GSEB Class 12 Maths Solutions Chapter 2 ત્રિકોણમિતીય પ્રતિવિધેયો Misc. Ques

GSEB Solutions Class 12 Maths Chapter 2 ત્રિમितीय પ્રતિવિધેયો Miscellaneous Exercise

Gujarat Board Textbook Solutions Class 12 Maths Chapter 2 ત્રિકોણમિતીય પ્રતિવિધેયો Miscellaneous Exercise

Question 1. \( \cos^{-1}\left(\cos \frac{13 \pi}{6}\right) \)

We know that the principal value branch for \( \cos^{-1}(\cos x) = x \) is when \( x \in [0, \pi] \). The given value is \( \frac{13\pi}{6} \). This is not within the range \( [0, \pi] \). We can rewrite \( \frac{13\pi}{6} \) as \( 2\pi + \frac{\pi}{6} \). Since \( \cos(2\pi + \theta) = \cos \theta \), we have \( \cos\left(\frac{13\pi}{6}\right) = \cos\left(2\pi + \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) \). Now, \( \frac{\pi}{6} \) is in the range \( [0, \pi] \). Therefore, \( \cos^{-1}\left(\cos \frac{13\pi}{6}\right) = \cos^{-1}\left(\cos \frac{\pi}{6}\right) = \frac{\pi}{6} \).
In simple words: The inverse cosine function gives an angle between 0 and π. Since 13π/6 is outside this range, we use the property that cos repeats every 2π. We adjust 13π/6 to π/6, which is inside the correct range, so the answer is π/6.

🎯 Exam Tip: Remember the principal value range for inverse trigonometric functions. Always adjust the angle to fit within this range using trigonometric identities to get the correct answer.

Question 2. \( \tan^{-1}\left(\tan \frac{7 \pi}{6}\right) \)

We know that the principal value branch for \( \tan^{-1}(\tan x) = x \) is when \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). The given value is \( \frac{7\pi}{6} \). This is not within the range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). We can rewrite \( \frac{7\pi}{6} \) as \( \pi + \frac{\pi}{6} \). Since \( \tan(\pi + \theta) = \tan \theta \), we have \( \tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) \). Now, \( \frac{\pi}{6} \) is in the range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Therefore, \( \tan^{-1}\left(\tan \frac{7\pi}{6}\right) = \tan^{-1}\left(\tan \frac{\pi}{6}\right) = \frac{\pi}{6} \).
In simple words: The inverse tangent function gives an angle between -π/2 and π/2. Since 7π/6 is outside this range, we use the property that tan repeats every π. We adjust 7π/6 to π/6, which is inside the correct range, so the answer is π/6.

🎯 Exam Tip: Similar to cosine, ensure the angle for inverse tangent is within its principal value range of \((-\frac{\pi}{2}, \frac{\pi}{2})\). Use identities like \( \tan(\pi + \theta) = \tan \theta \) for adjustments.

साबित करो :

Question 3. \( 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7} \)

Let's assume \( \theta = \sin^{-1}\frac{3}{5} \). This means \( \sin\theta = \frac{3}{5} \). The range for \( \theta \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज है जहाँ कोण θ है। इस त्रिभुज में, लंब (opposite side) 3 है और कर्ण (hypotenuse) 5 है। पायथागोरस प्रमेय का उपयोग करके आधार (adjacent side) 4 ज्ञात किया जा सकता है। From the right-angled triangle, if \( \sin\theta = \frac{3}{5} \) (opposite/hypotenuse), then the adjacent side is 4 (using Pythagoras theorem: \( \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \)). Therefore, \( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} \). We need to prove \( 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7} \). This is equivalent to proving \( 2\theta = \tan^{-1}\frac{24}{7} \), or \( \tan(2\theta) = \frac{24}{7} \). We know the double angle identity for tangent: \( \tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} \). Substitute the value of \( \tan\theta = \frac{3}{4} \): \( \tan(2\theta) = \frac{2\left(\frac{3}{4}\right)}{1 - \left(\frac{3}{4}\right)^2} \) \( = \frac{\frac{6}{4}}{1 - \frac{9}{16}} \) \( = \frac{\frac{3}{2}}{\frac{16 - 9}{16}} \) \( = \frac{\frac{3}{2}}{\frac{7}{16}} \) \( = \frac{3}{2} \times \frac{16}{7} \) \( = \frac{3 \times 8}{7} \) \( = \frac{24}{7} \) Thus, \( 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7} \) is proven.
In simple words: To prove this, we first change the sine inverse to a tangent. If sine is 3/5, then tangent is 3/4. Then, we use a formula for tan(2θ) with this tangent value. After calculating, we get 24/7, which matches the right side of the equation.

🎯 Exam Tip: When converting inverse trigonometric functions, drawing a right-angled triangle can be very helpful. Remember double angle formulas for tangent, like \( \tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} \), as they are frequently used in proofs.

