\right) \):
\( = \sin^{-1}\left(\frac{x}{a}\right) \)
In simple words: We used \( x = a\sin\theta \) to make the expression simpler. After replacing \( x \) with \( a\sin\theta \) and using basic rules of trigonometry like \( 1-\sin^2\theta = \cos^2\theta \), the complex fraction became \( \tan\theta \). Since the angle \( \theta \) was in the correct range, the inverse tangent cancelled out the tangent, leaving us with just \( \theta \). Finally, we put back the original expression for \( \theta \) to get the simple answer.

🎯 Exam Tip: When you encounter expressions with \( \sqrt{a^2-x^2} \), a substitution of \( x=a\sin\theta \) or \( x=a\cos\theta \) is generally most effective. Always remember to check the range of \( \theta \) to correctly handle absolute values like \( |\cos\theta| \) and the final simplification of \( \tan^{-1}(\tan\theta) \).

Question 10. tan⁻¹\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}
Answer:Let's make a substitution to simplify the expression.
Let \( x = a\tan\theta \).
From this, \( \tan\theta = \frac{x}{a} \), which means \( \theta = \tan^{-1}\left(\frac{x}{a}\right) \).
We are given the range for \( x \): \( \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}} \).
Divide by \( a \) (since \( a > 0 \), the inequalities remain the same):
\( \frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}} \)
Taking \( \tan^{-1} \) of all parts:
\( \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) < \tan^{-1}\left(\frac{x}{a}\right) < \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
\( -\frac{\pi}{6} < \theta < \frac{\pi}{6} \)
Now, substitute \( x = a\tan\theta \) into the original expression:
\( \tan^{-1}\left(\frac{3 a^2 (a\tan\theta)-(a\tan\theta)^3}{a^3-3 a (a\tan\theta)^2}\right) \)
\( = \tan^{-1}\left(\frac{3 a^3 \tan\theta-a^3 \tan^3\theta}{a^3-3 a^3 \tan^2\theta}\right) \)
Factor out \( a^3 \) from both the numerator and the denominator:
\( = \tan^{-1}\left(\frac{a^3(3 \tan\theta-\tan^3\theta)}{a^3(1-3 \tan^2\theta)}\right) \)
Cancel out \( a^3 \):
\( = \tan^{-1}\left(\frac{3 \tan\theta-\tan^3\theta}{1-3 \tan^2\theta}\right) \)
We recognize this as the formula for \( \tan(3\theta) \):
\( = \tan^{-1}(\tan(3\theta)) \)
Now, let's check the range of \( 3\theta \). Since \( -\frac{\pi}{6} < \theta < \frac{\pi}{6} \):
\( -\frac{3\pi}{6} < 3\theta < \frac{3\pi}{6} \)
\( -\frac{\pi}{2} < 3\theta < \frac{\pi}{2} \)
This range is exactly the principal value branch for \( \tan^{-1} \), so we can directly simplify:
\( = 3\theta \)
Substitute back \( \theta = \tan^{-1}\left(\frac{x}{a}\right) \):
\( = 3\tan^{-1}\left(\frac{x}{a}\right) \)
In simple words: To simplify this complex math problem, we replaced \( x \) with \( a\tan\theta \). After putting this into the equation and using a special rule for \( \tan(3\theta) \), the expression became much simpler, turning into \( \tan^{-1}(\tan(3\theta)) \). We also checked that our angles were in the correct range. This allowed us to get the final simple answer, which is three times \( \tan^{-1}(\frac{x}{a}) \).

🎯 Exam Tip: When you see expressions resembling \( \frac{3x-x^3}{1-3x^2} \) or \( \frac{3a^2x-x^3}{a^3-3ax^2} \), think of the \( \tan(3\theta) \) identity. A substitution of \( x=a\tan\theta \) is key. Always verify the range of \( 3\theta \) to ensure it lies within \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) for direct simplification of \( \tan^{-1}(\tan(3\theta)) \).

