Maharashtra Board Class 10 Maths Chapter 6 Trigonometry Set 6.1 Solutions

Question 1. If sin \( \theta = \frac{7}{25} \), find the values of cos \( \theta \) and tan \( \theta \).
Solution:
sin \( \theta = \frac{7}{25} \) [Given]
We know that,
\( \sin^2\theta + \cos^2\theta = 1 \)

\( \left(\frac{7}{25}\right)^2 + \cos^2\theta = 1 \)

\( \frac{49}{625} + \cos^2\theta = 1 \)

\( \cos^2\theta = 1 - \frac{49}{625} \)

\( \cos^2\theta = \frac{625-49}{625} \)

\( \cos^2\theta = \frac{576}{625} \)

\( \cos\theta = \frac{24}{25} \)
...[Taking square root of both sides] Now, tan \( \theta = \frac{\sin\theta}{\cos\theta} \)

tan \( \theta = \frac{\frac{7}{25}}{\frac{24}{25}} \)

\( = \frac{7}{25} \times \frac{25}{24} \)

\( \tan\theta = \frac{7}{24} \)

\( \cos\theta = \frac{24}{25} \) and tan \( \theta = \frac{7}{24} \)
Alternate Method:
sin \( \theta = \frac{7}{25} \) ...(i) [Given]
Consider \( \triangle\text{ABC} \), where \( \angle\text{ABC} = 90^\circ \) and \( \angle\text{ACB} = \theta \).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिखाया गया है, जहाँ कोण B समकोण (\(90^\circ\)) है और कोण C थीटा (\(\theta\)) है। भुजा AB की लंबाई 7k और भुजा AC की लंबाई 25k है।
Let AB = 7k and AC = 25k
In \( \triangle\text{ABC} \), \( \angle\text{B} = 90^\circ \)

\( \text{AB}^2 + \text{BC}^2 = \text{AC}^2 \) ... [Pythagoras theorem]

\( (7k)^2 + \text{BC}^2 = (25k)^2 \)

\( 49k^2 + \text{BC}^2 = 625k^2 \)

\( \text{BC}^2 = 625k^2 - 49k^2 \)

\( \text{BC}^2 = 576k^2 \)

\( \text{BC} = 24k \) ...[Taking square root of both sides]
Now, cos \( \theta = \frac{\text{BC}}{\text{AC}} \)

\( = \frac{24k}{25k} \)

\( \cos\theta = \frac{24}{25} \)
...[By definition]
Also, tan \( \theta = \frac{\text{AB}}{\text{BC}} \)

\( = \frac{7k}{24k} \)

\( \tan\theta = \frac{7}{24} \)
...[By definition]

\( \cos\theta = \frac{24}{25} \) and tan \( \theta = \frac{7}{24} \)
In simple words: Given sin \(\theta\), we use the identity \(\sin^2\theta + \cos^2\theta = 1\) to find cos \(\theta\), and then use \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) to find tan \(\theta\). An alternate method involves constructing a right-angled triangle and using Pythagoras theorem to find the sides and then the ratios.

🎯 Exam Tip: Remember the fundamental trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\) as it's crucial for solving many problems. Be careful with square roots, ensuring you take the positive value for acute angles.

Question 2. If tan \( \theta = \frac{3}{4} \), find the values of sec \( \theta \) and cos \( \theta \).
Solution:
tan \( \theta = \frac{3}{4} \) ...[Given]
We know that,
\( 1 + \tan^2\theta = \sec^2\theta \)

\( 1 + \left(\frac{3}{4}\right)^2 = \sec^2\theta \)

\( 1 + \frac{9}{16} = \sec^2\theta \)

\( \frac{16+9}{16} = \sec^2\theta \)

\( \sec^2\theta = \frac{25}{16} \)

\( \sec\theta = \frac{5}{4} \)
...[Taking square root of both sides]
Now, cos \( \theta = \frac{1}{\sec\theta} \)

\( = \frac{1}{\frac{5}{4}} \)

\( \cos\theta = \frac{4}{5} \)