Question 4. \( \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36} \)

Let \( A = \sin^{-1}\frac{8}{17} \) and \( B = \sin^{-1}\frac{3}{5} \). This means \( \sin A = \frac{8}{17} \) and \( \sin B = \frac{3}{5} \).
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला समकोण त्रिभुज कोण A के लिए है, जिसमें लंब 8 और कर्ण 17 है, जिससे आधार 15 प्राप्त होता है। दूसरा समकोण त्रिभुज कोण B के लिए है, जिसमें लंब 3 और कर्ण 5 है, जिससे आधार 4 प्राप्त होता है। Using a right-angled triangle: For A: If \( \sin A = \frac{8}{17} \) (opposite/hypotenuse), then the adjacent side is \( \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \). So, \( \tan A = \frac{8}{15} \). For B: If \( \sin B = \frac{3}{5} \) (opposite/hypotenuse), then the adjacent side is \( \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \). So, \( \tan B = \frac{3}{4} \). We need to prove \( A + B = \tan^{-1}\frac{77}{36} \). This means we need to find \( \tan(A+B) \). We use the formula \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). Substitute the values of \( \tan A \) and \( \tan B \): \( \tan(A+B) = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \times \frac{3}{4}} \) \( = \frac{\frac{32 + 45}{60}}{1 - \frac{24}{60}} \) \( = \frac{\frac{77}{60}}{\frac{60 - 24}{60}} \) \( = \frac{\frac{77}{60}}{\frac{36}{60}} \) \( = \frac{77}{36} \) Therefore, \( A + B = \tan^{-1}\frac{77}{36} \). This proves \( \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36} \).
In simple words: First, we change both sine inverse terms into tangent terms using right-angled triangles. Then, we use the formula for tan(A+B) with these tangent values. After calculations, the result is 77/36, which means the sum is tan inverse of 77/36.

🎯 Exam Tip: This type of problem often requires converting inverse sine/cosine to inverse tangent before using the tangent addition formula. Careful calculation with fractions is essential for accuracy.

Question 5. \( \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65} \)

Let \( A = \cos^{-1}\frac{4}{5} \) and \( B = \cos^{-1}\frac{12}{13} \). This means \( \cos A = \frac{4}{5} \) and \( \cos B = \frac{12}{13} \).
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला समकोण त्रिभुज कोण A के लिए है, जिसमें आधार 4 और कर्ण 5 है, जिससे लंब 3 प्राप्त होता है। दूसरा समकोण त्रिभुज कोण B के लिए है, जिसमें आधार 12 और कर्ण 13 है, जिससे लंब 5 प्राप्त होता है। Using a right-angled triangle: For A: If \( \cos A = \frac{4}{5} \) (adjacent/hypotenuse), then \( \sin A = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \). For B: If \( \cos B = \frac{12}{13} \) (adjacent/hypotenuse), then \( \sin B = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \). We need to prove \( A + B = \cos^{-1}\frac{33}{65} \). This means we need to find \( \cos(A+B) \). We use the formula \( \cos(A+B) = \cos A \cos B - \sin A \sin B \). Substitute the values: \( \cos(A+B) = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \) \( = \frac{48}{65} - \frac{15}{65} \) \( = \frac{48 - 15}{65} \) \( = \frac{33}{65} \) Therefore, \( A + B = \cos^{-1}\frac{33}{65} \). This proves \( \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65} \).
In simple words: We let the two inverse cosine terms be A and B. Then we find sin A and sin B from their cosine values. Using the cosine addition formula, cos(A+B) = cos A cos B - sin A sin B, we substitute the values. The calculation gives 33/65, which shows that the sum is cos inverse of 33/65.

🎯 Exam Tip: For inverse cosine addition problems, it's efficient to use the cosine addition formula \( \cos(A+B) = \cos A \cos B - \sin A \sin B \). Remember to find the corresponding sine values using \( \sin x = \sqrt{1 - \cos^2 x} \).

Question 6. \( \cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65} \)

Let \( A = \cos^{-1}\frac{12}{13} \) and \( B = \sin^{-1}\frac{3}{5} \). This means \( \cos A = \frac{12}{13} \) and \( \sin B = \frac{3}{5} \).
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला समकोण त्रिभुज कोण A के लिए है, जिसमें आधार 12 और कर्ण 13 है, जिससे लंब 5 प्राप्त होता है। दूसरा समकोण त्रिभुज कोण B के लिए है, जिसमें लंब 3 और कर्ण 5 है, जिससे आधार 4 प्राप्त होता है। Using a right-angled triangle: For A: If \( \cos A = \frac{12}{13} \), then \( \sin A = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \). For B: If \( \sin B = \frac{3}{5} \), then \( \cos B = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \). We need to prove \( A + B = \sin^{-1}\frac{56}{65} \). This means we need to find \( \sin(A+B) \). We use the formula \( \sin(A+B) = \sin A \cos B + \cos A \sin B \). Substitute the values: \( \sin(A+B) = \left(\frac{5}{13}\right)\left(\frac{4}{5}\right) + \left(\frac{12}{13}\right)\left(\frac{3}{5}\right) \) \( = \frac{20}{65} + \frac{36}{65} \) \( = \frac{20 + 36}{65} \) \( = \frac{56}{65} \) Therefore, \( A + B = \sin^{-1}\frac{56}{65} \). This proves \( \cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65} \).
In simple words: We convert both inverse cosine and inverse sine terms to their corresponding sine and cosine values. Then, we use the sine addition formula, sin(A+B) = sin A cos B + cos A sin B. After putting in the numbers, we get 56/65, which proves the statement.