કિંમત શોધો :

Question 11. tan⁻¹ \left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]
Answer:Let's solve the expression from the inside out:
First, find the value of \( \sin^{-1}\frac{1}{2} \).
We know that \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \).
So, \( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \).
Next, substitute this value into the expression:
\( 2 \sin^{-1}\frac{1}{2} = 2 \times \frac{\pi}{6} = \frac{\pi}{3} \)
Now, find the value of \( \cos\left(\frac{\pi}{3}\right) \).
We know that \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).
Substitute this back into the expression:
\( 2 \cos\left(\frac{\pi}{3}\right) = 2 \times \frac{1}{2} = 1 \)
Finally, we have the outermost function:
\( \tan^{-1}(1) \)
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \).
So, \( \tan^{-1}(1) = \frac{\pi}{4} \).
Thus, the value of the expression is \( \frac{\pi}{4} \).
In simple words: We started solving the problem from the innermost part. First, we found the angle for \( \sin^{-1}(\frac{1}{2}) \). Then we used that angle to find the cosine value. After multiplying by two, we were left with \( \tan^{-1}(1) \), which we knew equals \( \frac{\pi}{4} \).

🎯 Exam Tip: Always evaluate nested inverse trigonometric expressions from the innermost function outwards. Remember the principal value ranges for each inverse function to ensure the correct angle is selected (e.g., \( \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), \( \cos^{-1}x \in [0, \pi] \), \( \tan^{-1}x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \)).

Question 12. cot (tan⁻¹a + cot⁻¹a)
Answer:We need to find the value of the expression \( \cot (\tan^{-1}a + \cot^{-1}a) \).
We know a very important identity in inverse trigonometry:
\( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \) for any real number \( x \).
In this problem, \( x \) is replaced by \( a \).
So, \( \tan^{-1}a + \cot^{-1}a = \frac{\pi}{2} \).
Substitute this back into the expression:
\( \cot\left(\frac{\pi}{2}\right) \)
We know that the value of \( \cot\left(\frac{\pi}{2}\right) \) is 0.
Therefore, the value of the expression is 0.
In simple words: We used a simple math rule that says if you add \( \tan^{-1}a \) and \( \cot^{-1}a \), you always get \( \frac{\pi}{2} \). So, the problem became \( \cot(\frac{\pi}{2}) \). We know that \( \cot(90^\circ) \) is 0, so the final answer is 0.

🎯 Exam Tip: Memorize the fundamental inverse trigonometric identities, especially \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \), \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \), and \( \sec^{-1}x + \csc^{-1}x = \frac{\pi}{2} \). These identities frequently appear in simplification problems and can drastically reduce calculation steps.

Question 13. tan\frac{1}{2} \left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right], |x| < 1, y > 0 \text{ and } xy < 1.
Answer:Let's simplify each inverse trigonometric term separately.
For the first term, \( \sin^{-1}\frac{2x}{1+x^2} \):
Let \( x = \tan\theta \). Since \( |x| < 1 \), we know that \( -\frac{\pi}{4} < \theta < \frac{\pi}{4} \).
Substitute \( x = \tan\theta \):
\( \sin^{-1}\frac{2\tan\theta}{1+\tan^2\theta} \)
We know the identity \( \frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta) \).
So, this becomes \( \sin^{-1}(\sin(2\theta)) \).
Since \( -\frac{\pi}{4} < \theta < \frac{\pi}{4} \), then \( -\frac{\pi}{2} < 2\theta < \frac{\pi}{2} \). This is the principal value range for \( \sin^{-1} \).
Therefore, \( \sin^{-1}(\sin(2\theta)) = 2\theta \).
For the second term, \( \cos^{-1}\frac{1-y^2}{1+y^2} \):
Let \( y = \tan\phi \). Since \( y > 0 \) and \( |x|<1, xy<1 \), it implies \( y<1 \). So, \( 0 < \phi < \frac{\pi}{4} \).
Substitute \( y = \tan\phi \):
\( \cos^{-1}\frac{1-\tan^2\phi}{1+\tan^2\phi} \)
We know the identity \( \frac{1-\tan^2\phi}{1+\tan^2\phi} = \cos(2\phi) \).
So, this becomes \( \cos^{-1}(\cos(2\phi)) \).
Since \( 0 < \phi < \frac{\pi}{4} \), then \( 0 < 2\phi < \frac{\pi}{2} \). This is within the principal value range \( [0, \pi] \) for \( \cos^{-1} \).
Therefore, \( \cos^{-1}(\cos(2\phi)) = 2\phi \).
Now, substitute these simplified terms back into the original expression:
\( \tan\frac{1}{2} \left[2\theta + 2\phi\right] \)
\( = \tan\frac{1}{2} \left[2(\theta + \phi)\right] \)
\( = \tan(\theta + \phi) \)
Substitute back \( \theta = \tan^{-1}x \) and \( \phi = \tan^{-1}y \):
\( = \tan(\tan^{-1}x + \tan^{-1}y) \)
Using the identity \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \), which is valid because \( xy < 1 \):
\( = \tan\left(\tan^{-1}\left(\frac{x+y}{1-xy}\right)\right) \)
Finally, \( \tan(\tan^{-1}Z) = Z \):
\( = \frac{x+y}{1-xy} \)
In simple words: We broke down the problem into smaller parts. For the first part, we used \( x = \tan\theta \) to simplify \( \sin^{-1} \) to \( 2\theta \). For the second part, we used \( y = \tan\phi \) to simplify \( \cos^{-1} \) to \( 2\phi \). Then, we put these simplified parts back into the main problem, which became \( \tan(\theta+\phi) \). Using the \( \tan^{-1}x + \tan^{-1}y \) rule, the whole expression simplified to \( \frac{x+y}{1-xy} \).