\( \sec\theta = \frac{5}{4} \) and cos \( \theta = \frac{4}{5} \)
Alternate Method:
tan \( \theta = \frac{3}{4} \) ...(i)[Given]
Consider \( \triangle\text{ABC} \), where \( \angle\text{ABC} = 90^\circ \) and \( \angle\text{ACB} = \theta \).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिखाया गया है, जहाँ कोण B समकोण (\(90^\circ\)) है और कोण C थीटा (\(\theta\)) है। भुजा AB की लंबाई 3k और भुजा BC की लंबाई 4k है।
Let AB = 3k and BC = 4k
In \( \triangle\text{ABC} \), \( \angle\text{B} = 90^\circ \)

\( \text{AB}^2 + \text{BC}^2 = \text{AC}^2 \) ... [Pythagoras theorem]

\( (3k)^2 + (4k)^2 = \text{AC}^2 \)

\( 9k^2 + 16k^2 = \text{AC}^2 \)

\( \text{AC}^2 = 25k^2 \)

\( \text{AC} = 5k \) ...[Taking square root of both sides]
Now, sec \( \theta = \frac{\text{AC}}{\text{BC}} \)

\( = \frac{5k}{4k} \)

\( \sec\theta = \frac{5}{4} \)
...[By definition]
Also, cos \( \theta = \frac{\text{BC}}{\text{AC}} \)

\( = \frac{4k}{5k} \)

\( \cos\theta = \frac{4}{5} \)
...[By definition]

\( \sec\theta = \frac{5}{4} \) and cos \( \theta = \frac{4}{5} \)
In simple words: Given tan \(\theta\), we use the identity \(1 + \tan^2\theta = \sec^2\theta\) to find sec \(\theta\), and then use \(\cos\theta = \frac{1}{\sec\theta}\) to find cos \(\theta\). An alternate method uses a right-angled triangle to find all sides and then the required ratios.

🎯 Exam Tip: Mastering the three fundamental trigonometric identities (\(\sin^2\theta + \cos^2\theta = 1\), \(1 + \tan^2\theta = \sec^2\theta\), \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\)) is key. Clearly label steps, especially when deriving from identities or definitions.

Question 3. If cot \( \theta = \frac{40}{9} \), find the values of cosec \( \theta \) and sin \( \theta \).
Solution:
cot \( \theta = \frac{40}{9} \) ...[Given]
We know that,
\( 1 + \cot^2\theta = \operatorname{cosec}^2\theta \)

\( 1 + \left(\frac{40}{9}\right)^2 = \operatorname{cosec}^2\theta \)

\( 1 + \frac{1600}{81} = \operatorname{cosec}^2\theta \)

\( \frac{81+1600}{81} = \operatorname{cosec}^2\theta \)

\( \operatorname{cosec}^2\theta = \frac{1681}{81} \)

\( \operatorname{cosec}\theta = \frac{41}{9} \)
...[Taking square root of both sides]
Now, sin \( \theta = \frac{1}{\operatorname{cosec}\theta} \)

\( = \frac{1}{\frac{41}{9}} \)

\( \sin\theta = \frac{9}{41} \)

\( \operatorname{cosec}\theta = \frac{41}{9} \) and sin \( \theta = \frac{9}{41} \)
Alternate Method:
cot \( \theta = \frac{40}{9} \) ....(i) [Given]
Consider \( \triangle\text{ABC} \), where \( \angle\text{ABC} = 90^\circ \) and \( \angle\text{ACB} = \theta \)

ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिखाया गया है, जहाँ कोण B समकोण (\(90^\circ\)) है और कोण C थीटा (\(\theta\)) है। भुजा BC की लंबाई 40k और भुजा AB की लंबाई 9k है।
Let BC = 40k and AB = 9k
In \( \triangle\text{ABC} \), \( \angle\text{B} = 90^\circ \)

\( \text{AB}^2 + \text{BC}^2 = \text{AC}^2 \) ... [Pythagoras theorem]

\( (9k)^2 + (40k)^2 = \text{AC}^2 \)

\( 81k^2 + 1600k^2 = \text{AC}^2 \)