🎯 Exam Tip: For problems involving a mix of inverse cosine and inverse sine, convert both to sine and cosine values, then apply the sine addition formula \( \sin(A+B) = \sin A \cos B + \cos A \sin B \). Accuracy in finding the missing side of the right triangle is crucial.

Question 7. \( \tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} \)

Let's consider the right-hand side (RHS) of the equation. Let \( A = \sin^{-1}\frac{5}{13} \) and \( B = \cos^{-1}\frac{3}{5} \). This means \( \sin A = \frac{5}{13} \) and \( \cos B = \frac{3}{5} \).
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला समकोण त्रिभुज कोण A के लिए है, जिसमें लंब 5 और कर्ण 13 है, जिससे आधार 12 प्राप्त होता है। दूसरा समकोण त्रिभुज कोण B के लिए है, जिसमें आधार 3 और कर्ण 5 है, जिससे लंब 4 प्राप्त होता है। Using a right-angled triangle: For A: If \( \sin A = \frac{5}{13} \), then \( \cos A = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \). So, \( \tan A = \frac{\sin A}{\cos A} = \frac{5/13}{12/13} = \frac{5}{12} \). For B: If \( \cos B = \frac{3}{5} \), then \( \sin B = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \). So, \( \tan B = \frac{\sin B}{\cos B} = \frac{4/5}{3/5} = \frac{4}{3} \). Now we need to find \( \tan(A+B) \). We use the formula \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). Substitute the values: \( \tan(A+B) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}} \) \( = \frac{\frac{5 + 16}{12}}{1 - \frac{20}{36}} \) \( = \frac{\frac{21}{12}}{1 - \frac{5}{9}} \) \( = \frac{\frac{7}{4}}{\frac{9 - 5}{9}} \) \( = \frac{\frac{7}{4}}{\frac{4}{9}} \) \( = \frac{7}{4} \times \frac{9}{4} \) \( = \frac{63}{16} \) Therefore, \( A + B = \tan^{-1}\frac{63}{16} \). This proves \( \tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} \).
In simple words: We take the right side of the equation. We convert both inverse sine and inverse cosine terms into inverse tangent terms. Then, we use the formula for tan(A+B). After solving, we find the result is 63/16, which matches the left side of the equation.

🎯 Exam Tip: When the target is \( \tan^{-1} \), it's often easiest to convert all terms to \( \tan^{-1} \) using the right-triangle method, then apply the tangent addition formula repeatedly until simplified.

Question 8. \( \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4} \)

Let's take the left-hand side (LHS): \( \text{LHS} = \left(\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7}\right) + \left(\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8}\right) \) We use the formula \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \). For the first pair: \( \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} = \tan^{-1}\left(\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}}\right) \) \( = \tan^{-1}\left(\frac{\frac{7+5}{35}}{1 - \frac{1}{35}}\right) \) \( = \tan^{-1}\left(\frac{\frac{12}{35}}{\frac{34}{35}}\right) \) \( = \tan^{-1}\left(\frac{12}{34}\right) = \tan^{-1}\left(\frac{6}{17}\right) \) For the second pair: \( \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}}\right) \) \( = \tan^{-1}\left(\frac{\frac{8+3}{24}}{1 - \frac{1}{24}}\right) \) \( = \tan^{-1}\left(\frac{\frac{11}{24}}{\frac{23}{24}}\right) \) \( = \tan^{-1}\left(\frac{11}{23}\right) \) Now, add these two results: \( \text{LHS} = \tan^{-1}\left(\frac{6}{17}\right) + \tan^{-1}\left(\frac{11}{23}\right) \) Apply the formula again: \( = \tan^{-1}\left(\frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}}\right) \) \( = \tan^{-1}\left(\frac{\frac{6 \times 23 + 11 \times 17}{17 \times 23}}{1 - \frac{66}{17 \times 23}}\right) \) \( = \tan^{-1}\left(\frac{\frac{138 + 187}{391}}{\frac{391 - 66}{391}}\right) \) \( = \tan^{-1}\left(\frac{325}{325}\right) \) \( = \tan^{-1}(1) \) We know that \( \tan^{-1}(1) = \frac{\pi}{4} \). Therefore, \( \text{LHS} = \frac{\pi}{4} = \text{RHS} \).
In simple words: We group the four inverse tangent terms into two pairs. We use the addition formula for tan inverse for each pair to simplify them. Then, we use the addition formula one more time for the two simplified terms. This results in tan inverse of 1, which equals π/4.

