RBSE Solutions Class 10 Maths Chapter 6 त्रिकोणमितीय अनुपात More Ques

विविध प्रश्नमाला 6

वस्तुनिष्ठ प्रश्न (1 से 5 तक)

Question 1. \( \tan 60^\circ \) का मान है
(क) 3
(ख) \( \frac{1}{\sqrt{3}} \)
(ग) 1
(घ) \( \infty \)
Answer: (क) 3
In simple words: \( \tan 60^\circ \) is a standard trigonometric value. It represents the ratio of the side opposite to the 60-degree angle to the side adjacent to it in a right-angled triangle. This value is equal to \( \sqrt{3} \).

🎯 Exam Tip: Memorize the standard trigonometric values for \( 0^\circ, 30^\circ, 45^\circ, 60^\circ \), and \( 90^\circ \) as they appear frequently in questions.

Question 2. \( 2 \sin^2 60^\circ \cos 60^\circ \) का मान होगा
(क) \( \frac{4}{3} \)
(ख) \( \frac{3}{2} \)
(ग) \( \frac{3}{4} \)
(घ) \( \frac{1}{3} \)
Answer: (ग) \( \frac{3}{4} \)
In simple words: First, find the values of \( \sin 60^\circ \) and \( \cos 60^\circ \). Then square \( \sin 60^\circ \) and multiply all the numbers together. This calculation helps us find the final numerical value of the expression.

🎯 Exam Tip: Remember to apply the square to the entire trigonometric value, not just the angle, and follow the order of operations.

Question 3. यदि \( \operatorname{cosec} \theta = \frac{2}{\sqrt{3}} \) हो, तो \( \theta \) का मान है-
(क) \( \frac{\pi}{4} \)
(ख) \( \frac{\pi}{3} \)
(ग) \( \frac{\pi}{2} \)
(घ) \( \frac{\pi}{6} \)
Answer: (ख) \( \frac{\pi}{3} \)
In simple words: When \( \operatorname{cosec} \theta \) equals \( \frac{2}{\sqrt{3}} \), it means the angle \( \theta \) is \( 60^\circ \). In radians, \( 60^\circ \) is written as \( \frac{\pi}{3} \). This is a common angle from the trigonometric table.

🎯 Exam Tip: Be comfortable converting between degrees and radians, especially for standard angles. \( \pi \) radians equals \( 180^\circ \).

Question 4. \( \cos^2 45^\circ \) का मान होगा-
Answer: \( \frac{1}{2} \)
In simple words: To find the value, first remember that \( \cos 45^\circ \) is \( \frac{1}{\sqrt{2}} \). Then, simply square this value. Squaring \( \frac{1}{\sqrt{2}} \) gives \( \frac{1}{2} \).

🎯 Exam Tip: Squaring a fraction means squaring both the numerator and the denominator. Always simplify your final answer.

Question 5. यदि \( \theta = 45^\circ \) हो, तो \( \frac{1-\cos 2 \theta}{\sin 2 \theta} \) का मान है-
(क) 0
(ख) 1
(ग) 2
(घ) \( \infty \)
Answer: (ख) 1
In simple words: First, put \( \theta = 45^\circ \) into the expression to find \( 2\theta = 90^\circ \). Then, calculate \( \cos 90^\circ \) and \( \sin 90^\circ \). After putting these values, simplify the fraction. This will lead to the answer 1.

🎯 Exam Tip: Always calculate the interior angle \( 2\theta \) first before applying the trigonometric functions to avoid errors.

सिद्ध कीजिए (6 से 10 तक)

Question 6. सिद्ध कीजिए कि \( \cos 60^\circ = 2 \cos^2 30^\circ - 1 \)
Answer:
L.H.S. \( = \cos 60^\circ \)
\( = \frac{1}{2} \)

R.H.S. \( = 2 \cos^2 30^\circ - 1 \)
\( = 2 \left( \frac{\sqrt{3}}{2} \right)^2 - 1 \)
\( = 2 \left( \frac{3}{4} \right) - 1 \)
\( = \frac{3}{2} - 1 \)
\( = \frac{1}{2} \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: We need to show that both sides of the equation give the same value. Calculate the left side and the right side separately using the known values of cosine for \( 60^\circ \) and \( 30^\circ \). When both calculations result in \( \frac{1}{2} \), the proof is complete.

🎯 Exam Tip: Always show your steps clearly for both the Left Hand Side (L.H.S.) and the Right Hand Side (R.H.S.) to earn full marks in proof-based questions.

Question 7. सिद्ध कीजिए कि \( \sin 60^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \)
Answer:
L.H.S. \( = \sin 60^\circ \)
\( = \frac{\sqrt{3}}{2} \)

R.H.S. \( = \frac{2 \tan 30^\circ}{1+\tan^2 30^\circ} \)
\( = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 + \left( \frac{1}{\sqrt{3}} \right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} \)
\( = \frac{2}{\sqrt{3}} \times \frac{3}{4} \)
\( = \frac{6}{4\sqrt{3}} \)
\( = \frac{3}{2\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{2\sqrt{3}\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{2 \times 3} \)
\( = \frac{\sqrt{3}}{2} \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: To prove this, calculate the value of \( \sin 60^\circ \) for the left side. For the right side, substitute the value of \( \tan 30^\circ \) and simplify the complex fraction. Both sides will simplify to \( \frac{\sqrt{3}}{2} \). This type of identity is also related to the double angle formula for sine.

🎯 Exam Tip: When dealing with complex fractions, remember to multiply by the reciprocal of the denominator. Rationalizing the denominator can make the final comparison easier.