🎯 Exam Tip: Recognize standard forms like \( \frac{2x}{1+x^2} \) and \( \frac{1-y^2}{1+y^2} \). These often suggest substitutions like \( x=\tan\theta \) and \( y=\tan\phi \) because they relate to the double angle formulas for sine and cosine in terms of tangent. Always ensure the ranges of the substituted angles allow for direct simplification of \( \sin^{-1}(\sin(2\theta)) \) and \( \cos^{-1}(\cos(2\phi)) \).

Question 14. If sin(sin⁻¹\frac{1}{5} + cos⁻¹x) = 1, then find the value of x.
Answer:We are given the equation:
\( \sin\left(\sin^{-1}\frac{1}{5} + \cos^{-1}x\right) = 1 \)
We know that \( \sin\left(\frac{\pi}{2}\right) = 1 \).
So, the argument of the sine function must be equal to \( \frac{\pi}{2} \):
\( \sin^{-1}\frac{1}{5} + \cos^{-1}x = \frac{\pi}{2} \)
Now, we recall a fundamental identity in inverse trigonometry:
\( \sin^{-1}Z + \cos^{-1}Z = \frac{\pi}{2} \), which holds true for any \( Z \) in the range \( [-1, 1] \).
Comparing our equation \( \sin^{-1}\frac{1}{5} + \cos^{-1}x = \frac{\pi}{2} \) with this identity, we can see that:
\( Z = \frac{1}{5} \) and \( Z = x \)
Therefore, for the equation to be true, \( x \) must be equal to \( \frac{1}{5} \).
We must also check if \( x = \frac{1}{5} \) is in the valid domain for \( \cos^{-1}x \), which is \( [-1, 1] \). Since \( \frac{1}{5} \) is indeed within this range, our solution is valid.
The value of \( x \) is \( \frac{1}{5} \).
In simple words: The problem tells us that \( \sin \) of some angle sum is 1. We know that \( \sin(90^\circ) \) is 1, so the sum of the angles inside must be \( \frac{\pi}{2} \). There's a rule that says \( \sin^{-1}Z + \cos^{-1}Z \) is always \( \frac{\pi}{2} \). By comparing this rule to our problem, we easily find that \( x \) must be \( \frac{1}{5} \).

🎯 Exam Tip: When an equation involves a sum of inverse trigonometric functions equaling \( \frac{\pi}{2} \), immediately recall the complementary angle identities (e.g., \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \)). This is a common shortcut for solving such problems quickly. Always verify that the found value of \( x \) lies within the domain of the respective inverse trigonometric functions.

