GSEB Class 12 Maths Solutions Chapter 2 ત્રિકોણમિતીય પ્રતિવિધેયો Exercise 2.1

નીચેના પ્રતિવિધેય માટે તેની મુખ્ય કિંમત શોધો :

 

Question 1. \(\sin ^{-1}\left(\frac{-1}{2}\right)\)
Answer:Let's assume \(y = \sin ^{-1}\left(\frac{-1}{2}\right)\). So, \(\sin y = -\frac{1}{2}\). The principal value range for the \(\sin^{-1}\) function is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). We know that \(\sin\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac{1}{2}\). Thus, the principal value of \(\sin^{-1}\left(\frac{-1}{2}\right)\) is \(-\frac{\pi}{6}\).
In simple words: To find the main value, we set the expression equal to 'y'. Then we find the angle 'y' in the allowed range where the sine of 'y' gives the number inside the function.

🎯 Exam Tip: Always remember the principal value range for inverse trigonometric functions before finding the solution, as it is crucial for accuracy.

 

Question 2. \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
Answer:Let's assume \(y = \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\). So, \(\cos y = \frac{\sqrt{3}}{2}\). The principal value range for the \(\cos^{-1}\) function is \([0, \pi]\). We know that \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). Thus, the principal value of \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\) is \(\frac{\pi}{6}\).
In simple words: We set the expression to 'y'. Then we find the angle 'y' within the allowed range where the cosine of 'y' equals the given value.

🎯 Exam Tip: Knowing the standard trigonometric values for common angles (like \(\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}\)) is essential for quickly solving these problems.

 

Question 3. \(\text{cosec}^{-1}(2)\)
Answer:Let's assume \(y = \text{cosec}^{-1}(2)\). So, \(\text{cosec }y = 2\). The principal value range for the \(\text{cosec}^{-1}\) function is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\). We know that \(\text{cosec}\left(\frac{\pi}{6}\right) = 2\). Therefore, the principal value of \(\text{cosec}^{-1}(2)\) is \(\frac{\pi}{6}\).
In simple words: We let the expression be 'y', so cosec 'y' is 2. We find the angle 'y' in the correct range for which cosec 'y' is 2.

🎯 Exam Tip: Remember that \(\text{cosec}^{-1}(x)\) is defined for all real numbers except the interval \((-1, 1)\) and its range excludes 0.

 

Question 4. \(\tan^{-1}(-\sqrt{3})\)
Answer:Let's assume \(y = \tan^{-1}(-\sqrt{3})\). So, \(\tan y = -\sqrt{3}\). The principal value range for the \(\tan^{-1}\) function is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). We know that \(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\). Therefore, the principal value of \(\tan^{-1}(-\sqrt{3})\) is \(-\frac{\pi}{3}\).
In simple words: We set the function to 'y', so tan 'y' is -\(\sqrt{3}\). We then find the angle 'y' in the principal range for which tan 'y' is -\(\sqrt{3}\).

🎯 Exam Tip: The range of \(\tan^{-1}\) is an open interval, meaning the endpoints are not included. Pay attention to the sign of the value inside the function.

 

Question 5. \(\cos^{-1}\left(\frac{-1}{2}\right)\)
Answer:Let's assume \(y = \cos^{-1}\left(\frac{-1}{2}\right)\). So, \(\cos y = \frac{-1}{2}\). The principal value range for the \(\cos^{-1}\) function is \([0, \pi]\). We know that \(\cos\left(\pi-\frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}\). So, \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\). Therefore, the principal value of \(\cos^{-1}\left(\frac{-1}{2}\right)\) is \(\frac{2\pi}{3}\).
In simple words: We set 'y' to the function. Since the value is negative, we find the angle in the second quadrant (for cosine) that gives this value.

🎯 Exam Tip: When the value inside \(\cos^{-1}\) is negative, the angle will be in the second quadrant, typically calculated as \(\pi - \theta\), where \(\theta\) is the positive angle for the absolute value.