\( \text{AC}^2 = 1681k^2 \)

\( \text{AC} = 41k \) ... [Taking square root of both sides]
Now, cosec \( \theta = \frac{\text{AC}}{\text{AB}} \)

\( = \frac{41k}{9k} \)

\( \operatorname{cosec}\theta = \frac{41}{9} \)
...[By definition]
Also, sin \( \theta = \frac{\text{AB}}{\text{AC}} \)

\( = \frac{9k}{41k} \)

\( \sin\theta = \frac{9}{41} \)
...[By definition]

\( \operatorname{cosec}\theta = \frac{41}{9} \) and sin \( \theta = \frac{9}{41} \)
In simple words: Given cot \(\theta\), we use the identity \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\) to find cosec \(\theta\), and then use \(\sin\theta = \frac{1}{\operatorname{cosec}\theta}\) to find sin \(\theta\). A right-angled triangle can also be used to find the sides and subsequently the required trigonometric ratios.

🎯 Exam Tip: Practice square root calculations for perfect squares to avoid errors. The relationship between reciprocal trigonometric ratios (e.g., sin \(\theta\) and cosec \(\theta\)) is fundamental for efficiency.

Question 4. If 5 sec \( \theta \) – 12 cosec \( \theta \) = 0, find the values of sec \( \theta \), cos \( \theta \) and sin \( \theta \).
Solution:
5 sec \( \theta \) – 12 cosec \( \theta \) = 0 ...[Given]

5 sec \( \theta \) = 12 cosec \( \theta \)

\( \frac{5}{\cos\theta} = \frac{12}{\sin\theta} \) ...[ \( \sec\theta = \frac{1}{\cos\theta} \), \( \operatorname{cosec}\theta = \frac{1}{\sin\theta} \) ]

\( \frac{\sin\theta}{\cos\theta} = \frac{12}{5} \)

\( \tan\theta = \frac{12}{5} \)
We know that,
\( 1 + \tan^2\theta = \sec^2\theta \)

\( 1 + \left(\frac{12}{5}\right)^2 = \sec^2\theta \)

\( 1 + \frac{144}{25} = \sec^2\theta \)

\( \frac{25+144}{25} = \sec^2\theta \)

\( \sec^2\theta = \frac{169}{25} \)

\( \sec\theta = \frac{13}{5} \)
...[Taking square root of both sides]
Now, cos \( \theta = \frac{1}{\sec\theta} \)

\( = \frac{1}{\frac{13}{5}} \)

\( \cos\theta = \frac{5}{13} \)
We know that, \( \sin^2\theta + \cos^2\theta = 1 \)

\( \sin^2\theta + \left(\frac{5}{13}\right)^2 = 1 \)

\( \sin^2\theta + \frac{25}{169} = 1 \)

\( \sin^2\theta = 1 - \frac{25}{169} \)

\( \sin^2\theta = \frac{169-25}{169} \)

\( \sin^2\theta = \frac{144}{169} \)

\( \sin\theta = \frac{12}{13} \)
...[Taking square root of both sides]

\( \sec\theta = \frac{13}{5} \), cos \( \theta = \frac{5}{13} \), sin \( \theta = \frac{12}{13} \)
In simple words: First, simplify the given equation to find tan \(\theta\). Then use the identity \(1 + \tan^2\theta = \sec^2\theta\) to find sec \(\theta\). From sec \(\theta\), find cos \(\theta\) using the reciprocal relationship. Finally, use \(\sin^2\theta + \cos^2\theta = 1\) to find sin \(\theta\).

🎯 Exam Tip: When given an equation relating two trigonometric ratios, simplify it to find one ratio (like tan \(\theta\)) and then use identities to find the others. Ensure all calculated values are positive for angles in the first quadrant.