Question 8. सिद्ध कीजिए कि \( \cos 60^{\circ}=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \)
Answer:
L.H.S. \( = \cos 60^\circ \)
\( = \frac{1}{2} \)

R.H.S. \( = \frac{1-\tan^2 30^\circ}{1+\tan^2 30^\circ} \)
\( = \frac{1-\left(\frac{1}{\sqrt{3}}\right)^2}{1+\left(\frac{1}{\sqrt{3}}\right)^2} \)
\( = \frac{1-\frac{1}{3}}{1+\frac{1}{3}} \)
\( = \frac{\frac{3-1}{3}}{\frac{3+1}{3}} \)
\( = \frac{\frac{2}{3}}{\frac{4}{3}} \)
\( = \frac{2}{3} \times \frac{3}{4} \)
\( = \frac{1}{2} \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: This proof involves showing that the values of \( \cos 60^\circ \) and the given expression with \( \tan 30^\circ \) are the same. After putting in the value of \( \tan 30^\circ \), simplify the numerator and the denominator separately. This identity is also a form of the double angle formula for cosine.

🎯 Exam Tip: Remember that \( \tan^2 \theta \) means \( (\tan \theta)^2 \). Simplify the numerator and denominator before dividing them.

Question 9. सिद्ध कीजिए कि \( (\sqrt{2})^2 = 2 \)
Answer:
L.H.S. \( = (\sqrt{2})^2 \)
\( = 2 \)

R.H.S. \( = 2 \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: To prove this, we simply square the number \( \sqrt{2} \). When you square a square root, you get the number inside the root. In this case, \( \sqrt{2} \times \sqrt{2} = 2 \).

🎯 Exam Tip: The square root symbol \( \sqrt{} \) indicates the principal (positive) square root. Squaring a number and taking its square root are inverse operations.

Question 10. सिद्ध कीजिए कि \( 4 \tan 30^\circ \sin 45^\circ \sin 60^\circ \sin 90^\circ = \sqrt{2} \)
Answer:
L.H.S. \( = 4 \tan 30^\circ \sin 45^\circ \sin 60^\circ \sin 90^\circ \)
\( = 4 \times \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \times 1 \)
\( = \frac{4 \times 1 \times 1 \times \sqrt{3} \times 1}{\sqrt{3} \times \sqrt{2} \times 2} \)
\( = \frac{4\sqrt{3}}{2\sqrt{6}} \)
\( = \frac{2}{\sqrt{2}} \)
\( = \frac{2\sqrt{2}}{\sqrt{2}\sqrt{2}} \)
\( = \frac{2\sqrt{2}}{2} \)
\( = \sqrt{2} \)
R.H.S. \( = \sqrt{2} \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: Substitute the known trigonometric values for each angle into the expression. Then, multiply all the numbers in the numerator and denominator. Simplify the resulting fraction by rationalizing the denominator if needed. The final result will be \( \sqrt{2} \).

🎯 Exam Tip: Pay close attention to simplifying square roots in the denominator by multiplying both the numerator and denominator by the radical.

Question 11. \( \sin^2 60^\circ \cot 60^\circ \) का मान ज्ञात कीजिए।
Answer:
\( \sin^2 60^\circ \cot 60^\circ \)
\( = \left( \frac{\sqrt{3}}{2} \right)^2 \times \frac{1}{\sqrt{3}} \)
\( = \frac{3}{4} \times \frac{1}{\sqrt{3}} \)
\( = \frac{3}{4\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{4\sqrt{3}\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{4 \times 3} \)
\( = \frac{\sqrt{3}}{4} \)
In simple words: First, substitute the values of \( \sin 60^\circ \) and \( \cot 60^\circ \). Remember to square the sine value. Then multiply the resulting numbers and simplify the fraction. Rationalize the denominator to get the final answer.

🎯 Exam Tip: Always double-check your trigonometric values from the standard table. A small mistake in the initial value can lead to an incorrect final answer.

Question 12. \( 4 \cos^3 30^\circ – 3 \cos 30^\circ \) का मान ज्ञात कीजिए।
Answer:
\( 4 \cos^3 30^\circ – 3 \cos 30^\circ \)
\( = 4 \left( \frac{\sqrt{3}}{2} \right)^3 - 3 \left( \frac{\sqrt{3}}{2} \right) \)
\( = 4 \left( \frac{3\sqrt{3}}{8} \right) - \frac{3\sqrt{3}}{2} \)
\( = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} \)
\( = 0 \)
In simple words: Substitute the value of \( \cos 30^\circ \) into the expression. Calculate the cubic term and the simple term separately. Then subtract them to find the final answer. This is an identity related to \( \cos 3\theta \).

🎯 Exam Tip: When calculating higher powers of trigonometric values, remember to apply the power to both the numerator and the denominator of the fraction.

Question 13. यदि \( \cot \theta = \frac{1}{\sqrt{3}} \) हो, तो सिद्ध कीजिए \( \frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{3}{5} \)
Answer:
दिया है \( \cot \theta = \frac{1}{\sqrt{3}} \)
\( \implies \cot \theta = \cot 60^\circ \)
\( \implies \theta = 60^\circ \)
अब, L.H.S. \( = \frac{1-\cos^2 \theta}{2-\sin^2 \theta} \)
\( = \frac{1-\cos^2 60^\circ}{2-\sin^2 60^\circ} \)
\( = \frac{1-\left(\frac{1}{2}\right)^2}{2-\left(\frac{\sqrt{3}}{2}\right)^2} \)
\( = \frac{1-\frac{1}{4}}{2-\frac{3}{4}} \)
\( = \frac{\frac{4-1}{4}}{\frac{8-3}{4}} \)
\( = \frac{\frac{3}{4}}{\frac{5}{4}} \)
\( = \frac{3}{4} \times \frac{4}{5} \)
\( = \frac{3}{5} \)
R.H.S. \( = \frac{3}{5} \)
अतः, L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: First, find the angle \( \theta \) using the given value of \( \cot \theta \). Then, put this angle into the left side of the equation. Calculate the values of \( \sin \theta \) and \( \cos \theta \) and simplify the fraction. The result will match the right side, \( \frac{3}{5} \).

🎯 Exam Tip: When \( \sin^2 \theta \) or \( \cos^2 \theta \) appears, remember that it's \( (\sin \theta)^2 \) or \( (\cos \theta)^2 \). This is crucial for correct substitution and calculation.

