UP Board Solutions Class 10 Maths Chapter 10 Trigonometrical Ratios and Identities Ex 103

Ex 10.3 Trigonometrical Ratios and Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

प्रश्न 1. sin(90° – θ)
Answer: \( \sin(90^{\circ} - \theta) = \cos \theta \)
Here, when the angle changes from \( (90^{\circ} - \theta) \) to just \( \theta \), the sine function changes to cosine. This is a fundamental trigonometric identity.
In simple words: जब कोण \( 90^{\circ} - \theta \) होता है, तो उसका साइन (sin) हमेशा \( \theta \) के कोसाइन (cos) के बराबर होता है।

🎯 Exam Tip: Remember the basic complementary angle identities: \( \sin(90^{\circ} - \theta) = \cos \theta \), \( \cos(90^{\circ} - \theta) = \sin \theta \), and \( \tan(90^{\circ} - \theta) = \cot \theta \).

प्रश्न 2. tan(90° + θ)
Answer: \( \tan(90^{\circ} + \theta) = - \cot \theta \)
The tangent function changes to cotangent when the angle is \( (90^{\circ} + \theta) \). Since \( (90^{\circ} + \theta) \) lies in the second quadrant, where tangent is negative, the result is also negative.
In simple words: जब कोण \( 90^{\circ} + \theta \) होता है, तो उसका टैन (tan) हमेशा \( \theta \) के कॉट (cot) के नकारात्मक मान के बराबर होता है।

🎯 Exam Tip: For angles like \( (90^{\circ} + \theta) \), first identify the quadrant to determine the sign of the trigonometric ratio, and then apply the appropriate function change (e.g., tan to cot, sin to cos).

प्रश्न 3. cos (180° – θ)
Answer: \( \cos(180^{\circ} - \theta) = - \cos \theta \)
When the angle is \( (180^{\circ} - \theta) \), the cosine function remains cosine, but because \( (180^{\circ} - \theta) \) falls in the second quadrant where cosine values are negative, we add a minus sign.
In simple words: यदि कोण \( 180^{\circ} - \theta \) है, तो उसका कोसाइन (cos) \( \theta \) के कोसाइन (cos) के नकारात्मक मान के बराबर होगा।

🎯 Exam Tip: For angles like \( (180^{\circ} - \theta) \) or \( (180^{\circ} + \theta) \), the trigonometric function usually does not change (cos remains cos, sin remains sin). Only the sign changes based on the quadrant.

प्रश्न 4. cos37° cosec53°
Answer:
\( \cos 37^{\circ} \text{ cosec} 53^{\circ} \)
\( = \cos 37^{\circ} \cdot \text{cosec}(90^{\circ} - 37^{\circ}) \)
\( = \cos 37^{\circ} \cdot \sec 37^{\circ} \)
\( = \cos 37^{\circ} \cdot \frac{1}{\cos 37^{\circ}} \)
\( = 1 \)
The key here is that \( \text{cosec}(90^{\circ} - \theta) \) transforms to \( \sec \theta \), which is the reciprocal of \( \cos \theta \).
In simple words: इस सवाल को हल करने के लिए, हम \( \text{cosec} 53^{\circ} \) को \( \text{cosec}(90^{\circ} - 37^{\circ}) \) में बदलते हैं, जो \( \sec 37^{\circ} \) बन जाता है। क्योंकि \( \sec \) और \( \cos \) एक-दूसरे के उल्टे होते हैं, वे कट जाते हैं और उत्तर 1 आता है।

🎯 Exam Tip: Look for angles that add up to \( 90^{\circ} \) (complementary angles) when you see products of different trigonometric ratios. This often simplifies the expression using identities.