🎯 Exam Tip: When summing multiple \( \tan^{-1}x \) terms, group them in pairs and apply the \( \tan^{-1}x + \tan^{-1}y \) formula iteratively. This systematically simplifies the expression. Watch out for arithmetic errors in fractions.

Question 9. \( \tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) x \in [0, 1] \)

Let's consider the right-hand side (RHS): \( \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) \). Let \( x = \tan^2\theta \). Then \( \sqrt{x} = \tan\theta \). From this, \( \theta = \tan^{-1}\sqrt{x} \). Substitute \( x = \tan^2\theta \) into the RHS: \( \text{RHS} = \frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) \) We know the trigonometric identity \( \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos(2\theta) \). So, \( \text{RHS} = \frac{1}{2}\cos^{-1}(\cos(2\theta)) \) Since \( x \in [0, 1] \), \( \tan^2\theta \in [0, 1] \). This means \( \tan\theta \in [0, 1] \) (since \( \sqrt{x} \) is positive). If \( \tan\theta \in [0, 1] \), then \( \theta \in \left[0, \frac{\pi}{4}\right] \). Consequently, \( 2\theta \in \left[0, \frac{\pi}{2}\right] \). Within this range, \( \cos^{-1}(\cos(2\theta)) = 2\theta \). So, \( \text{RHS} = \frac{1}{2}(2\theta) = \theta \). Substitute back \( \theta = \tan^{-1}\sqrt{x} \): \( \text{RHS} = \tan^{-1}\sqrt{x} \). This matches the left-hand side (LHS). Therefore, \( \tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) \) is proven.
In simple words: To prove this, we assume \( x = \tan^2\theta \). We substitute this into the right side of the equation. Using a trigonometric identity, the expression simplifies to \( \frac{1}{2}\cos^{-1}(\cos(2\theta)) \), which becomes \( \frac{1}{2}(2\theta) = \theta \). Replacing \( \theta \) with \( \tan^{-1}\sqrt{x} \) gives us the left side.

🎯 Exam Tip: For expressions involving \( \frac{1-x}{1+x} \), try substitutions like \( x = \tan^2\theta \) or \( x = \cos(2\theta) \) to simplify. Always verify the range of the substituted variable to ensure the inverse function properties are correctly applied.

Question 10. \( \cot^{-1} \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} = \frac{x}{2}, x \in \left(0, \frac{\pi}{4}\right) \)

Let's take the left-hand side (LHS): \( \cot^{-1} \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \). We know that \( 1 + \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2 \). And \( 1 - \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2 \). Given \( x \in \left(0, \frac{\pi}{4}\right) \), this implies \( \frac{x}{2} \in \left(0, \frac{\pi}{8}\right) \). In this interval, \( \cos\frac{x}{2} > \sin\frac{x}{2} > 0 \). So, \( \sqrt{1+\sin x} = \sqrt{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2} = \cos\frac{x}{2} + \sin\frac{x}{2} \). And \( \sqrt{1-\sin x} = \sqrt{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2} = \cos\frac{x}{2} - \sin\frac{x}{2} \). Substitute these into the expression: \( \text{LHS} = \cot^{-1} \frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) + \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) - \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)} \) \( = \cot^{-1} \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} \) \( = \cot^{-1}\left(\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\right) \) \( = \cot^{-1}\left(\cot\frac{x}{2}\right) \) Since \( \frac{x}{2} \in \left(0, \frac{\pi}{8}\right) \), it is in the principal value branch of \( \cot^{-1}(y) \) which is \( (0, \pi) \). Therefore, \( \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2} \). This matches the right-hand side (RHS).
In simple words: We rewrite \( 1+\sin x \) and \( 1-\sin x \) as perfect squares involving \( \cos\frac{x}{2} \) and \( \sin\frac{x}{2} \). Then we substitute these into the cot inverse expression. After simplifying, the expression becomes \( \cot^{-1}(\cot\frac{x}{2}) \). Since x is in the given range, this simplifies to \( \frac{x}{2} \).

🎯 Exam Tip: When dealing with \( \sqrt{1 \pm \sin x} \), use the identities \( 1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} \) and \( \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} \) to form perfect squares. Always check the interval of \( x \) to decide the sign when taking the square root.

Question 11. \( \tan^{-1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}x, -\frac{1}{\sqrt{2}} \leq x \leq 1 \)

(सूयन : x = cos2θ सो.)