 

Question 6. \(\tan^{-1}(-1)\)
Answer:Let's assume \(y = \tan^{-1}(-1)\). So, \(\tan y = -1\). The principal value range for the \(\tan^{-1}\) function is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). We know that \(\tan\left(-\frac{\pi}{4}\right) = -1\). Therefore, the principal value of \(\tan^{-1}(-1)\) is \(-\frac{\pi}{4}\).
In simple words: We set 'y' to the function. We then find the angle 'y' in the allowed range where the tangent of 'y' is -1.

🎯 Exam Tip: For negative values in \(\tan^{-1}\), the principal value will be a negative angle in the fourth quadrant (between \(-\frac{\pi}{2}\) and \(0\)).

 

Question 7. \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\)
Answer:Let's assume \(y = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\). So, \(\sec y = \frac{2}{\sqrt{3}}\). The principal value range for the \(\sec^{-1}\) function is \([0, \pi] - \left\{\frac{\pi}{2}\right\}\). We know that \(\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}\). Therefore, the principal value of \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\) is \(\frac{\pi}{6}\).
In simple words: We set the function to 'y', so sec 'y' is \(\frac{2}{\sqrt{3}}\). We find the angle 'y' in its specific range that satisfies this.

🎯 Exam Tip: Be careful with the range of \(\sec^{-1}\), which excludes \(\frac{\pi}{2}\) because secant is undefined at that angle.

 

Question 8. \(\cot^{-1}(\sqrt{3})\)
Answer:Let's assume \(y = \cot^{-1}(\sqrt{3})\). So, \(\cot y = \sqrt{3}\). The principal value range for the \(\cot^{-1}\) function is \((0, \pi)\). We know that \(\cot\left(\frac{\pi}{6}\right) = \sqrt{3}\). Therefore, the principal value of \(\cot^{-1}(\sqrt{3})\) is \(\frac{\pi}{6}\).
In simple words: We set 'y' to the function. We then find the angle 'y' in the allowed range where the cotangent of 'y' is \(\sqrt{3}\).

🎯 Exam Tip: The range of \(\cot^{-1}\) is an open interval \((0, \pi)\), which means the angle will always be positive and less than \(\pi\).

 

Question 9. \(\cos^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)
Answer:Let's assume \(y = \cos^{-1}\left(\frac{-1}{\sqrt{2}}\right)\). So, \(\cos y = -\frac{1}{\sqrt{2}}\). The principal value range for the \(\cos^{-1}\) function is \([0, \pi]\). We know that \(\cos\left(\pi-\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}\). So, \(\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}\). Therefore, the principal value of \(\cos^{-1}\left(\frac{-1}{\sqrt{2}}\right)\) is \(\frac{3\pi}{4}\).
In simple words: We set 'y' to the function. Since the value is negative, we look for an angle in the second quadrant where its cosine is -\(\frac{1}{\sqrt{2}}\).

🎯 Exam Tip: For negative arguments in \(\cos^{-1}\), always use the formula \(\pi - \theta\) to find the correct principal value in the range \([0, \pi]\).

 

Question 10. \(\text{cosec}^{-1}(-\sqrt{2})\)
Answer:Let's assume \(y = \text{cosec}^{-1}(-\sqrt{2})\). So, \(\text{cosec }y = -\sqrt{2}\). The principal value range for the \(\text{cosec}^{-1}\) function is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\). We know that \(\text{cosec}\left(-\frac{\pi}{4}\right) = -\sqrt{2}\). Therefore, the principal value of \(\text{cosec}^{-1}(-\sqrt{2})\) is \(-\frac{\pi}{4}\).
In simple words: We set 'y' to the function. Since it's negative, we find the negative angle in the fourth quadrant whose cosecant is -\(\sqrt{2}\).

🎯 Exam Tip: For negative values in \(\text{cosec}^{-1}\), the principal value will be a negative angle in the fourth quadrant, similar to \(\sin^{-1}\).