प्रश्न 5. tan210°
Answer:
\( \tan 210^{\circ} \)
\( = \tan(180^{\circ} + 30^{\circ}) \)
\( = \tan 30^{\circ} \)
\( = \frac{1}{\sqrt{3}} \)
Here, \( (180^{\circ} + 30^{\circ}) \) lies in the third quadrant, where the tangent function is positive, so the sign remains unchanged.
In simple words: \( \tan 210^{\circ} \) को \( \tan(180^{\circ} + 30^{\circ}) \) के रूप में लिखा जा सकता है। तीसरे चतुर्थांश में टैन धनात्मक होता है, इसलिए यह \( \tan 30^{\circ} \) के बराबर है, जिसका मान \( \frac{1}{\sqrt{3}} \) होता है।

🎯 Exam Tip: Break down angles greater than \( 90^{\circ} \) into forms like \( (180^{\circ} \pm \theta) \) or \( (360^{\circ} \pm \theta) \) to use trigonometric identities and find their values.

प्रश्न 6. \(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\)
Answer:
\( \frac{\cos 55^{\circ}}{\sin 35^{\circ}} \)
\( = \frac{\cos (90^{\circ} - 35^{\circ})}{\sin 35^{\circ}} \)
\( = \frac{\sin 35^{\circ}}{\sin 35^{\circ}} \)
\( = 1 \)
By using the complementary angle identity \( \cos(90^{\circ} - \theta) = \sin \theta \), the numerator transforms into the denominator, simplifying the expression to 1.
In simple words: \( \cos 55^{\circ} \) को \( \sin 35^{\circ} \) में बदल सकते हैं, क्योंकि \( 55^{\circ} \) और \( 35^{\circ} \) मिलकर \( 90^{\circ} \) बनाते हैं। इसलिए ऊपर और नीचे की संख्या समान हो जाती है, जिससे उत्तर 1 आता है।

🎯 Exam Tip: When you see a fraction with sine and cosine functions and their angles add up to \( 90^{\circ} \), apply complementary angle identities to simplify it to 1.

प्रश्न 7. sin 225°
Answer:
\( \sin 225^{\circ} \)
\( = \sin(180^{\circ} + 45^{\circ}) \)
\( = - \sin 45^{\circ} \)
\( = - \frac{1}{\sqrt{2}} \)
The angle \( (180^{\circ} + 45^{\circ}) \) falls in the third quadrant, where the sine function is negative, so the result carries a minus sign.
In simple words: \( \sin 225^{\circ} \) को \( \sin(180^{\circ} + 45^{\circ}) \) के रूप में लिखा जाता है। क्योंकि यह तीसरे चतुर्थांश में है जहाँ साइन नकारात्मक होता है, उत्तर \( -\sin 45^{\circ} \) है, जो \( -\frac{1}{\sqrt{2}} \) होता है।

🎯 Exam Tip: Always determine the quadrant of the angle after using \( (180^{\circ} \pm \theta) \) or \( (360^{\circ} \pm \theta) \) to correctly apply the positive or negative sign to the trigonometric ratio.

Ex 10.3 Trigonometrical Ratios and Identities लघु उत्तरीय प्रश्न-I (Short Answer Type Questions-I)

प्रश्न 8. sin 27° sec63° का मान ज्ञात कीजिए|
Answer:
\( \sin 27^{\circ} \cdot \sec 63^{\circ} \)
\( = \sin 27^{\circ} \cdot \sec(90^{\circ} - 27^{\circ}) \)
\( = \sin 27^{\circ} \cdot \text{cosec} 27^{\circ} \)
\( = \sin 27^{\circ} \cdot \frac{1}{\sin 27^{\circ}} \)
\( = 1 \)
Since \( \sec(90^{\circ} - \theta) = \text{cosec} \theta \) and \( \text{cosec} \theta = \frac{1}{\sin \theta} \), these identities help in simplifying the expression.
In simple words: \( \sec 63^{\circ} \) को \( \sec(90^{\circ} - 27^{\circ}) \) के रूप में लिखकर उसे \( \text{cosec} 27^{\circ} \) में बदल देते हैं। क्योंकि \( \text{cosec} \) और \( \sin \) एक-दूसरे के उल्टे होते हैं, वे गुणा होकर 1 बन जाते हैं।

🎯 Exam Tip: When evaluating expressions with products of different trigonometric functions, try to convert one of the functions using complementary angle identities to make it the reciprocal of the other.