Question 14. सिद्ध कीजिए कि \( 3(\tan^2 30^\circ + \cot^2 30^\circ) – 8(\sin^2 45^\circ + \cos^2 30^\circ) = 0 \)
Answer:
L.H.S. \( = 3(\tan^2 30^\circ + \cot^2 30^\circ) – 8(\sin^2 45^\circ + \cos^2 30^\circ) \)
\( = 3 \left( \left(\frac{1}{\sqrt{3}}\right)^2 + (\sqrt{3})^2 \right) - 8 \left( \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \right) \)
\( = 3 \left( \frac{1}{3} + 3 \right) - 8 \left( \frac{1}{2} + \frac{3}{4} \right) \)
\( = 3 \left( \frac{1+9}{3} \right) - 8 \left( \frac{2+3}{4} \right) \)
\( = 3 \left( \frac{10}{3} \right) - 8 \left( \frac{5}{4} \right) \)
\( = 10 - 10 \)
\( = 0 \)
R.H.S. \( = 0 \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: Replace all trigonometric functions with their known values for \( 30^\circ \) and \( 45^\circ \). Carefully perform the squaring and additions inside the brackets first. Then, multiply by 3 and 8 respectively, and finally subtract the two main terms. The result should be zero.

🎯 Exam Tip: Be careful with the order of operations: powers first, then multiplication, then addition/subtraction. Double-check fraction arithmetic.

Question 15. सिद्ध कीजिए कि \( 4(\sin^4 30^\circ + \cos^4 60^\circ) - 3(\cos^2 45^\circ - \sin^2 90^\circ) = \frac{15}{4} \)
Answer:
L.H.S. \( = 4(\sin^4 30^\circ + \cos^4 60^\circ) - 3(\cos^2 45^\circ - \sin^2 90^\circ) \)
\( = 4 \left[ \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4 \right] - 3 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 - (1)^2 \right] \)
\( = 4 \left[ \frac{1}{16} + \frac{1}{16} \right] - 3 \left[ \frac{1}{2} - 1 \right] \)
\( = 4 \left[ \frac{2}{16} \right] - 3 \left[ -\frac{1}{2} \right] \)
\( = 4 \left( \frac{1}{8} \right) + \frac{3}{2} \)
\( = \frac{1}{2} + \frac{3}{2} \)
\( = \frac{1+3}{2} \)
\( = \frac{4}{2} \)
\( = 2 \)
The given identity \( = \frac{15}{4} \) is incorrect based on calculations. The correct result is \( 2 \).
In simple words: Replace the sine and cosine values with their fractions and 1. Perform the powers and then add or subtract within the brackets. Then multiply by 4 and 3. Finally, add the two main parts. The calculation shows the expression equals 2.

🎯 Exam Tip: Double-check the powers, especially \( \sin^4 30^\circ \) and \( \cos^4 60^\circ \). Sometimes, if an identity doesn't match the expected value, it might indicate a typo in the question itself. Present your calculation accurately.

Question 16. सिद्ध कीजिए कि \( \frac{\cos 30^\circ + \sin 60^\circ}{1+ \cos 60^\circ + \sin 30^\circ} = \frac{\sqrt{3}}{2} \)
Answer:
L.H.S. \( = \frac{\cos 30^\circ + \sin 60^\circ}{1+ \cos 60^\circ + \sin 30^\circ} \)
\( = \frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}}{1 + \frac{1}{2} + \frac{1}{2}} \)
\( = \frac{\frac{2\sqrt{3}}{2}}{1 + 1} \)
\( = \frac{\sqrt{3}}{2} \)
R.H.S. \( = \frac{\sqrt{3}}{2} \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: Substitute the standard trigonometric values into the expression. Simplify the numerator by adding the terms and simplify the denominator by adding the terms. The final fraction will directly give \( \frac{\sqrt{3}}{2} \).

🎯 Exam Tip: When simplifying fractions with multiple terms in the numerator or denominator, treat them as separate calculations before combining.

Question 17. सिद्ध कीजिए कि \( 2(\cos^2 45^\circ + \tan^2 60^\circ) – 6(\sin^2 45^\circ – \tan^2 30^\circ) = 6 \)
Answer:
L.H.S. \( = 2(\cos^2 45^\circ + \tan^2 60^\circ) – 6(\sin^2 45^\circ – \tan^2 30^\circ) \)
\( = 2 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 + (\sqrt{3})^2 \right] - 6 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2 \right] \)
\( = 2 \left[ \frac{1}{2} + 3 \right] - 6 \left[ \frac{1}{2} - \frac{1}{3} \right] \)
\( = 2 \left[ \frac{1+6}{2} \right] - 6 \left[ \frac{3-2}{6} \right] \)
\( = 2 \left( \frac{7}{2} \right) - 6 \left( \frac{1}{6} \right) \)
\( = 7 - 1 \)
\( = 6 \)
R.H.S. \( = 6 \)
Since L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: Replace all trigonometric terms with their correct values and perform the squares. Then simplify the expressions inside each bracket. Multiply the first bracket by 2 and the second by 6. Finally, subtract the two results to get 6.

🎯 Exam Tip: Be careful with negative signs, especially when subtracting a term like \( 1 - 1/2 \). Always perform calculations step-by-step to avoid errors.

अन्य महत्त्वपूर्ण प्रश्न

वस्तुनिष्ठ प्रश्न

Question 1. \( 2 \operatorname{cosec} 30^\circ \sec 30^\circ \) बराबर है-
(A) \( \frac{2}{\sqrt{3}} \)
(B) \( \frac{8}{\sqrt{3}} \)
(C) 8
(D) \( \frac{\sqrt{3}}{2} \)
Answer: (B) \( \frac{8}{\sqrt{3}} \)
In simple words: First, find the values of \( \operatorname{cosec} 30^\circ \) and \( \sec 30^\circ \). Then multiply these two values by 2. This will give you the answer \( \frac{8}{\sqrt{3}} \).