Question 5. If tan \( \theta \) = 1, then find the value of \( \frac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta} \).
Solution:
tan \( \theta \) = 1 ...[Given]
We know that, tan 45° = 1

tan \( \theta \) = tan 45°

\( \theta \) = 45°
Now, sin \( \theta \) = sin 45° \( = \frac{1}{\sqrt{2}} \)
cos \( \theta \) = cos 45° \( = \frac{1}{\sqrt{2}} \)
sec \( \theta \) = sec 45° \( = \sqrt{2} \)
cosec \( \theta \) = cosec 45° \( = \sqrt{2} \)

\( \frac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta} = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\sqrt{2} + \sqrt{2}} \)

\( = \frac{\frac{2}{\sqrt{2}}}{2\sqrt{2}} \)

\( = \frac{2}{\sqrt{2}} \times \frac{1}{2\sqrt{2}} \)

\( = \frac{1}{2} \)

\( \frac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta} = \frac{1}{2} \)
In simple words: Since tan \(\theta = 1\), we identify \(\theta\) as 45 degrees. Then, we substitute the known values of sin 45°, cos 45°, sec 45°, and cosec 45° into the given expression and simplify to find the final numerical value.

🎯 Exam Tip: Recognizing special angles (like 0°, 30°, 45°, 60°, 90°) and their trigonometric ratios is essential for quickly solving such problems. Simplify fractions involving square roots carefully.

Question 6. Prove that:

(i) \( \frac{\sin^2\theta}{\cos\theta} + \cos\theta = \sec\theta \)
Solution:
L.H.S. \( = \frac{\sin^2\theta}{\cos\theta} + \cos\theta \)

\( = \frac{\sin^2\theta + \cos^2\theta}{\cos\theta} \)

\( = \frac{1}{\cos\theta} \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \) ]

\( = \sec\theta \)
\( = \text{R.H.S.} \)
In simple words: To prove the identity, combine the terms on the left-hand side by finding a common denominator, then use the fundamental identity \(\sin^2\theta + \cos^2\theta = 1\) and the definition \(\frac{1}{\cos\theta} = \sec\theta\).

🎯 Exam Tip: Always start from the more complex side of the identity (usually L.H.S.) and simplify it step-by-step towards the R.H.S. Using fundamental identities is crucial.

(ii) \( \cos^2\theta (1 + \tan^2\theta) = 1 \)
Solution:
L.H.S. \( = \cos^2\theta(1 + \tan^2\theta) \)

\( = \cos^2\theta \sec^2\theta \) ...[ \( 1 + \tan^2\theta = \sec^2\theta \) ]

\( = \cos^2\theta \cdot \frac{1}{\cos^2\theta} \) ...[ \( \sec\theta = \frac{1}{\cos\theta} \) ]

\( = 1 \)
\( = \text{R.H.S.} \)
In simple words: To prove this identity, substitute the identity \(1 + \tan^2\theta = \sec^2\theta\) into the expression, then use the reciprocal identity \(\sec\theta = \frac{1}{\cos\theta}\) and simplify.

🎯 Exam Tip: Recognize and apply the Pythagorean identities immediately. When you have a product like \(\cos^2\theta \sec^2\theta\), consider if they are reciprocals to simplify to 1.

(iii) \( \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} = \sec\theta - \tan\theta \)
Solution:
L.H.S. \( = \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} \)

\( = \sqrt{\frac{1-\sin\theta}{1+\sin\theta} \times \frac{1-\sin\theta}{1-\sin\theta}} \)
On rationalising the denominator

\( = \sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}} \)

\( = \sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}} \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \implies 1-\sin^2\theta = \cos^2\theta \) ]

\( = \frac{1-\sin\theta}{\cos\theta} \)

\( = \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta} \)

\( = \sec\theta - \tan\theta \)
\( = \text{R.H.S.} \)
In simple words: To prove this, rationalize the denominator of the expression inside the square root by multiplying by the conjugate. This simplifies the denominator to \(\cos^2\theta\), allowing you to take the square root and then split the fraction to get \(\sec\theta - \tan\theta\).

🎯 Exam Tip: Rationalization is a common technique when dealing with square roots and sums/differences in the denominator. Remember the identity \(1-\sin^2\theta = \cos^2\theta\) for simplification.