प्रश्न 9. tan 37° tan 53° का मान ज्ञात कीजिए|
Answer:
\( \tan 37^{\circ} \cdot \tan 53^{\circ} \)
\( = \tan 37^{\circ} \cdot \tan(90^{\circ} - 37^{\circ}) \)
\( = \tan 37^{\circ} \cdot \cot 37^{\circ} \)
\( = \tan 37^{\circ} \cdot \frac{1}{\tan 37^{\circ}} \)
\( = 1 \)
The identity \( \tan(90^{\circ} - \theta) = \cot \theta \) is used here. Also, \( \cot \theta \) is the reciprocal of \( \tan \theta \), leading to the product being 1.
In simple words: \( \tan 53^{\circ} \) को \( \tan(90^{\circ} - 37^{\circ}) \) के रूप में लिखकर उसे \( \cot 37^{\circ} \) में बदलते हैं। चूंकि \( \tan \) और \( \cot \) एक-दूसरे के उल्टे होते हैं, वे गुणा होकर 1 बन जाते हैं।

🎯 Exam Tip: Remember that \( \tan \theta \cdot \cot \theta = 1 \). If you have two tan functions whose angles add up to \( 90^{\circ} \), you can convert one to cot and get 1.

प्रश्न 10. सिद्ध कीजिए: sin (180° + θ) = cos(90° + θ)
Answer:
LHS: \( \sin(180^{\circ} + \theta) \)
\( \implies = - \sin \theta \)
RHS: \( \cos(90^{\circ} + \theta) \)
\( \implies = - \sin \theta \)
Since LHS = RHS, the identity is proven. Both expressions simplify to \( - \sin \theta \) due to angle transformations in their respective quadrants.
In simple words: हम बाएं पक्ष (LHS) और दाएं पक्ष (RHS) को अलग-अलग हल करते हैं। \( \sin(180^{\circ} + \theta) \) तीसरे चतुर्थांश में \( -\sin \theta \) होता है, और \( \cos(90^{\circ} + \theta) \) दूसरे चतुर्थांश में भी \( -\sin \theta \) होता है। इस तरह, दोनों पक्ष बराबर सिद्ध होते हैं।

🎯 Exam Tip: When proving identities, always simplify the Left Hand Side (LHS) and Right Hand Side (RHS) separately. Show each step clearly and verify if they simplify to the same expression.

प्रश्न 11. सिद्ध कीजिए: tan 45° – cot 45° = 0
Answer:
LHS \( = \tan 45^{\circ} - \cot 45^{\circ} \)
\( = 1 - 1 \)
\( = 0 \)
\( = \text{RHS} \)
The values of \( \tan 45^{\circ} \) and \( \cot 45^{\circ} \) are standard trigonometric values that are easy to remember.
In simple words: हम जानते हैं कि \( \tan 45^{\circ} \) का मान 1 होता है और \( \cot 45^{\circ} \) का मान भी 1 होता है। जब हम 1 में से 1 घटाते हैं, तो उत्तर 0 आता है, जो हमें सिद्ध करना था।

🎯 Exam Tip: Memorize the standard trigonometric values for angles like \( 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ} \), as they are frequently used in problems.

प्रश्न 12. सिद्ध कीजिए: \(\frac{\cos 21^{\circ}}{\sin 69^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\circ}}\) = 2
Answer:
LHS \( = \frac{\cos 21^{\circ}}{\sin 69^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\circ}} \)
\( = \frac{\cos 21^{\circ}}{\sin(90^{\circ}-21^{\circ})} + \frac{\sin 59^{\circ}}{\cos(90^{\circ}-59^{\circ})} \)
\( = \frac{\cos 21^{\circ}}{\cos 21^{\circ}} + \frac{\sin 59^{\circ}}{\sin 59^{\circ}} \)
\( = 1 + 1 \)
\( = 2 \)
\( = \text{RHS} \)
In this problem, the angles in each fraction are complementary (add up to \( 90^{\circ} \)), allowing the use of \( \sin(90^{\circ} - \theta) = \cos \theta \) and \( \cos(90^{\circ} - \theta) = \sin \theta \).
In simple words: पहले भाग में, \( \sin 69^{\circ} \) को \( \cos 21^{\circ} \) में बदल देते हैं, और दूसरे भाग में, \( \cos 31^{\circ} \) को \( \sin 59^{\circ} \) में बदल देते हैं। इससे दोनों भिन्न 1 बन जाती हैं, और 1 + 1 का जोड़ 2 होता है, जो हमें सिद्ध करना था।

🎯 Exam Tip: Always check if the angles in trigonometric fractions are complementary. If they are, you can use the \( 90^{\circ} - \theta \) identities to simplify the terms.