🎯 Exam Tip: Remember that \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \). Use these reciprocal relationships when values are not directly memorized.

Question 2. यदि \( \sin \theta = \frac{\sqrt{3}}{2} \) हो, तो \( \theta \) का मान है-
Answer: \( 60^\circ \)
In simple words: The value \( \frac{\sqrt{3}}{2} \) is a common trigonometric value. When \( \sin \theta \) equals this value, the angle \( \theta \) is \( 60^\circ \). You can find this by looking at a standard trigonometric table.

🎯 Exam Tip: Recognizing common trigonometric values for angles like \( 30^\circ, 45^\circ, 60^\circ \) is fundamental for quick calculations.

Question 3. यदि \( \theta = 30^\circ \) हो, तो \( \frac{1-\sin ^{2} 2 \theta}{\cos 2 \theta} \) का मान होगा-
(A) 1
(B) -1
(C) 2
(D) \( \frac{1}{2} \)
Answer: (D) \( \frac{1}{2} \)
In simple words: First, find \( 2\theta \) by putting \( \theta = 30^\circ \). This makes \( 2\theta = 60^\circ \). Then, use the identity \( 1-\sin^2 A = \cos^2 A \) to simplify the top part of the fraction. This leaves \( \cos 60^\circ \) as the answer.

🎯 Exam Tip: Using trigonometric identities like \( \sin^2 A + \cos^2 A = 1 \) can greatly simplify expressions before substituting values.

Question 4. \( \cot 30^\circ \) का मान है-
(A) \( \frac{1}{\sqrt{3}} \)
(B) \( \sqrt{3} \)
(C) \( \frac{\sqrt{3}}{2} \)
(D) \( \sqrt{2} \)
Answer: (B) \( \sqrt{3} \)
In simple words: \( \cot 30^\circ \) is the reciprocal of \( \tan 30^\circ \). Since \( \tan 30^\circ \) is \( \frac{1}{\sqrt{3}} \), its reciprocal will be \( \sqrt{3} \).

🎯 Exam Tip: Remember the reciprocal relationships: \( \cot \theta = \frac{1}{\tan \theta} \), \( \sec \theta = \frac{1}{\cos \theta} \), \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \).

Question 5. \( \sec 45^\circ \) का मान है-
(A) \( \sqrt{2} \)
(B) 1
(C) \( \frac{1}{\sqrt{3}} \)
(D) \( \sqrt{3} \)
Answer: (A) \( \sqrt{2} \)
In simple words: \( \sec 45^\circ \) is the reciprocal of \( \cos 45^\circ \). Since \( \cos 45^\circ \) is \( \frac{1}{\sqrt{2}} \), its reciprocal is \( \sqrt{2} \).

🎯 Exam Tip: Always relate secant to cosine for easier recall of values. \( \sec 45^\circ \) is a frequently tested value.

Question 6. \( \sin^2 45^\circ \) का मान है-
(A) \( \frac{1}{\sqrt{2}} \)
(B) \( \frac{\sqrt{3}}{2} \)
(C) \( \frac{1}{2} \)
(D) \( \frac{1}{\sqrt{3}} \)
Answer: (C) \( \frac{1}{2} \)
In simple words: First, find the value of \( \sin 45^\circ \), which is \( \frac{1}{\sqrt{2}} \). Then, square this value. Squaring \( \frac{1}{\sqrt{2}} \) gives \( \frac{1}{2} \).

🎯 Exam Tip: Remember that squaring a fraction means squaring both the top and bottom parts. \( (\frac{1}{\sqrt{2}})^2 = \frac{1^2}{(\sqrt{2})^2} = \frac{1}{2} \).

Question 7. \( \cot^2 60^\circ \) का मान होता है-
(A) \( \frac{1}{3} \)
(B) 3
(C) 1
(D) \( \infty \)
Answer: (A) \( \frac{1}{3} \)
In simple words: To find this, first recall that \( \cot 60^\circ \) is \( \frac{1}{\sqrt{3}} \). Then, square this value. Squaring \( \frac{1}{\sqrt{3}} \) results in \( \frac{1}{3} \).

🎯 Exam Tip: Make sure the square is applied to the entire value of the trigonometric function, not just a part of it.

Question 9. यदि \( \sec \theta=\frac{2}{\sqrt{3}} \), तो \( \theta \) का मान है-
(A) \( \frac{\pi}{3} \)
(B) \( \frac{\pi}{6} \)
(C) \( \frac{\pi}{2} \)
(D) \( \frac{\pi}{4} \)
Answer: (B) \( \frac{\pi}{6} \)
In simple words: If \( \sec \theta \) is \( \frac{2}{\sqrt{3}} \), it means \( \cos \theta \) is \( \frac{\sqrt{3}}{2} \). This is the value for \( \cos 30^\circ \). In radians, \( 30^\circ \) is \( \frac{\pi}{6} \).

🎯 Exam Tip: Knowing the reciprocal identities is important. If you know \( \cos \theta \), you can easily find \( \sec \theta \).

Question 10. \( 2 \sin 30^\circ \cos 30^\circ \) का मान है-
(A) \( \frac{\sqrt{3}}{4} \)
(B) \( \frac{3}{2} \)
(C) \( \frac{\sqrt{3}}{2} \)
(D) \( \frac{3}{4} \)
Answer: (C) \( \frac{\sqrt{3}}{2} \)
In simple words: Substitute the values for \( \sin 30^\circ \) and \( \cos 30^\circ \) into the expression. Then multiply the numbers. This specific expression is also equal to \( \sin(2 \times 30^\circ) \) or \( \sin 60^\circ \).

🎯 Exam Tip: Recognize the double angle formula \( 2 \sin A \cos A = \sin 2A \). This can save time in calculations.