(iv) \( (\sec\theta - \cos\theta)(\cot\theta + \tan\theta) = \tan\theta \cdot \sec\theta \)
Solution:
L.H.S. \( = (\sec\theta - \cos\theta)(\cot\theta + \tan\theta) \)

\( = \left(\frac{1}{\cos\theta} - \cos\theta\right) \left(\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\right) \)

\( = \left(\frac{1-\cos^2\theta}{\cos\theta}\right) \left(\frac{\cos^2\theta + \sin^2\theta}{\sin\theta \cos\theta}\right) \)

\( = \left(\frac{\sin^2\theta}{\cos\theta}\right) \left(\frac{1}{\sin\theta \cos\theta}\right) \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \), \( \sin^2\theta = 1-\cos^2\theta \) ]

\( = \frac{\sin^2\theta}{\cos^2\theta \sin\theta} \)

\( = \frac{\sin\theta}{\cos^2\theta} \)

\( = \frac{\sin\theta}{\cos\theta} \cdot \frac{1}{\cos\theta} \)

\( = \tan\theta \cdot \sec\theta \)
\( = \text{R.H.S.} \)
In simple words: Convert all terms in the L.H.S. to sin \(\theta\) and cos \(\theta\). Simplify each bracket separately by finding a common denominator, then multiply the resulting fractions and use fundamental identities like \(\sin^2\theta + \cos^2\theta = 1\) to reach the R.H.S.

🎯 Exam Tip: When faced with an identity involving multiple ratios, converting all terms to sine and cosine is often the most reliable strategy. Keep track of common denominators and cancellations.

(v) \( \cot\theta + \tan\theta = \operatorname{cosec}\theta \cdot \sec\theta \)
Solution:
L.H.S. \( = \cot\theta + \tan\theta \)

\( = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} \)

\( = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta \cos\theta} \)

\( = \frac{1}{\sin\theta \cos\theta} \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \) ]

\( = \frac{1}{\sin\theta} \cdot \frac{1}{\cos\theta} \)

\( = \operatorname{cosec}\theta \cdot \sec\theta \)
\( = \text{R.H.S.} \)
In simple words: Convert cot \(\theta\) and tan \(\theta\) into their sine and cosine forms. Combine the fractions using a common denominator, then apply the identity \(\sin^2\theta + \cos^2\theta = 1\) and express the result in terms of cosec \(\theta\) and sec \(\theta\).

🎯 Exam Tip: This is a very common identity. Practice converting tan \(\theta\) and cot \(\theta\) to sin/cos and recognizing the result as cosec \(\theta\) and sec \(\theta\).

(vi) \( \frac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta \)
Solution:
L.H.S. \( = \frac{1}{\sec\theta - \tan\theta} \)

\( = \frac{1}{\sec\theta - \tan\theta} \times \frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta} \)
...[On rationalising the denominator]

\( = \frac{\sec\theta + \tan\theta}{\sec^2\theta - \tan^2\theta} \)

\( = \frac{\sec\theta + \tan\theta}{1} \) ...[ \( 1+\tan^2\theta = \sec^2\theta \implies \sec^2\theta - \tan^2\theta = 1 \) ]

\( = \sec\theta + \tan\theta \)
\( = \text{R.H.S.} \)
In simple words: To prove this identity, rationalize the denominator of the left-hand side by multiplying the numerator and denominator by the conjugate term, \(\sec\theta + \tan\theta\). This uses the difference of squares formula and the identity \(\sec^2\theta - \tan^2\theta = 1\).

🎯 Exam Tip: Rationalization with conjugates is a powerful technique for simplifying expressions involving trigonometric sums/differences in the denominator. Always look for the Pythagorean identity \(\sec^2\theta - \tan^2\theta = 1\).