Ex 10.3 Trigonometrical Ratios and Identities लघु उत्तरीय प्रश्न-II (Short Answer Type Questions-II)

प्रश्न 13. सिद्ध कीजिए कि (i) sin 210°= – 1/2 (ii) tan 225°= 1 (iii) sec 210° = \(-\frac{2}{\sqrt{3}}\) (iv) cosec 225°= \(-\sqrt{2}\)
Answer:
(i) LHS \( = \sin 210^{\circ} \)
\( = \sin(180^{\circ} + 30^{\circ}) \)
\( = - \sin 30^{\circ} \)
\( = - \frac{1}{2} \)
\( = \text{RHS} \)
(ii) LHS \( = \tan 225^{\circ} \)
\( = \tan(180^{\circ} + 45^{\circ}) \)
\( = \tan 45^{\circ} \)
\( = 1 \)
\( = \text{RHS} \)
(iii) LHS \( = \sec 210^{\circ} \)
\( = \sec(180^{\circ} + 30^{\circ}) \)
\( = - \sec 30^{\circ} \)
\( = - \frac{2}{\sqrt{3}} \)
\( = \text{RHS} \)
(iv) LHS \( = \text{cosec} 225^{\circ} \)
\( = \text{cosec}(180^{\circ} + 45^{\circ}) \)
\( = - \text{cosec} 45^{\circ} \)
\( = - \sqrt{2} \)
\( = \text{RHS} \)
All these transformations involve reducing the angle to an acute angle using \( (180^{\circ} + \theta) \) and determining the correct sign based on the quadrant.
In simple words: प्रत्येक भाग में, हम कोण को \( 180^{\circ} + \theta \) के रूप में लिखते हैं। फिर, तीसरे चतुर्थांश में साइन, टैन, सेक और कोसेक के नियमों का उपयोग करते हुए, हम उनके मान ज्ञात करते हैं और सिद्ध करते हैं कि वे दिए गए मानों के बराबर हैं।

🎯 Exam Tip: For angles greater than \( 90^{\circ} \), visualize the angle in the unit circle to correctly identify its quadrant. This helps determine if the trigonometric ratio is positive or negative after applying the reduction formula.

प्रश्न 14. \(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}+\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\) – 4cos60° = 0
Answer:
LHS \( = \frac{\sin 35^{\circ}}{\cos 55^{\circ}}+\frac{\cos 55^{\circ}}{\sin 35^{\circ}} - 4 \cos 60^{\circ} \)
\( = \frac{\sin 35^{\circ}}{\cos(90^{\circ}-35^{\circ})} + \frac{\cos 55^{\circ}}{\sin(90^{\circ}-55^{\circ})} - 4 \times \frac{1}{2} \)
\( = \frac{\sin 35^{\circ}}{\sin 35^{\circ}} + \frac{\cos 55^{\circ}}{\cos 55^{\circ}} - 2 \)
\( = 1 + 1 - 2 \)
\( = 0 \)
\( = \text{RHS} \)
This problem combines complementary angle identities with a standard trigonometric value. Always address each term in the expression separately before combining them.
In simple words: पहले दो भिन्नों में, \( \cos 55^{\circ} \) को \( \sin 35^{\circ} \) और \( \sin 35^{\circ} \) को \( \cos 55^{\circ} \) में बदल देते हैं, जिससे प्रत्येक भिन्न 1 बन जाती है। फिर \( 4 \cos 60^{\circ} \) का मान \( 4 \times \frac{1}{2} = 2 \) होता है। अंत में, \( 1 + 1 - 2 = 0 \) आता है, जो हमें सिद्ध करना था।

🎯 Exam Tip: When an expression involves multiple terms, simplify each term individually using relevant identities or known values, then combine them to reach the final result.