अतिलघूत्तरात्मक प्रश्न

Question 1. यदि \( \tan \theta = 1 \) हो, तो \( \sec \theta \) का मान ज्ञात कीजिए।
Answer:
दिया है, \( \tan \theta = 1 \)
\( \implies \tan \theta = \tan 45^\circ \)
\( \implies \theta = 45^\circ \)
अतः, \( \sec \theta = \sec 45^\circ \)
\( = \sqrt{2} \)
In simple words: First, find the angle \( \theta \) using the given value of \( \tan \theta \). Since \( \tan 45^\circ = 1 \), \( \theta \) is \( 45^\circ \). Then, find the value of \( \sec 45^\circ \), which is \( \sqrt{2} \).

🎯 Exam Tip: Remember the relationship between tangent and angle. You can also use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \) to find \( \sec \theta \) directly.

Question 2. यदि \( \sin \theta = \cos \theta \) हो, तो \( \theta \) का मान ज्ञात कीजिए।
Answer:
दिया है, \( \sin \theta = \cos \theta \)
दोनों ओर \( \cos \theta \) से भाग करने पर:
\( \frac{\sin \theta}{\cos \theta} = 1 \)
\( \implies \tan \theta = 1 \)
\( \implies \tan \theta = \tan 45^\circ \)
\( \implies \theta = 45^\circ \)
In simple words: If \( \sin \theta \) and \( \cos \theta \) are equal, it means their ratio \( \frac{\sin \theta}{\cos \theta} \) is 1. Since \( \frac{\sin \theta}{\cos \theta} \) is \( \tan \theta \), we know \( \tan \theta = 1 \). This happens when \( \theta \) is \( 45^\circ \). This is true for \( 0^\circ \le \theta < 90^\circ \).

🎯 Exam Tip: Dividing both sides by \( \cos \theta \) is a standard technique when \( \sin \theta = \cos \theta \). Make sure \( \cos \theta \ne 0 \) for the division to be valid.

Question 4. \( \sec^2 30^\circ + \operatorname{cosec}^2 60^\circ \) का मान ज्ञात कीजिए।
Answer:
\( \sec^2 30^\circ + \operatorname{cosec}^2 60^\circ \)
\( = \left( \frac{2}{\sqrt{3}} \right)^2 + \left( \frac{2}{\sqrt{3}} \right)^2 \)
\( = \frac{4}{3} + \frac{4}{3} \)
\( = \frac{4+4}{3} \)
\( = \frac{8}{3} \)
In simple words: First, substitute the values for \( \sec 30^\circ \) and \( \operatorname{cosec} 60^\circ \). Square both values and then add the resulting fractions. This will give you \( \frac{8}{3} \).

🎯 Exam Tip: Note that \( \sec 30^\circ = \operatorname{cosec} 60^\circ \) because \( 30^\circ + 60^\circ = 90^\circ \). This relationship can simplify the thought process.

Question 5. \( \sin^2 60^\circ + \cos^2 30^\circ \) का मान लिखिए।
Answer:
\( \sin^2 60^\circ + \cos^2 30^\circ \)
\( = \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 \)
\( = \frac{3}{4} + \frac{3}{4} \)
\( = \frac{3+3}{4} \)
\( = \frac{6}{4} \)
\( = \frac{3}{2} \)
In simple words: Substitute the known values of \( \sin 60^\circ \) and \( \cos 30^\circ \). Square both terms and then add the two fractions together. The final sum will be \( \frac{3}{2} \).

🎯 Exam Tip: Remember that \( \sin 60^\circ = \cos 30^\circ \). This means the expression is simply \( 2 \times (\sin 60^\circ)^2 \).

Question 6. \( \tan^2 60^\circ + 3 \cos^2 30^\circ \) का मान ज्ञात कीजिए।
Answer:
\( \tan^2 60^\circ + 3 \cos^2 30^\circ \)
\( = (\sqrt{3})^2 + 3 \left( \frac{\sqrt{3}}{2} \right)^2 \)
\( = 3 + 3 \left( \frac{3}{4} \right) \)
\( = 3 + \frac{9}{4} \)
\( = \frac{12+9}{4} \)
\( = \frac{21}{4} \)
In simple words: Substitute the values of \( \tan 60^\circ \) and \( \cos 30^\circ \). Square the first term and square the second term before multiplying it by 3. Then, add the resulting numbers by finding a common denominator.

🎯 Exam Tip: Be careful with the multiplication before addition when dealing with a coefficient like '3' in front of a squared trigonometric term.

Question 7. \( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \) का मान ज्ञात कीजिए।
Answer:
\( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \)
\( = \left( \frac{\sqrt{3}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \right) \left( \frac{1}{2} \right) \)
\( = \frac{3}{4} + \frac{1}{4} \)
\( = \frac{3+1}{4} \)
\( = \frac{4}{4} \)
\( = 1 \)
In simple words: This expression is a formula for \( \sin(A+B) \). Substitute the values of \( \sin \) and \( \cos \) for \( 60^\circ \) and \( 30^\circ \). Multiply the terms and then add them. The result is 1.

🎯 Exam Tip: Recognize the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \). This allows for a quicker solution if you notice it. Here, \( A=60^\circ, B=30^\circ \).

Question 8. यदि \( \tan \theta = \sqrt{3} \) हो, तो \( \theta \) का मान लिखिए।
Answer:
दिया है, \( \tan \theta = \sqrt{3} \)
\( \implies \tan \theta = \tan 60^\circ \)
\( \implies \theta = 60^\circ \)
In simple words: When \( \tan \theta \) equals \( \sqrt{3} \), the angle \( \theta \) is \( 60^\circ \). This is a standard trigonometric value.

🎯 Exam Tip: Keep the basic trigonometric values for \( 30^\circ, 45^\circ \), and \( 60^\circ \) readily in mind. They are key for many problems.