Question 15. If tan⁻¹ \frac{x-1}{x-2} + tan⁻¹\frac{x+1}{x+2}=\frac{\pi}{4}, then find the value of x.
Answer:We are given the equation:
\( \tan^{-1} \frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2}=\frac{\pi}{4} \)
We use the sum formula for inverse tangents:
\( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \)
Here, \( A = \frac{x-1}{x-2} \) and \( B = \frac{x+1}{x+2} \).
Applying the formula:
\( \tan^{-1}\left(\frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \frac{x-1}{x-2} \times \frac{x+1}{x+2}}\right) = \frac{\pi}{4} \)
Now, take the tangent of both sides:
\( \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \frac{x-1}{x-2} \times \frac{x+1}{x+2}} = \tan\left(\frac{\pi}{4}\right) \)
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \).
So, \( \frac{\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2) - (x-1)(x+1)}{(x-2)(x+2)}} = 1 \)
The common denominator \( (x-2)(x+2) \) cancels out.
Numerator:
\( (x-1)(x+2) + (x+1)(x-2) \)
\( = (x^2+2x-x-2) + (x^2-2x+x-2) \)
\( = (x^2+x-2) + (x^2-x-2) \)
\( = 2x^2-4 \)
Denominator:
\( (x-2)(x+2) - (x-1)(x+1) \)
\( = (x^2-4) - (x^2-1) \)
\( = x^2-4-x^2+1 \)
\( = -3 \)
Now, set the simplified fraction equal to 1:
\( \frac{2x^2-4}{-3} = 1 \)
Multiply both sides by -3:
\( 2x^2-4 = -3 \)
Add 4 to both sides:
\( 2x^2 = 1 \)
Divide by 2:
\( x^2 = \frac{1}{2} \)
Take the square root of both sides:
\( x = \pm\sqrt{\frac{1}{2}} \)
\( x = \pm\frac{1}{\sqrt{2}} \)
We check the condition for the formula \( AB < 1 \).
\( AB = \frac{x-1}{x-2} \times \frac{x+1}{x+2} = \frac{x^2-1}{x^2-4} \)
Substitute \( x^2 = \frac{1}{2} \):
\( AB = \frac{\frac{1}{2}-1}{\frac{1}{2}-4} = \frac{-\frac{1}{2}}{-\frac{7}{2}} = \frac{1}{7} \)
Since \( \frac{1}{7} < 1 \), both values of \( x \) are valid.
The values of \( x \) are \( \frac{1}{\sqrt{2}} \) and \( -\frac{1}{\sqrt{2}} \).
In simple words: We used a special rule to combine the two inverse tangent terms on the left side of the equation into a single inverse tangent. Then, we took \( \tan \) of both sides, remembering that \( \tan(\frac{\pi}{4}) \) is 1. We simplified the big fraction we got, which led to a simple equation for \( x^2 \). Solving this, we found the two possible values for \( x \).

🎯 Exam Tip: When solving equations involving sums of inverse tangents, apply the \( \tan^{-1}A + \tan^{-1}B \) formula. Be careful with algebraic simplification of the complex fraction and always verify that the condition \( AB < 1 \) for the formula's validity is met by your solution(s) to avoid extraneous roots.

પ્રશ્ન ક્રમાંક 16 થી 18 ની અભિવ્યક્તિની કિંમત શોધો :