Let's consider the left-hand side (LHS): \( \tan^{-1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \). As per the hint, let \( x = \cos(2\theta) \). This implies \( 2\theta = \cos^{-1}x \), so \( \theta = \frac{1}{2}\cos^{-1}x \). Substitute \( x = \cos(2\theta) \) into the expression: \( \text{LHS} = \tan^{-1}\frac{\sqrt{1+\cos(2\theta)}-\sqrt{1-\cos(2\theta)}}{\sqrt{1+\cos(2\theta)}+\sqrt{1-\cos(2\theta)}} \) We use the identities: \( 1+\cos(2\theta) = 2\cos^2\theta \) \( 1-\cos(2\theta) = 2\sin^2\theta \) So, \( \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2\theta} = \sqrt{2}|\cos\theta| \). And \( \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2\theta} = \sqrt{2}|\sin\theta| \). Given the range \( -\frac{1}{\sqrt{2}} \leq x \leq 1 \). If \( x = \cos(2\theta) \), then \( -\frac{1}{\sqrt{2}} \leq \cos(2\theta) \leq 1 \). This means \( 0 \leq 2\theta \leq \frac{3\pi}{4} \). So, \( 0 \leq \theta \leq \frac{3\pi}{8} \). In this range, \( \cos\theta \geq 0 \) and \( \sin\theta \geq 0 \). Thus, \( |\cos\theta| = \cos\theta \) and \( |\sin\theta| = \sin\theta \). Substitute these into the LHS: \( \text{LHS} = \tan^{-1}\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta} \) \( = \tan^{-1}\frac{\sqrt{2}(\cos\theta - \sin\theta)}{\sqrt{2}(\cos\theta + \sin\theta)} \) \( = \tan^{-1}\left(\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}\right) \) Now, divide the numerator and denominator by \( \cos\theta \): \( = \tan^{-1}\left(\frac{1 - \frac{\sin\theta}{\cos\theta}}{1 + \frac{\sin\theta}{\cos\theta}}\right) \) \( = \tan^{-1}\left(\frac{1 - \tan\theta}{1 + \tan\theta}\right) \) We know that \( \frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right) \). So, \( \text{LHS} = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) \). Since \( 0 \leq \theta \leq \frac{3\pi}{8} \), then \( -\frac{3\pi}{8} \leq -\theta \leq 0 \). Adding \( \frac{\pi}{4} \): \( \frac{\pi}{4} - \frac{3\pi}{8} \leq \frac{\pi}{4} - \theta \leq \frac{\pi}{4} \). So, \( -\frac{\pi}{8} \leq \frac{\pi}{4} - \theta \leq \frac{\pi}{4} \). This range is within \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Therefore, \( \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta \). Substitute back \( \theta = \frac{1}{2}\cos^{-1}x \): \( \text{LHS} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x \). This matches the right-hand side (RHS).
In simple words: We substitute \( x = \cos(2\theta) \) into the left side. Using trigonometric identities, we simplify \( \sqrt{1+x} \) and \( \sqrt{1-x} \). After simplifying the fraction, it becomes \( \tan^{-1}(\frac{1-\tan\theta}{1+\tan\theta}) \), which further simplifies to \( \tan^{-1}(\tan(\frac{\pi}{4}-\theta)) \). This equals \( \frac{\pi}{4}-\theta \). Finally, replacing \( \theta \) with \( \frac{1}{2}\cos^{-1}x \) gives the right side.

🎯 Exam Tip: The substitution \( x = \cos(2\theta) \) is key when dealing with terms like \( \sqrt{1 \pm x} \). Remember the formulas \( 1+\cos(2\theta) = 2\cos^2\theta \) and \( 1-\cos(2\theta) = 2\sin^2\theta \). Be careful with the range of \( \theta \) to correctly remove absolute values and simplify \( \tan^{-1}(\tan y) = y \).

Question 12. \( \frac{9 \pi}{8}-\frac{9}{4} \sin^{-1} \frac{1}{3}=\frac{9}{4} \sin^{-1} \frac{2 \sqrt{2}}{3} \)