लघूत्तरात्मक प्रश्न

Question 1. सिद्ध कीजिए \( 3 \tan^2 30^\circ - \frac{4}{3} \sin^2 60^\circ - \frac{1}{2} \operatorname{cosec}^2 45^\circ + \frac{4}{3} \sin^2 90^\circ = \frac{1}{3} \)
Answer:
L.H.S. \( = 3 \tan^2 30^\circ - \frac{4}{3} \sin^2 60^\circ - \frac{1}{2} \operatorname{cosec}^2 45^\circ + \frac{4}{3} \sin^2 90^\circ \)
\( = 3 \left( \frac{1}{\sqrt{3}} \right)^2 - \frac{4}{3} \left( \frac{\sqrt{3}}{2} \right)^2 - \frac{1}{2} (\sqrt{2})^2 + \frac{4}{3} (1)^2 \)
\( = 3 \left( \frac{1}{3} \right) - \frac{4}{3} \left( \frac{3}{4} \right) - \frac{1}{2} (2) + \frac{4}{3} (1) \)
\( = 1 - 1 - 1 + \frac{4}{3} \)
\( = -1 + \frac{4}{3} \)
\( = \frac{-3+4}{3} \)
\( = \frac{1}{3} \)
R.H.S. \( = \frac{1}{3} \)
अतः, L.H.S. \( = \) R.H.S., the identity is proven.
In simple words: Replace each trigonometric function with its exact numerical value. Carefully perform all the squaring, multiplication, and subtraction steps. Simplify the expression part by part. The final simplified value on the left side will be \( \frac{1}{3} \). This type of problem tests your precision with calculations.

🎯 Exam Tip: When an expression has many terms, evaluate each term separately first. Then combine them using addition and subtraction to minimize errors.

Question 2. यदि \( \tan 3x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ \) हो, तो \( x \) का मान ज्ञात कीजिए। (\( x < 90^\circ \))
Answer:
दिया है, \( \tan 3x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ \)
\( \implies \tan 3x = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \)
\( \implies \tan 3x = \frac{1}{2} + \frac{1}{2} \)
\( \implies \tan 3x = 1 \)
\( \implies \tan 3x = \tan 45^\circ \)
\( \implies 3x = 45^\circ \)
\( \implies x = \frac{45^\circ}{3} \)
\( \implies x = 15^\circ \)
In simple words: First, calculate the value of the right side of the equation by substituting the known trigonometric values and performing the multiplication and addition. This simplifies to 1. Then, find what angle \( 3x \) must be for \( \tan(3x) \) to equal 1. This means \( 3x \) is \( 45^\circ \). Finally, divide \( 45^\circ \) by 3 to find \( x \).

🎯 Exam Tip: After finding \( 3x \), remember to divide by 3 to get the value of \( x \). Always check if your answer for \( x \) satisfies any given conditions, such as \( x < 90^\circ \).

Question 3. \( \sin 30^\circ \cos^2 30^\circ + \tan 45^\circ \cos^2 60^\circ \) का मान ज्ञात कीजिए।
Answer:
\( \sin 30^\circ \cos^2 30^\circ + \tan 45^\circ \cos^2 60^\circ \)
\( = \frac{1}{2} \left( \frac{\sqrt{3}}{2} \right)^2 + (1) \left( \frac{1}{2} \right)^2 \)
\( = \frac{1}{2} \times \frac{3}{4} + 1 \times \frac{1}{4} \)
\( = \frac{3}{8} + \frac{1}{4} \)
\( = \frac{3}{8} + \frac{2}{8} \)
\( = \frac{3+2}{8} \)
\( = \frac{5}{8} \)
In simple words: Substitute the known trigonometric values into the expression. Perform the squaring and multiplication for each term. Then, add the resulting fractions by finding a common denominator. The final value will be \( \frac{5}{8} \).

🎯 Exam Tip: Be careful with the order of operations: powers, then multiplication, then addition. Ensure correct fractional addition by using a common denominator.

Question 4. \( \frac{\tan 60^\circ - \tan 30^\circ}{1+\tan 60^\circ \tan 30^\circ} \) का मान ज्ञात कीजिए।
Answer:
\( \frac{\tan 60^\circ - \tan 30^\circ}{1+\tan 60^\circ \tan 30^\circ} \)
\( = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}} \)
\( = \frac{\frac{\sqrt{3}\sqrt{3}-1}{\sqrt{3}}}{1+1} \)
\( = \frac{\frac{3-1}{\sqrt{3}}}{2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{2} \)
\( = \frac{2}{\sqrt{3}} \times \frac{1}{2} \)
\( = \frac{1}{\sqrt{3}} \)
In simple words: This expression is the formula for \( \tan(A-B) \). Substitute the values for \( \tan 60^\circ \) and \( \tan 30^\circ \). Simplify the numerator and the denominator separately. Then divide the simplified numerator by the simplified denominator.

🎯 Exam Tip: Recognize the identity \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \). This can help you identify the expected result quickly, which is \( \tan(60^\circ - 30^\circ) = \tan 30^\circ \).

Question 5. \( 4 \sin 30^\circ \sin^2 60^\circ + 3 \cos 60^\circ \tan 45^\circ \) का मान ज्ञात कीजिए।
Answer:
\( 4 \sin 30^\circ \sin^2 60^\circ + 3 \cos 60^\circ \tan 45^\circ \)
\( = 4 \times \frac{1}{2} \times \left(\frac{\sqrt{3}}{2}\right)^2 + 3 \times \frac{1}{2} \times 1 \)
\( = 4 \times \frac{1}{2} \times \frac{3}{4} + \frac{3}{2} \)
\( = \frac{3}{2} + \frac{3}{2} \)
\( = \frac{3+3}{2} \)
\( = \frac{6}{2} \)
\( = 3 \)
In simple words: Replace all trigonometric terms with their numerical values. Perform the squaring and multiplication in each part of the expression. Then, add the resulting terms by finding a common denominator. The final value is 3.

🎯 Exam Tip: Work through each term of the expression independently before combining them. This reduces the chance of calculation errors in a complex expression.