(vii) \( \sin^4\theta - \cos^4\theta = 1 - 2\cos^2\theta \)
Solution:
L.H.S. \( = \sin^4\theta - \cos^4\theta \)

\( = (\sin^2\theta)^2 - (\cos^2\theta)^2 \)

\( = (\sin^2\theta + \cos^2\theta)(\sin^2\theta - \cos^2\theta) \)

\( = (1)(\sin^2\theta - \cos^2\theta) \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \) ]

\( = \sin^2\theta - \cos^2\theta \)

\( = (1 - \cos^2\theta) - \cos^2\theta \) ...[ \( \sin^2\theta = 1 - \cos^2\theta \) ]

\( = 1 - 2\cos^2\theta \)
\( = \text{R.H.S.} \)
In simple words: Start by factoring the left-hand side as a difference of squares. Then apply the fundamental identity \(\sin^2\theta + \cos^2\theta = 1\) and replace \(\sin^2\theta\) with \(1 - \cos^2\theta\) to simplify the expression to the right-hand side.

🎯 Exam Tip: Recognizing algebraic identities like \(a^2 - b^2 = (a-b)(a+b)\) in trigonometric expressions is very helpful. Aim to express everything in terms of the ratio present in the R.H.S., which is \(\cos\theta\) here.

(viii) \( \sec\theta + \tan\theta = \frac{\cos\theta}{1-\sin\theta} \)
Solution:
L.H.S. \( = \sec\theta + \tan\theta \)

\( = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \)

\( = \frac{1 + \sin\theta}{\cos\theta} \)

\( = \frac{1 + \sin\theta}{\cos\theta} \times \frac{1 - \sin\theta}{1 - \sin\theta} \)

\( = \frac{1^2 - \sin^2\theta}{\cos\theta(1 - \sin\theta)} \)

\( = \frac{1 - \sin^2\theta}{\cos\theta(1 - \sin\theta)} \)

\( = \frac{\cos^2\theta}{\cos\theta(1 - \sin\theta)} \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \implies 1-\sin^2\theta = \cos^2\theta \) ]

\( = \frac{\cos\theta}{1 - \sin\theta} \)
\( = \text{R.H.S.} \)
In simple words: Convert sec \(\theta\) and tan \(\theta\) to their sin/cos forms and combine them. Then, multiply the numerator and denominator by the conjugate of the numerator, \(1 - \sin\theta\), to create \(\cos^2\theta\) in the numerator which then simplifies with the denominator.

🎯 Exam Tip: When the target R.H.S. involves \(1 \pm \sin\theta\) or \(1 \pm \cos\theta\), consider multiplying by the conjugate to transform expressions using Pythagorean identities.

(ix) If \( \tan\theta + \frac{1}{\tan\theta} = 2 \), then show that \( \tan^2\theta + \frac{1}{\tan^2\theta} = 2 \)
Solution:
If \( \tan\theta + \frac{1}{\tan\theta} = 2 \) ...[Given]

\( \left(\tan\theta + \frac{1}{\tan\theta}\right)^2 = 4 \)
[Squaring both sides]

\( \tan^2\theta + 2(\tan\theta)\left(\frac{1}{\tan\theta}\right) + \frac{1}{\tan^2\theta} = 4 \)

\( \tan^2\theta + 2 + \frac{1}{\tan^2\theta} = 4 \)

\( \tan^2\theta + \frac{1}{\tan^2\theta} = 4 - 2 \)

\( \tan^2\theta + \frac{1}{\tan^2\theta} = 2 \)
In simple words: Start with the given equation and square both sides. Apply the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\) to expand the left side. Since \( \tan\theta \cdot \frac{1}{\tan\theta} = 1 \), simplify the middle term and rearrange to get the desired result.

🎯 Exam Tip: This problem is more algebraic in nature, relying on squaring binomials. Recognize that \(\tan\theta\) and \(\frac{1}{\tan\theta}\) are reciprocals, making their product 1.