નીચેની અભિવ્યક્તિઓનું મૂલ્ય મેળવો :

 

Question 11. \(\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)\)
Answer:We know the principal value ranges: For \(\tan^{-1}\): \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). So, \(\tan^{-1}(1) = \frac{\pi}{4}\). For \(\cos^{-1}\): \([0, \pi]\). So, \(\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\). For \(\sin^{-1}\): \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). So, \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\). Now, we add these values: \(\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right)\) \(= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}\) To sum these fractions, we find a common denominator, which is 12: \(= \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12}\) \(= \frac{3\pi + 8\pi - 2\pi}{12}\) \(= \frac{9\pi}{12}\) \(= \frac{3\pi}{4}\)
In simple words: First, find the main value for each inverse trigonometric function separately. Then, add these values together by finding a common denominator for the fractions.

🎯 Exam Tip: Break down complex problems into smaller parts. Calculate each inverse function's principal value correctly before performing the final arithmetic operation.

 

Question 12. \(\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)\)
Answer:We know the principal value ranges: For \(\cos^{-1}\): \([0, \pi]\). So, \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\). For \(\sin^{-1}\): \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). So, \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\). Now, substitute these values into the expression: \(\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2\left(\frac{\pi}{6}\right)\) \(= \frac{\pi}{3} + \frac{2\pi}{6}\) \(= \frac{\pi}{3} + \frac{\pi}{3}\) \(= \frac{2\pi}{3}\)
In simple words: Find the principal value of each inverse function. Then, put these values into the expression and solve it step-by-step.

🎯 Exam Tip: Pay close attention to coefficients (like '2' in front of \(\sin^{-1}\)) and perform multiplication before addition to avoid errors.

પ્રશ્નો 13 તથા 14 માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી યોગ્ય વિકલ્પ પસંદ કરો :

 

Question 13. જો \(\sin^{-1}x = y\) હોય, તો
(A) \(0 \leq y \leq \pi\)
(B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
(C) \(0 < y < \pi\)
(D) \(-\frac{\pi}{2} < y < \frac{\pi}{2}\)
Answer: (B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
We know that if \(\sin^{-1}x = y\), then \(x = \sin y\). The principal value range for the \(\sin^{-1}\) function is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Therefore, the correct option is (B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).
In simple words: The question asks for the allowed range of 'y' when 'y' is the output of the \(\sin^{-1}x\) function. This range is a fixed rule for inverse sine.

🎯 Exam Tip: This question tests your fundamental knowledge of the principal value ranges of inverse trigonometric functions. Memorizing these ranges is critical.

 

Question 14. \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2)\) નું મૂલ્ય
(A) \(\pi\)
(B) \(-\frac{\pi}{3}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{2\pi}{3}\)
Answer: (B) \(-\frac{\pi}{3}\)
We know the principal value ranges: For \(\tan^{-1}\): \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). So, \(\tan^{-1}\sqrt{3} = \frac{\pi}{3}\). For \(\sec^{-1}\): \([0, \pi] - \left\{\frac{\pi}{2}\right\}\). So, \(\sec^{-1}(-2) = \pi - \sec^{-1}(2)\) since \(\sec^{-1}(-x) = \pi - \sec^{-1}(x)\). We know \(\sec^{-1}(2) = \frac{\pi}{3}\). Therefore, \(\sec^{-1}(-2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\). Now, substitute these values into the expression: \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3}\) \(= \frac{\pi - 2\pi}{3}\) \(= -\frac{\pi}{3}\) Thus, option (B) is correct.
In simple words: First, find the main value for \(\tan^{-1}\sqrt{3}\). Then, find the main value for \(\sec^{-1}(-2)\). Subtract the second value from the first to get the final answer.

🎯 Exam Tip: Be mindful of the negative argument in inverse functions. For \(\sec^{-1}(-x)\), remember to use the property \(\pi - \sec^{-1}(x)\) to find the correct principal value.