प्रश्न 15. \(\frac{\cos 51^{\circ}}{\sin 39^{\circ}}+\frac{\tan 38^{\circ}}{\cot 52^{\circ}}+\frac{\sec 47^{\circ}}{\text{cosec } 43^{\circ}}\) = 3
Answer:
LHS \( = \frac{\cos 51^{\circ}}{\sin 39^{\circ}}+\frac{\tan 38^{\circ}}{\cot 52^{\circ}}+\frac{\sec 47^{\circ}}{\text{cosec } 43^{\circ}} \)
\( = \frac{\cos(90^{\circ}-39^{\circ})}{\sin 39^{\circ}}+\frac{\tan 38^{\circ}}{\cot(90^{\circ}-38^{\circ})}+\frac{\sec(90^{\circ}-43^{\circ})}{\text{cosec } 43^{\circ}} \)
\( = \frac{\sin 39^{\circ}}{\sin 39^{\circ}}+\frac{\tan 38^{\circ}}{\tan 38^{\circ}}+\frac{\text{cosec } 43^{\circ}}{\text{cosec } 43^{\circ}} \)
\( = 1 + 1 + 1 \)
\( = 3 \)
\( = \text{RHS} \)
Each term in the sum can be simplified to 1 by applying complementary angle identities correctly.
In simple words: इस सवाल में, तीनों भिन्नों में कोणों का जोड़ \( 90^{\circ} \) है। हम \( \cos 51^{\circ} \) को \( \sin 39^{\circ} \), \( \cot 52^{\circ} \) को \( \tan 38^{\circ} \), और \( \sec 47^{\circ} \) को \( \text{cosec } 43^{\circ} \) में बदल देते हैं। इससे प्रत्येक भिन्न 1 बन जाती है, और \( 1 + 1 + 1 \) का जोड़ 3 होता है, जो हमें सिद्ध करना था।

🎯 Exam Tip: This type of problem frequently appears. Always check if the angles in the numerator and denominator are complementary; this is the key to simplifying the expression.

Ex 10.3 Trigonometrical Ratios and Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

प्रश्न 16. सिद्ध कीजिए कि \(\frac{\sin(90^{\circ}+\theta)}{\cos \theta} + \frac{\sin (180^{\circ}+\theta)}{\sin \theta} - \frac{\tan \theta}{\cot (90^{\circ}+\theta)}\) = 3
Answer:
LHS \( = \frac{\sin(90^{\circ}+\theta)}{\cos \theta} + \frac{\sin (180^{\circ}+\theta)}{\sin \theta} - \frac{\tan \theta}{\cot (90^{\circ}+\theta)} \)
\( = \frac{\cos \theta}{\cos \theta} + \frac{- \sin \theta}{\sin \theta} - \frac{\tan \theta}{-\tan \theta} \)
\( = 1 + (-1) - (-1) \)
\( = 1 - 1 + 1 \)
\( = 1 \)
It seems there might be a typo in the question's RHS, as the steps lead to 1, not 3. However, based on the source's provided solution (which also shows 1+1+1=3 for a slightly different problem), we will adjust the final step if the intent was 3. If we interpret the source's solution `1+1+1=3` as the *desired* outcome, the given expression simplifies to 1. The solution provided by the source shows `cose/cose -sine/sine -tane/-tane = 1+1+1=3`. This implies a different problem or calculation. Sticking to the question as written:
LHS \( = \frac{\sin(90^{\circ}+\theta)}{\cos \theta} + \frac{\sin (180^{\circ}+\theta)}{\sin \theta} - \frac{\tan \theta}{\cot (90^{\circ}+\theta)} \)
\( = \frac{\cos \theta}{\cos \theta} + \frac{- \sin \theta}{\sin \theta} - \frac{\tan \theta}{-\tan \theta} \)
\( = 1 - 1 - (-1) \)
\( = 1 - 1 + 1 \)
\( = 1 \)
The question as written simplifies to 1. If the target was 3, the problem statement would need adjustment.
In simple words: हम दिए गए समीकरण के बाएं पक्ष (LHS) को सरल करते हैं। \( \sin(90^{\circ}+\theta) \) को \( \cos \theta \) में, \( \sin(180^{\circ}+\theta) \) को \( -\sin \theta \) में, और \( \cot(90^{\circ}+\theta) \) को \( -\tan \theta \) में बदलते हैं। इससे समीकरण \( 1 + (-1) - (-1) \) बन जाता है, जिसका हल 1 आता है।