Question 16. sin⁻¹\left(\sin \frac{2 \pi}{3}\right)
Answer:We want to find the value of \( \sin^{-1}\left(\sin \frac{2 \pi}{3}\right) \).
We know that \( \sin^{-1}(\sin x) = x \) only if the angle \( x \) is within the principal value range of \( \sin^{-1} \), which is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
The angle \( \frac{2 \pi}{3} \) is equivalent to \( 120^\circ \), which is not in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) (i.e., \( [-90^\circ, 90^\circ] \)).
So, we need to find an angle \( y \) in the principal value range such that \( \sin y = \sin\left(\frac{2 \pi}{3}\right) \).
We can rewrite \( \sin\left(\frac{2 \pi}{3}\right) \) using the property \( \sin(\pi - \alpha) = \sin\alpha \):
\( \sin\left(\frac{2 \pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) \)
\( = \sin\left(\frac{3\pi - \pi}{3}\right) \)
\( = \sin\left(\frac{\pi}{3}\right) \)
Now, substitute this back into the original expression:
\( \sin^{-1}\left(\sin \frac{2 \pi}{3}\right) = \sin^{-1}\left(\sin \frac{\pi}{3}\right) \)
Since \( \frac{\pi}{3} \) (which is \( 60^\circ \)) is in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we can simplify it directly:
\( = \frac{\pi}{3} \)
The value of the expression is \( \frac{\pi}{3} \).
In simple words: We need to simplify the expression. The angle \( \frac{2\pi}{3} \) is not in the normal range for \( \sin^{-1} \). So, we found another angle \( \frac{\pi}{3} \) that has the same sine value as \( \frac{2\pi}{3} \) and is in the correct range. This way, \( \sin^{-1} \) and \( \sin \) cancelled out, leaving us with \( \frac{\pi}{3} \).

🎯 Exam Tip: When evaluating \( \sin^{-1}(\sin x) \), always check if \( x \) lies within the principal value branch \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). If not, use trigonometric identities (like \( \sin(\pi-x) = \sin x \) or \( \sin(2\pi+x) = \sin x \)) to find an equivalent angle within this range before simplifying.

Question 17. tan⁻¹\left(\tan \frac{3 \pi}{4}\right)
Answer:We want to find the value of \( \tan^{-1}\left(\tan \frac{3 \pi}{4}\right) \).
We know that \( \tan^{-1}(\tan x) = x \) only if the angle \( x \) is within the principal value range of \( \tan^{-1} \), which is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
The angle \( \frac{3 \pi}{4} \) is equivalent to \( 135^\circ \), which is not in the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) (i.e., \( (-90^\circ, 90^\circ) \)).
So, we need to find an angle \( y \) in the principal value range such that \( \tan y = \tan\left(\frac{3 \pi}{4}\right) \).
We can rewrite \( \tan\left(\frac{3 \pi}{4}\right) \) using the property \( \tan(\pi - \alpha) = -\tan\alpha \):
\( \tan\left(\frac{3 \pi}{4}\right) = \tan\left(\pi - \frac{\pi}{4}\right) \)
\( = -\tan\left(\frac{\pi}{4}\right) \)
\( = -1 \)
Now, we need to find \( \tan^{-1}(-1) \).
The angle whose tangent is -1 and lies in the principal value range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) is \( -\frac{\pi}{4} \).
So, \( \tan^{-1}\left(\tan \frac{3 \pi}{4}\right) = \tan^{-1}(-1) = -\frac{\pi}{4} \).
The value of the expression is \( -\frac{\pi}{4} \).
In simple words: We need to simplify this expression. The angle \( \frac{3\pi}{4} \) is outside the main range for \( \tan^{-1} \). So, we found its equivalent tangent value, which is -1. Then we looked for an angle in the correct range for \( \tan^{-1} \) that has a tangent of -1. That angle is \( -\frac{\pi}{4} \).

🎯 Exam Tip: For \( \tan^{-1}(\tan x) \), always first adjust \( x \) into the principal value branch \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) using identities like \( \tan(\pi+x) = \tan x \) or \( \tan(\pi-x) = -\tan x \). This is critical for getting the correct simplified angle.