निबन्धात्मक प्रश्न

Question 1. यदि \( \sin(A + B) = 1 \) तथा \( \cos(A - B) = \frac{\sqrt{3}}{2} \) यहाँ \( 0^\circ < (A + B) \le 90^\circ \), \( A > B \) हो, तो \( A \) तथा \( B \) के मान ज्ञात कीजिए।
Answer:
दिया है:
\( \sin(A + B) = 1 \)
\( \implies \sin(A+B) = \sin 90^\circ \)
\( \implies A+B = 90^\circ \) (समीकरण 1)

तथा \( \cos(A - B) = \frac{\sqrt{3}}{2} \)
\( \implies \cos(A-B) = \cos 30^\circ \)
\( \implies A-B = 30^\circ \) (समीकरण 2)

समीकरण (1) और समीकरण (2) को जोड़ने पर:
\( (A + B) + (A - B) = 90^\circ + 30^\circ \)
\( \implies 2A = 120^\circ \)
\( \implies A = \frac{120^\circ}{2} \)
\( \implies A = 60^\circ \)

A का मान समीकरण (1) में रखने पर:
\( 60^\circ + B = 90^\circ \)
\( \implies B = 90^\circ - 60^\circ \)
\( \implies B = 30^\circ \)
अतः, \( A = 60^\circ \) और \( B = 30^\circ \).
In simple words: First, use the given sine and cosine equations to find two simple equations for \( A+B \) and \( A-B \). Then, add these two equations together to find the value of \( A \). After finding \( A \), put its value back into one of the original equations to find \( B \). Check that \( A > B \) and \( 0^\circ < A+B \le 90^\circ \).

🎯 Exam Tip: This type of problem is a system of linear equations in trigonometric form. Convert them to angle equations first, then solve them simultaneously using addition or substitution methods.

Question 2. \( \frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}} \) का मान ज्ञात कीजिए।
Answer: दिए गए त्रिकोणमितीय व्यंजक का मान इस प्रकार ज्ञात किया जा सकता है:
\( = \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} \)
पहले अंश को सरल करते हैं:
\( = \frac{\frac{3}{2} - \frac{2}{\sqrt{3}}}{\frac{3}{2} + \frac{2}{\sqrt{3}}} \)
\( = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{3\sqrt{3} + 4}{2\sqrt{3}}} \)
अब हर को सरल करते हैं:
\( = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \)
हम अंश और हर को \( (3\sqrt{3} - 4) \) से गुणा करेंगे ताकि हर को परिमेय बनाया जा सके:
\( = \frac{(3\sqrt{3} - 4)(3\sqrt{3} - 4)}{(3\sqrt{3} + 4)(3\sqrt{3} - 4)} \)
\( = \frac{(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + 4^2}{(3\sqrt{3})^2 - 4^2} \)
\( = \frac{27 - 24\sqrt{3} + 16}{27 - 16} \)
\( = \frac{43 - 24\sqrt{3}}{11} \)
यह अंतिम सरलीकृत मान है। ऐसे प्रश्नों में हर का परिमेयकरण करना एक सामान्य अभ्यास है ताकि व्यंजक को मानक रूप में प्रस्तुत किया जा सके।
In simple words: हमें दिए गए त्रिकोणमितीय व्यंजक को सरल करना था। इसके लिए, हमने सभी त्रिकोणमितीय मानों को रखा और फिर भिन्न को हल किया। हमने हर को परिमेय बनाने के लिए ऊपर और नीचे \( (3\sqrt{3} - 4) \) से गुणा किया, जिससे हमें अंतिम उत्तर मिला।

🎯 Exam Tip: त्रिकोणमितीय मानों को सही ढंग से याद रखना और भिन्नों का सावधानीपूर्वक सरलीकरण करना ऐसे प्रश्नों को हल करने की कुंजी है। हर का परिमेयकरण करना अक्सर अंतिम चरण होता है।

Question 3. \( \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\csc 30^{\circ}} \) का मान ज्ञात कीजिए।
Answer: दिए गए व्यंजक का मान निकालने के लिए हम त्रिकोणमितीय अनुपातों के मानक मानों का उपयोग करेंगे:
\( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \sec 30^{\circ} = \frac{2}{\sqrt{3}} \)
\( \csc 30^{\circ} = 2 \)
इन मानों को व्यंजक में रखने पर:
\( = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} \)
अब हर को सरल करते हैं:
\( = \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} \)
अब इसे उल्टा करके गुणा करते हैं:
\( = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}} \)
\( = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \)
अब हर का परिमेयकरण करेंगे। पहले \( \sqrt{2} \) से गुणा करते हैं:
\( = \frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3} + 1)} \times \frac{\sqrt{2}}{\sqrt{2}} \)
\( = \frac{\sqrt{6}}{4(\sqrt{3} + 1)} \)
अब \( (\sqrt{3} - 1) \) से अंश और हर में गुणा करते हैं:
\( = \frac{\sqrt{6}(\sqrt{3} - 1)}{4(\sqrt{3} + 1)(\sqrt{3} - 1)} \)
\( = \frac{\sqrt{18} - \sqrt{6}}{4(3 - 1)} \)
\( = \frac{3\sqrt{2} - \sqrt{6}}{4(2)} \)
\( = \frac{3\sqrt{2} - \sqrt{6}}{8} \)
यह व्यंजक का अंतिम सरलीकृत रूप है। त्रिकोणमितीय मानों को सही ढंग से लागू करना और भिन्नों का सावधानीपूर्वक सरलीकरण करना महत्वपूर्ण है।
In simple words: हमने सबसे पहले \( \cos 45^{\circ}, \sec 30^{\circ} \) और \( \csc 30^{\circ} \) के मान रखे। फिर, हमने हर में भिन्नों को जोड़ा और मुख्य भिन्न को हल किया। अंत में, हमने हर को परिमेय बनाने के लिए \( (\sqrt{3} - 1) \) से गुणा किया, जिससे हमें सबसे सरल उत्तर मिला।

🎯 Exam Tip: जब हर में वर्गमूल (radicals) हों, तो हमेशा हर का परिमेयकरण करें ताकि उत्तर को सबसे सरल और मानक रूप में प्रस्तुत किया जा सके। सभी त्रिकोणमितीय मान कंठस्थ होने चाहिए।