(x) \( \frac{\tan A}{(1+\tan^2A)^2} + \frac{\cot A}{(1+\cot^2A)^2} = \sin A \cos A \)
Solution:
L.H.S. \( = \frac{\tan A}{(1+\tan^2A)^2} + \frac{\cot A}{(1+\cot^2A)^2} \)

\( = \frac{\tan A}{(\sec^2A)^2} + \frac{\cot A}{(\operatorname{cosec}^2A)^2} \) ...[ \( 1+\tan^2\theta = \sec^2\theta \), \( 1+\cot^2\theta = \operatorname{cosec}^2\theta \) ]

\( = \frac{\tan A}{\sec^4A} + \frac{\cot A}{\operatorname{cosec}^4A} \)

\( = \frac{\sin A}{\cos A} \cdot \cos^4A + \frac{\cos A}{\sin A} \cdot \sin^4A \)

\( = \sin A \cos^3A + \cos A \sin^3A \)

\( = \sin A \cos A (\cos^2A + \sin^2A) \)

\( = \sin A \cos A (1) \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \) ]

\( = \sin A \cos A \)
\( = \text{R.H.S.} \)
In simple words: Replace \(1+\tan^2A\) with \(\sec^2A\) and \(1+\cot^2A\) with \(\operatorname{cosec}^2A\). Then convert all terms to sin A and cos A. Simplify the powers and factor out \(\sin A \cos A\), using the identity \(\sin^2A + \cos^2A = 1\).

🎯 Exam Tip: When dealing with higher powers of trigonometric ratios, simplify using fundamental identities first. Look for common factors to extract and reduce the expression to its simplest form.

(xi) \( \sec^4A (1 - \sin^4A) - 2\tan^2A = 1 \)
Solution:
L.H.S. \( = \sec^4A (1 - \sin^4A) - 2\tan^2A \)

\( = \sec^4A [1^2 - (\sin^2A)^2] - 2\tan^2A \)

\( = \sec^4A (1 - \sin^2A) (1 + \sin^2A) - 2\tan^2A \)

\( = \sec^4A \cos^2A (1 + \sin^2A) - 2\tan^2A \) ...[ \( \sin^2\theta + \cos^2\theta = 1 \implies 1-\sin^2\theta = \cos^2\theta \) ]

\( = \frac{1}{\cos^4A} \cos^2A (1 + \sin^2A) - 2\tan^2A \)

\( = \frac{1}{\cos^2A} (1 + \sin^2A) - 2\tan^2A \)

\( = \frac{1}{\cos^2A} + \frac{\sin^2A}{\cos^2A} - 2\tan^2A \)

\( = \sec^2A + \tan^2A - 2\tan^2A \)

\( = \sec^2A - \tan^2A \)

\( = 1 \) ...[ \( \sec^2\theta - \tan^2\theta = 1 \) ]
\( = \text{R.H.S.} \)
In simple words: Factor \(1 - \sin^4A\) using the difference of squares. Use the identities \(1 - \sin^2A = \cos^2A\) and \(\sec A = \frac{1}{\cos A}\) to simplify the first term. Express the terms in sec A and tan A, then use the identity \(\sec^2A - \tan^2A = 1\) for the final simplification.

🎯 Exam Tip: This problem combines multiple identities and algebraic manipulations. Break down complex terms like \(\sin^4A\) into squares to apply identities effectively. Pay attention to exponent rules.

(xii) \( \frac{\tan\theta}{\sec\theta-1} = \frac{\tan\theta + \sec\theta + 1}{\tan\theta + \sec\theta - 1} \)
Solution:
L.H.S. \( = \frac{\tan\theta}{\sec\theta-1} \)

\( = \frac{\tan\theta}{\sec\theta-1} \times \frac{\sec\theta+1}{\sec\theta+1} \)
...[On rationalising the denominator]

\( = \frac{\tan\theta(\sec\theta+1)}{\sec^2\theta-1} \)

\( = \frac{\tan\theta(\sec\theta+1)}{\tan^2\theta} \) ...[ \( 1+\tan^2\theta = \sec^2\theta \implies \sec^2\theta - 1 = \tan^2\theta \) ]

\( = \frac{\sec\theta+1}{\tan\theta} \)
By theorem on equal ratios,

\( \frac{\tan\theta}{\sec\theta-1} = \frac{\sec\theta+1}{\tan\theta} = \frac{\tan\theta+(\sec\theta+1)}{(\sec\theta-1)+\tan\theta} \)

\( = \frac{\tan\theta+\sec\theta+1}{\tan\theta+\sec\theta-1} \)
\( = \text{R.H.S.} \)
In simple words: Start with the L.H.S. and rationalize the denominator by multiplying by the conjugate \(\sec\theta+1\). Use the identity \(\sec^2\theta - 1 = \tan^2\theta\) to simplify. Then, apply the property of equal ratios, \( \frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d} \), to transform the expression into the R.H.S.