Question 18. tan\left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)
Answer:Let the expression be \( \tan(A+B) \), where:
\( A = \sin^{-1}\frac{3}{5} \)
\( B = \cot^{-1}\frac{3}{2} \)
From \( A = \sin^{-1}\frac{3}{5} \):
This means \( \sin A = \frac{3}{5} \).
Since \( \sin A \) is positive and \( A \) is from \( \sin^{-1} \), \( A \) is an angle in the first quadrant.
We can imagine a right-angled triangle where the opposite side is 3 and the hypotenuse is 5.
Using Pythagoras theorem, the adjacent side is \( \sqrt{5^2-3^2} = \sqrt{25-9} = \sqrt{16} = 4 \).
So, \( \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} \).
From \( B = \cot^{-1}\frac{3}{2} \):
This means \( \cot B = \frac{3}{2} \).
Since \( \cot B \) is positive and \( B \) is from \( \cot^{-1} \), \( B \) is also an angle in the first quadrant.
We know that \( \tan B = \frac{1}{\cot B} \).
So, \( \tan B = \frac{1}{3/2} = \frac{2}{3} \).
Now, we need to calculate \( \tan(A+B) \). We use the tangent addition formula:
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Substitute the values of \( \tan A \) and \( \tan B \):
\( = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}} \)
Calculate the numerator:
\( \frac{3}{4} + \frac{2}{3} = \frac{(3 \times 3) + (2 \times 4)}{4 \times 3} = \frac{9 + 8}{12} = \frac{17}{12} \)
Calculate the denominator:
\( 1 - \frac{3 \times 2}{4 \times 3} = 1 - \frac{6}{12} = 1 - \frac{1}{2} = \frac{1}{2} \)
Now, divide the numerator by the denominator:
\( = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \times 2 \)
\( = \frac{17}{6} \)
The value of the expression is \( \frac{17}{6} \).
In simple words: We separated the problem into two parts, calling the angles A and B. For angle A, we used the given \( \sin A \) to find \( \tan A \) by drawing a small triangle. For angle B, we used the given \( \cot B \) to find \( \tan B \). Then, we used a special rule to add the tangents of A and B together, and after some simple fraction math, we got our final answer.

🎯 Exam Tip: When faced with a sum or difference of different inverse trigonometric functions inside a direct trigonometric function (e.g., \( \tan(\sin^{-1}x + \cot^{-1}y) \)), convert all inverse functions to the same type (usually \( \tan^{-1} \)) or directly find their tangent values by constructing right-angled triangles. This makes applying sum/difference identities straightforward. Ensure to consider the quadrant of the angles to get the correct signs for trigonometric ratios.

પ્રશ્નો 19 થી 21 માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી યોગ્ય વિકલ્પ પસંદ કરો :

Question 19. cos⁻¹\left(\cos \frac{7 \pi}{6}\right) = .........
Answer:We want to find the value of \( \cos^{-1}\left(\cos \frac{7 \pi}{6}\right) \).
We know that \( \cos^{-1}(\cos x) = x \) only if the angle \( x \) is within the principal value range of \( \cos^{-1} \), which is \( [0, \pi] \).
The angle \( \frac{7 \pi}{6} \) is equivalent to \( 210^\circ \), which is not in the range \( [0, \pi] \) (i.e., \( [0^\circ, 180^\circ] \)).
So, we need to find an angle \( y \) in the principal value range \( [0, \pi] \) such that \( \cos y = \cos\left(\frac{7 \pi}{6}\right) \).
We can rewrite \( \cos\left(\frac{7 \pi}{6}\right) \) using the property \( \cos(\pi + \alpha) = -\cos\alpha \):
\( \cos\left(\frac{7 \pi}{6}\right) = \cos\left(\pi + \frac{\pi}{6}\right) \)
\( = -\cos\left(\frac{\pi}{6}\right) \)
Now, substitute this back into the original expression:
\( \cos^{-1}\left(\cos \frac{7 \pi}{6}\right) = \cos^{-1}\left(-\cos\frac{\pi}{6}\right) \)
We know that \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \).
So, \( = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \)
Using the property \( \cos^{-1}(-Z) = \pi - \cos^{-1}Z \):
\( = \pi - \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
We know that \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \).
So, \( = \pi - \frac{\pi}{6} \)
\( = \frac{6\pi - \pi}{6} \)
\( = \frac{5\pi}{6} \)
This angle \( \frac{5\pi}{6} \) (which is \( 150^\circ \)) is indeed in the principal value range \( [0, \pi] \).
The correct option is (B).
(A) \( \frac{7 \pi}{6} \)
(B) \( \frac{5 \pi}{6} \)
(C) \( \frac{\pi}{3} \)
(D) \( \frac{\pi}{6} \)
Answer: (B) \frac{5 \pi}{6}
In simple words: We need to simplify this expression. The angle \( \frac{7\pi}{6} \) is outside the normal range for \( \cos^{-1} \). So, we found its cosine value, which is \( -\cos(\frac{\pi}{6}) \). Then, using a rule for inverse cosine of a negative number, we found the angle \( \pi - \frac{\pi}{6} \), which gives \( \frac{5\pi}{6} \). This new angle is in the correct range for \( \cos^{-1} \).