We need to prove \( \frac{9 \pi}{8}-\frac{9}{4} \sin^{-1} \frac{1}{3}=\frac{9}{4} \sin^{-1} \frac{2 \sqrt{2}}{3} \). Let's rearrange the equation to bring all \( \sin^{-1} \) terms to one side: \( \frac{9 \pi}{8} = \frac{9}{4} \sin^{-1} \frac{2 \sqrt{2}}{3} + \frac{9}{4} \sin^{-1} \frac{1}{3} \) Divide both sides by \( \frac{9}{4} \): \( \frac{9 \pi}{8} \times \frac{4}{9} = \sin^{-1} \frac{2 \sqrt{2}}{3} + \sin^{-1} \frac{1}{3} \) \( \frac{\pi}{2} = \sin^{-1} \frac{2 \sqrt{2}}{3} + \sin^{-1} \frac{1}{3} \) Now, we need to prove this new equation: \( \sin^{-1} \frac{2 \sqrt{2}}{3} + \sin^{-1} \frac{1}{3} = \frac{\pi}{2} \). Let \( \theta = \sin^{-1}\frac{1}{3} \). This means \( \sin\theta = \frac{1}{3} \). Using the identity \( \cos^2\theta = 1 - \sin^2\theta \): \( \cos^2\theta = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \). Since \( \theta = \sin^{-1}\frac{1}{3} \), \( \theta \) is in the first quadrant, so \( \cos\theta \) is positive. \( \cos\theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \). This implies \( \theta = \cos^{-1}\frac{2\sqrt{2}}{3} \). So, we can rewrite the equation as: \( \sin^{-1} \frac{2 \sqrt{2}}{3} + \cos^{-1} \frac{2 \sqrt{2}}{3} = \frac{\pi}{2} \) We know the identity \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \). Here, \( x = \frac{2\sqrt{2}}{3} \). So, the identity holds true. Therefore, the original equation is proven.
In simple words: First, we move all the inverse sine terms to one side and simplify the equation to show that \( \sin^{-1}\frac{2\sqrt{2}}{3} + \sin^{-1}\frac{1}{3} = \frac{\pi}{2} \). Then, we let \( \sin^{-1}\frac{1}{3} \) be \( \theta \), which means \( \sin\theta = \frac{1}{3} \). From this, we find \( \cos\theta = \frac{2\sqrt{2}}{3} \). So, \( \theta \) is also \( \cos^{-1}\frac{2\sqrt{2}}{3} \). The equation becomes \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \), which is a known identity.

🎯 Exam Tip: When proving identities, sometimes it's easier to rearrange the equation first. The identity \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) is very useful for simplifying expressions. Remember to correctly find cosine from sine (or vice versa) using \( \sin^2\theta + \cos^2\theta = 1 \).

નીચેનાં સમીકરણ ઉકેલો :

Question 13. \( 2 \tan^{-1} (\cos x) = \tan^{-1} (2\operatorname{cosec} x) \)

We are given the equation \( 2 \tan^{-1} (\cos x) = \tan^{-1} (2\operatorname{cosec} x) \). Using the formula \( 2\tan^{-1}A = \tan^{-1}\left(\frac{2A}{1-A^2}\right) \), where \( A = \cos x \): \( \tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1} (2\operatorname{cosec} x) \) Since \( \tan^{-1} \) is a one-to-one function, we can equate the arguments: \( \frac{2\cos x}{1-\cos^2 x} = 2\operatorname{cosec} x \) We know \( 1-\cos^2 x = \sin^2 x \) and \( \operatorname{cosec} x = \frac{1}{\sin x} \). \( \frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x} \) Since \( \sin x \neq 0 \) (because \( \operatorname{cosec} x \) is defined, and also \( 1-\cos^2 x \) is in the denominator, so \( \sin x \neq 0 \)), we can multiply both sides by \( \sin x \) and divide by 2: \( \frac{\cos x}{\sin x} = 1 \) \( \cot x = 1 \) This means \( \tan x = 1 \). The general solution for \( \tan x = 1 \) is \( x = n\pi + \frac{\pi}{4} \), where \( n \in \mathbb{Z} \). However, we need to consider the domain of the original equation. For \( \operatorname{cosec} x \) to be defined, \( \sin x \neq 0 \). Also, for \( \tan^{-1}(A) \) to be defined, A can be any real number. Also, the argument of \( \tan^{-1}\left(\frac{2A}{1-A^2}\right) \) must be defined, so \( 1-A^2 \neq 0 \), i.e., \( \cos^2 x \neq 1 \), so \( \sin x \neq 0 \). If \( x = \frac{\pi}{4} \), then \( \cos x = \frac{1}{\sqrt{2}} \) and \( \operatorname{cosec} x = \sqrt{2} \). LHS: \( 2\tan^{-1}(\cos x) = 2\tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \) \( = \tan^{-1}\left(\frac{2 \times \frac{1}{\sqrt{2}}}{1 - \left(\frac{1}{\sqrt{2}}\right)^2}\right) = \tan^{-1}\left(\frac{\sqrt{2}}{1 - \frac{1}{2}}\right) = \tan^{-1}\left(\frac{\sqrt{2}}{\frac{1}{2}}\right) = \tan^{-1}(2\sqrt{2}) \). RHS: \( \tan^{-1}(2\operatorname{cosec} x) = \tan^{-1}(2\sqrt{2}) \). LHS = RHS, so \( x = \frac{\pi}{4} \) is a solution. Also, note that for \( x = 0 \), \( \cos(0) = 1 \) and \( \operatorname{cosec}(0) \) is undefined, so \( x=0 \) is not a solution. If \( x = \frac{\pi}{4} \), the equation holds true. Therefore, \( x = \frac{\pi}{4} \) is the solution to the equation.
In simple words: We start by using a formula to change \( 2\tan^{-1}(\cos x) \) into \( \tan^{-1} \) of a fraction. Then we set the inside parts of the \( \tan^{-1} \) functions equal to each other. After simplifying using trigonometric identities, we get \( \cot x = 1 \), which means \( x = \frac{\pi}{4} \). We also check that \( x=0 \) is not a solution because \( \operatorname{cosec} x \) would be undefined.