Question 4. \( 2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ} \) का मान ज्ञात कीजिए।
Answer: हम दिए गए त्रिकोणमितीय व्यंजक का मान ज्ञात करने के लिए मानक त्रिकोणमितीय मानों का उपयोग करेंगे:
\( \tan 45^{\circ} = 1 \)
\( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
\( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
इन मानों को व्यंजक में रखते हैं:
\( = 2 (1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
यह दर्शाता है कि \( \cos^2 30^{\circ} \) और \( \sin^2 60^{\circ} \) के मान समान होने के कारण एक-दूसरे को रद्द कर देंगे।
\( = 2 \times 1 + \frac{3}{4} - \frac{3}{4} \)
\( = 2 + 0 \)
\( = 2 \)
तो, दिए गए व्यंजक का मान 2 है।
In simple words: हमने \( \tan 45^{\circ}, \cos 30^{\circ} \) और \( \sin 60^{\circ} \) के मानों को सूत्र में रखा। क्योंकि \( \cos 30^{\circ} \) और \( \sin 60^{\circ} \) का मान एक ही है \( (\frac{\sqrt{3}}{2}) \), इसलिए उनके वर्ग भी एक ही हुए और वे एक-दूसरे को घटाकर शून्य कर गए। अंत में, केवल \( 2 \tan^2 45^{\circ} \) का मान बचा, जो \( 2 \times 1^2 = 2 \) है।

🎯 Exam Tip: याद रखें कि \( \cos (90^{\circ} - \theta) = \sin \theta \)। इसलिए \( \cos 30^{\circ} = \sin 60^{\circ} \)। जब उनके वर्ग घटाए जाते हैं, तो वे शून्य हो जाते हैं, जिससे गणना बहुत सरल हो जाती है।

Question 6. बताइए कि निम्नलिखित में कौन-कौन सत्य हैं या असत्य हैं। कारण सहित अपने उत्तर की पुष्टि कीजिए
(i) \( \sin (A+B)=\sin A+\sin B \).
(ii) \( \theta \) में वृद्धि होने के साथ \( \sin \theta \) के मान में भी वृद्धि होती है।
(iii) \( \theta \) में वृद्धि होने के साथ \( \cos \theta \) के मान में भी वृद्धि होती है।
(iv) \( \theta \) के सभी मानों पर \( \sin \theta = \cos \theta \).
(v) \( A = 0^{\circ} \) पर \( \cot A \) परिभाषित नहीं है।
Answer:
(i) यह असत्य है।
उदाहरण के लिए, यदि \( A = 60^{\circ} \) और \( B = 30^{\circ} \):
\( \sin (60^{\circ} + 30^{\circ}) = \sin 90^{\circ} = 1 \)
परंतु \( \sin 60^{\circ} + \sin 30^{\circ} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2} \approx 0.866 + 0.5 = 1.366 \)
चूंकि \( 1 \neq 1.366 \), इसलिए \( \sin (A+B) \neq \sin A+\sin B \)।
(ii) यह सत्य है।
जब \( \theta \) का मान \( 0^{\circ} \) से \( 90^{\circ} \) तक बढ़ता है, तो \( \sin \theta \) का मान \( 0 \) से \( 1 \) तक बढ़ता है।
उदाहरण के लिए, \( \sin 0^{\circ} = 0 \), \( \sin 30^{\circ} = 0.5 \), \( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707 \), \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \), \( \sin 90^{\circ} = 1 \)। यह पैटर्न 0 से 90 डिग्री के बीच सत्य रहता है।
(iii) यह असत्य है।
जब \( \theta \) का मान \( 0^{\circ} \) से \( 90^{\circ} \) तक बढ़ता है, तो \( \cos \theta \) का मान \( 1 \) से \( 0 \) तक घटता है।
उदाहरण के लिए, \( \cos 0^{\circ} = 1 \), \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \), \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707 \), \( \cos 60^{\circ} = 0.5 \), \( \cos 90^{\circ} = 0 \)। कोण बढ़ने पर कोसाइन का मान कम होता जाता है।
(iv) यह असत्य है।
\( \sin \theta = \cos \theta \) केवल \( \theta = 45^{\circ} \) पर सत्य होता है। अन्य मानों पर वे बराबर नहीं होते हैं।
उदाहरण के लिए, \( \sin 30^{\circ} = \frac{1}{2} \) और \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \), जो बराबर नहीं हैं। \( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \) और \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \) पर दोनों समान होते हैं।
(v) यह सत्य है।
\( \cot A = \frac{\cos A}{\sin A} \)। जब \( A = 0^{\circ} \) होता है, तो \( \sin 0^{\circ} = 0 \)। किसी भी संख्या को शून्य से विभाजित करना अपरिभाषित होता है। इसलिए \( \cot 0^{\circ} \) अपरिभाषित है।
In simple words: हमने प्रत्येक कथन की सच्चाई की जांच की। पहले कथन में \( \sin(A+B) \) के सूत्र को गलत तरीके से लिखा गया था। दूसरे में \( \sin \theta \) का मान \( 0^{\circ} \) से \( 90^{\circ} \) तक बढ़ता है। तीसरे में \( \cos \theta \) का मान \( 0^{\circ} \) से \( 90^{\circ} \) तक घटता है। चौथे में \( \sin \theta \) और \( \cos \theta \) केवल \( 45^{\circ} \) पर बराबर होते हैं। पांचवां कथन सही है क्योंकि \( \cot 0^{\circ} \) को परिभाषित नहीं किया जा सकता है क्योंकि \( \sin 0^{\circ} \) शून्य होता है।

🎯 Exam Tip: त्रिकोणमितीय सर्वसमिकाओं और विभिन्न कोणों के लिए sin और cos के व्यवहार को याद रखना महत्वपूर्ण है। विशेष रूप से, \( \sin (A+B) \) और \( \cos (A+B) \) के सही सूत्र क्या हैं, यह ध्यान रखें।