🎯 Exam Tip: For \( \cos^{-1}(\cos x) \), always ensure the argument \( x \) is adjusted to fall within the principal value branch \( [0, \pi] \). Use identities like \( \cos(2\pi-x) = \cos x \) or \( \cos(\pi+x) = -\cos x \), and the property \( \cos^{-1}(-y) = \pi - \cos^{-1}y \). Failing to adjust the angle is a common mistake that leads to incorrect answers.

Question 20.sin\(\left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)\) =
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{4}\)
(D) 1
Answer: (D) 1
Answer: To solve this, first remember that \(\sin^{-1}(-x) = -\sin^{-1}(x)\). So, \(\sin^{-1}\left(-\frac{1}{2}\right)\) becomes \(-\sin^{-1}\left(\frac{1}{2}\right)\). We know that \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\), which means \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\).
Therefore, \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\).
Now, substitute this back into the main expression:
\(\sin\left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right)\)
\(= \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right)\)
\(= \sin\left(\frac{2\pi}{6} + \frac{\pi}{6}\right)\)
\(= \sin\left(\frac{3\pi}{6}\right)\)
\(= \sin\left(\frac{\pi}{2}\right)\)
\(= 1\)
In simple words: We first find the value of \(\sin^{-1}\left(-\frac{1}{2}\right)\), which is \(-\frac{\pi}{6}\). Then we add this to \(\frac{\pi}{3}\) and find the sine of the resulting angle, which is 1.

🎯 Exam Tip: Remember the basic properties of inverse trigonometric functions, especially for negative arguments, as they are crucial for simplifying these expressions correctly.

Question 21.tan\(^{-1}\)\(\sqrt{3}\) - cot\(^{-1}\)\((-\sqrt{3})\) =
(Α) \(\pi\)
(B) \(-\frac{\pi}{2}\)
(D) \(2\sqrt{3}\)
Answer: (B) \(-\frac{\pi}{2}\)
Answer: Let's evaluate each term separately.
First, for \(\tan^{-1}\left(\sqrt{3}\right)\): We know that \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\), so \(\tan^{-1}\left(\sqrt{3}\right) = \frac{\pi}{3}\).
Next, for \(\cot^{-1}\left(-\sqrt{3}\right)\): We use the property \(\cot^{-1}(-x) = \pi - \cot^{-1}(x)\).
So, \(\cot^{-1}\left(-\sqrt{3}\right) = \pi - \cot^{-1}\left(\sqrt{3}\right)\).
We know that \(\cot\left(\frac{\pi}{6}\right) = \sqrt{3}\), so \(\cot^{-1}\left(\sqrt{3}\right) = \frac{\pi}{6}\).
Substituting this back, we get \(\cot^{-1}\left(-\sqrt{3}\right) = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}\).
Now, substitute both values back into the original expression:
\(\tan^{-1}\left(\sqrt{3}\right) - \cot^{-1}\left(-\sqrt{3}\right) = \frac{\pi}{3} - \frac{5\pi}{6}\)
To subtract, find a common denominator, which is 6:
\(= \frac{2\pi}{6} - \frac{5\pi}{6}\)
\(= \frac{2\pi - 5\pi}{6}\)
\(= \frac{-3\pi}{6}\)
\(= -\frac{\pi}{2}\)
In simple words: We calculate the value of \(\tan^{-1}\left(\sqrt{3}\right)\) and \(\cot^{-1}\left(-\sqrt{3}\right)\) separately. Then we subtract the second value from the first to get the final answer.

🎯 Exam Tip: Pay close attention to the properties of inverse trigonometric functions, especially when the input is negative, as \(\tan^{-1}(-x) = -\tan^{-1}(x)\) but \(\cot^{-1}(-x) = \pi - \cot^{-1}(x)\).