🎯 Exam Tip: When solving equations with inverse trigonometric functions, convert both sides to the same inverse function if possible. Remember to consider the domain restrictions of the original functions (e.g., \( \operatorname{cosec} x \) is undefined for \( \sin x = 0 \)) to validate the solutions.

Question 14. \( \tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x \)

We are given the equation \( \tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x \). We know the formula \( \tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}(1) - \tan^{-1}(x) \). So, the LHS becomes: \( \tan^{-1}(1) - \tan^{-1}(x) = \frac{1}{2} \tan^{-1}(x) \) We know \( \tan^{-1}(1) = \frac{\pi}{4} \). \( \frac{\pi}{4} - \tan^{-1}(x) = \frac{1}{2} \tan^{-1}(x) \) Move \( \tan^{-1}(x) \) to the RHS: \( \frac{\pi}{4} = \frac{1}{2} \tan^{-1}(x) + \tan^{-1}(x) \) \( \frac{\pi}{4} = \left(\frac{1}{2} + 1\right) \tan^{-1}(x) \) \( \frac{\pi}{4} = \frac{3}{2} \tan^{-1}(x) \) Now, solve for \( \tan^{-1}(x) \): \( \tan^{-1}(x) = \frac{\pi}{4} \times \frac{2}{3} \) \( \tan^{-1}(x) = \frac{\pi}{6} \) To find \( x \), take the tangent of both sides: \( x = \tan\left(\frac{\pi}{6}\right) \) We know \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \). So, \( x = \frac{1}{\sqrt{3}} \).
In simple words: We use a formula to change the left side, \( \tan^{-1}(\frac{1-x}{1+x}) \), into \( \tan^{-1}(1) - \tan^{-1}(x) \). Then, we substitute \( \tan^{-1}(1) \) with \( \frac{\pi}{4} \). We rearrange the equation to solve for \( \tan^{-1}(x) \), which gives \( \frac{\pi}{6} \). Finally, we find \( x \) by taking the tangent of \( \frac{\pi}{6} \), which is \( \frac{1}{\sqrt{3}} \).

🎯 Exam Tip: The identity \( \tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}a - \tan^{-1}b \) is very helpful here. Recognizing that \( \tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}(1) - \tan^{-1}(x) \) simplifies the problem significantly. Pay attention to basic algebraic manipulation.

પ્રશ્નો 15 થી 17 માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી સૌગ્ય વિકલ્પ પસંદ કરો :

Question 15. \( \sin(\tan^{-1} x), |x| < 1 = \dots \)
(A) \( \frac{x}{\sqrt{1-x^2}} \)
(B) \( \frac{1}{\sqrt{1-x^2}} \)
(C) \( \frac{1}{\sqrt{1+x^2}} \)
(D) \( \frac{x}{\sqrt{1+x^2}} \)

Let \( y = \sin(\tan^{-1} x) \). Let \( \theta = \tan^{-1} x \). This means \( \tan\theta = x \). Since \( \tan\theta = x = \frac{x}{1} \), we can form a right-angled triangle where the opposite side is \( x \) and the adjacent side is \( 1 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज है जहाँ कोण θ है। इस त्रिभुज में, लंब x है और आधार 1 है। पायथागोरस प्रमेय का उपयोग करके कर्ण \( \sqrt{x^2 + 1^2} = \sqrt{x^2+1} \) ज्ञात किया जा सकता है। The hypotenuse will be \( \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{x^2 + 1^2} = \sqrt{1+x^2} \). Now, we need to find \( \sin\theta \). \( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1+x^2}} \). So, \( y = \sin(\tan^{-1} x) = \sin\theta = \frac{x}{\sqrt{1+x^2}} \). Therefore, the correct option is (D).
In simple words: We let \( \tan^{-1}x \) be \( \theta \), which means \( \tan\theta = x \). We then draw a right-angled triangle with the opposite side as x and the adjacent side as 1. The hypotenuse is \( \sqrt{1+x^2} \). From this triangle, we find that \( \sin\theta \) is \( \frac{x}{\sqrt{1+x^2}} \).

🎯 Exam Tip: For expressions like \( \sin(\tan^{-1} x) \) or \( \cos(\sin^{-1} x) \), use a right-angled triangle. Let the inverse function be \( \theta \), draw the triangle with the known ratio, find the third side using Pythagoras theorem, and then determine the desired trigonometric ratio.