🎯 Exam Tip: Be very careful with signs when using angle transformation formulas. Always simplify each term separately before combining them. If the result doesn't match the RHS, double-check your sign conventions and identities.

प्रश्न 17. \(\frac{2 \tan 43^{\circ}}{\cot 47^{\circ}} - \frac{\tan 70^{\circ}}{\cot 20^{\circ}}\) = 1
Answer:
LHS \( = \frac{2 \tan 43^{\circ}}{\cot 47^{\circ}} - \frac{\tan 70^{\circ}}{\cot 20^{\circ}} \)
\( = \frac{2 \tan 43^{\circ}}{\cot(90^{\circ}-43^{\circ})} - \frac{\tan 70^{\circ}}{\cot(90^{\circ}-70^{\circ})} \)
\( = \frac{2 \tan 43^{\circ}}{\tan 43^{\circ}} - \frac{\tan 70^{\circ}}{\tan 70^{\circ}} \)
\( = 2 - 1 \)
\( = 1 \)
\( = \text{RHS} \)
This problem utilizes the complementary angle identity \( \cot(90^{\circ} - \theta) = \tan \theta \) to simplify both fractions.
In simple words: पहले भिन्न में, \( \cot 47^{\circ} \) को \( \tan 43^{\circ} \) में बदलते हैं, और दूसरे भिन्न में, \( \cot 20^{\circ} \) को \( \tan 70^{\circ} \) में बदलते हैं। इससे पहली भिन्न 2 बन जाती है और दूसरी भिन्न 1 बन जाती है। अंत में, \( 2 - 1 \) का हल 1 आता है, जो हमें सिद्ध करना था।

🎯 Exam Tip: Remember that \( \cot \theta = \tan(90^{\circ} - \theta) \). This identity is crucial for simplifying expressions involving cotangent and tangent of complementary angles.

प्रश्न 18. tan240° + sin120° = \(\frac{3 \sqrt{3}}{2}\)
Answer:
LHS \( = \tan 240^{\circ} + \sin 120^{\circ} \)
\( = \tan(180^{\circ} + 60^{\circ}) + \sin(180^{\circ} - 60^{\circ}) \)
\( = \tan 60^{\circ} + \sin 60^{\circ} \)
\( = \sqrt{3} + \frac{\sqrt{3}}{2} \)
\( = \frac{2\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \)
\( = \frac{2\sqrt{3} + \sqrt{3}}{2} \)
\( = \frac{3\sqrt{3}}{2} \)
\( = \text{RHS} \)
This problem involves transforming angles into acute angles and then using known standard trigonometric values for \( 60^{\circ} \).
In simple words: हम \( \tan 240^{\circ} \) को \( \tan(180^{\circ} + 60^{\circ}) \) के रूप में लिखते हैं, जो \( \tan 60^{\circ} \) के बराबर है। फिर, \( \sin 120^{\circ} \) को \( \sin(180^{\circ} - 60^{\circ}) \) के रूप में लिखते हैं, जो \( \sin 60^{\circ} \) के बराबर है। \( \tan 60^{\circ} \) का मान \( \sqrt{3} \) और \( \sin 60^{\circ} \) का मान \( \frac{\sqrt{3}}{2} \) होता है। इन दोनों को जोड़ने पर \( \frac{3\sqrt{3}}{2} \) आता है, जो हमें सिद्ध करना था।

🎯 Exam Tip: When dealing with angles outside the first quadrant, use the reference angle and the CAST rule (or similar quadrant rule) to determine the sign and value of the trigonometric function. Break down larger angles into \( (180^{\circ} \pm \theta) \) or \( (360^{\circ} \pm \theta) \).