🎯 Exam Tip: This problem uses both rationalization and the "theorem on equal ratios" (\( \frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d} \)). Remember to look for these advanced techniques when direct simplification is difficult.

Question 1. Fill in the blanks with reference to the figure given below. (Textbook pg. no. 124)

ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिया गया है, जिसमें कोण B पर समकोण (90 डिग्री) है और कोण C को θ (थीटा) दर्शाया गया है। भुजा AB कोण θ के सामने की भुजा है, भुजा BC कोण θ के संलग्न भुजा है, और भुजा AC कर्ण (hypotenuse) है। इस त्रिभुज के आधार पर त्रिकोणमितीय अनुपात पूछे गए हैं।
Answer:
(i) sin θ = \(\frac{AB}{AC}\)
(ii) cos θ = \(\frac{BC}{AC}\)
(iii) tan θ = \(\frac{AB}{BC}\)
In simple words: This question asks you to recall the basic trigonometric ratios (sine, cosine, tangent) in a right-angled triangle, relating them to the sides (opposite, adjacent, hypotenuse) with respect to a given angle θ.

🎯 Exam Tip: Understanding and correctly identifying the opposite, adjacent, and hypotenuse sides relative to the angle θ is crucial for accurate trigonometric ratio calculations.

Question 2. Complete the relations in ratios given below. (Textbook pg, no. 124)

(i) sin θ / cos θ = _____
(ii) sin θ = cos (90 - _____)
(iii) cos θ = sin (90 - _____)
(iv) tan θ x tan (90 - θ) = _____
Answer:
(i) sin θ / cos θ = [tan θ]
(ii) sin θ = cos (90 - θ)
(iii) cos θ = (90 - θ)
(iv) tan θ x tan (90 - θ) = 1
In simple words: This question tests your knowledge of fundamental trigonometric identities and complementary angle relationships, specifically how sine and cosine, and tangent and cotangent, relate to each other for angles summing to 90 degrees. For (iii), the blank is filled with θ to complete the identity cos θ = sin(90 - θ).

🎯 Exam Tip: Memorizing basic trigonometric identities and complementary angle formulas is essential for quickly solving such fill-in-the-blank questions.

Question 3. Complete the equation. (Textbook pg. no, 124)

sin² θ + cos² θ = _____
Answer:
sin² θ + cos² θ = [1]
In simple words: This is the most fundamental Pythagorean identity in trigonometry, stating that the sum of the squares of sine and cosine of any angle θ is always equal to 1.

🎯 Exam Tip: The identity \( \sin^2 \theta + \cos^2 \theta = 1 \) is a cornerstone of trigonometry and is frequently used to simplify expressions and prove other identities.

Question 4. Write the values of the following trigonometric ratios. (Textbook pg. no. 124)

(i) sin 30° = _____
(ii) cos 30° = _____
(iii) tan 30° = _____
(iv) sin 60° = _____
(v) cos 45° = _____
(vi) tan 45° = _____
Answer:
(i) sin 30° = \(\frac{1}{2}\)
(ii) cos 30° = \(\frac{\sqrt{3}}{2}\)
(iii) tan 30° = \(\frac{1}{\sqrt{3}}\)
(iv) sin 60° = \(\frac{\sqrt{3}}{2}\)
(v) cos 45° = \(\frac{1}{\sqrt{2}}\)
(vi) tan 45° = 1
In simple words: This question requires you to recall the standard trigonometric values for common angles (30°, 45°, 60°), which are often derived from special right-angled triangles.

🎯 Exam Tip: It is highly recommended to memorize the trigonometric values for 0°, 30°, 45°, 60°, and 90° as they are fundamental for solving a wide range of problems in trigonometry.