Question 16. sin\(^{-1}\)(1-x)-2sin\(^{-1}\)x=\(\frac{\pi}{2}\), तो x = .........
(A) 0, \(\frac{1}{2}\)
(B) 1, \(\frac{1}{2}\)
(C) 0
(D) \(\frac{1}{2}\)
Answer: (C) 0
We are given the equation: sin\(^{-1}\)(1-x) - 2sin\(^{-1}\)x = \(\frac{\pi}{2}\).
To find the value of x, we can test the options provided.
First, let's try x = 0.
Substitute x = 0 into the equation:
sin\(^{-1}\)(1-0) - 2sin\(^{-1}\)(0)
= sin\(^{-1}\)(1) - 2(0)
= \(\frac{\pi}{2}\) - 0
= \(\frac{\pi}{2}\)
Since the left side equals the right side (\(\frac{\pi}{2}\)), x = 0 is a correct solution.
Next, let's try x = \(\frac{1}{2}\).
Substitute x = \(\frac{1}{2}\) into the equation:
sin\(^{-1}\)(1-\(\frac{1}{2}\)) - 2sin\(^{-1}\)(\(\frac{1}{2}\))
= sin\(^{-1}\)(\(\frac{1}{2}\)) - 2sin\(^{-1}\)(\(\frac{1}{2}\))
= -sin\(^{-1}\)(\(\frac{1}{2}\))
= -\(\frac{\pi}{6}\)
Since -\(\frac{\pi}{6}\) is not equal to \(\frac{\pi}{2}\), x = \(\frac{1}{2}\) is not a solution.
Therefore, the only correct value for x from the options is 0.
In simple words: We check which given value of x makes the equation true. Putting x = 0 into the equation gives \(\frac{\pi}{2}\) on both sides, so it works. Putting x = \(\frac{1}{2}\) does not give \(\frac{\pi}{2}\) on the left side, so it is not correct.

🎯 Exam Tip: For multiple-choice questions involving equations, a quick and effective strategy is to substitute the given options into the equation to verify which one satisfies it. This can save time compared to solving the equation algebraically.

Question 17. tan\(^{-1}\)\(\frac{x}{y}\) - tan\(^{-1}\)\(\frac{x-y}{x+y}\) = .........
(A) \(\frac{\pi}{2}\)
(B) \(\frac{\pi}{3}\)
(C) \(\frac{\pi}{4}\)
(D) \(\frac{3\pi}{4}\)
Answer: (C) \(\frac{\pi}{4}\)
Let's simplify the given expression:
tan\(^{-1}\)\(\frac{x}{y}\) - tan\(^{-1}\)\(\frac{x-y}{x+y}\)
We can divide the numerator and denominator of the second term by y:
= tan\(^{-1}\)\(\frac{x}{y}\) - tan\(^{-1}\)\(\frac{\frac{x}{y}-\frac{y}{y}}{\frac{x}{y}+\frac{y}{y}}\)
= tan\(^{-1}\)\(\frac{x}{y}\) - tan\(^{-1}\)\(\frac{\frac{x}{y}-1}{1+\frac{x}{y}}\)
Now, we use the inverse trigonometric identity: tan\(^{-1}\)A - tan\(^{-1}\)B = tan\(^{-1}\)\(\frac{A-B}{1+AB}\).
Also, we know that tan\(^{-1}\)\(\frac{A-B}{1+AB}\) can be written as tan\(^{-1}\)A - tan\(^{-1}\)B.
In our expression, let A = \(\frac{x}{y}\) and B = 1.
So, tan\(^{-1}\)\(\frac{\frac{x}{y}-1}{1+\frac{x}{y}}\) can be rewritten as tan\(^{-1}\)\(\frac{x}{y}\) - tan\(^{-1}\)(1).
Substitute this back into the original expression:
= tan\(^{-1}\)\(\frac{x}{y}\) - (tan\(^{-1}\)\(\frac{x}{y}\) - tan\(^{-1}\)(1))
= tan\(^{-1}\)\(\frac{x}{y}\) - tan\(^{-1}\)\(\frac{x}{y}\) + tan\(^{-1}\)(1)
= tan\(^{-1}\)(1)
We know that tan\(^{-1}\)(1) = \(\frac{\pi}{4}\).
Therefore, the expression simplifies to \(\frac{\pi}{4}\).
In simple words: We have two inverse tangent terms. We change the second term by dividing its top and bottom parts by y. This makes it look like the formula for tan\(^{-1}\)A - tan\(^{-1}\)B. When we apply this, the first part of the expression cancels out, leaving us with just tan\(^{-1}\)(1). We know that tan\(^{-1}\)(1) is equal to \(\frac{\pi}{4}\).

🎯 Exam Tip: Always look for ways to apply fundamental inverse trigonometric identities, especially the sum/difference formulas for tan\(^{-1}\), as they can simplify complex expressions into much simpler forms. Recognizing \(\frac{A-B}{1+AB}\) as tan\(^{-1}\)A - tan\(^{-1}\)B is key here.