Maharashtra Board Class 10 Maths Chapter 6 Trigonometry Set 6 Solutions

Question 1. Choose the correct alternative answer for the following questions.
(i) sin θ.cosec θ = ?
(A) 1
(B) 0
(C) \( \frac{1}{2} \)
(D) \( \sqrt{2} \)
Answer: (A) 1
In simple words: The product of sine and cosecant of an angle is always 1, as cosecant is the reciprocal of sine.

🎯 Exam Tip: Remember basic trigonometric identities. The reciprocal identities are fundamental for solving many problems.

(ii) cosec 45° = ?
(A) \( \frac{1}{\sqrt{2}} \)
(B) \( \sqrt{2} \)
(C) \( \frac{\sqrt{3}}{1} \)
(D) \( \frac{1}{\sqrt{3}} \)
Answer: (B) \( \sqrt{2} \)
In simple words: Cosec 45° is the reciprocal of sin 45°, and since sin 45° is \( \frac{1}{\sqrt{2}} \), cosec 45° is \( \sqrt{2} \).

🎯 Exam Tip: Memorize the trigonometric ratios for standard angles like 0°, 30°, 45°, 60°, and 90° and their reciprocal functions.

(iii) 1 + tan² θ = ?
(A) cot² θ
(B) cosec² θ
(C) sec² θ
(D) tan² θ
Answer: (C) sec² θ
In simple words: This is one of the fundamental Pythagorean identities in trigonometry, which states that 1 plus tangent squared of an angle equals secant squared of that angle.

🎯 Exam Tip: Understanding and recalling the three Pythagorean identities (\( \sin^2\theta + \cos^2\theta = 1 \), \( 1 + \tan^2\theta = \sec^2\theta \), \( 1 + \cot^2\theta = \operatorname{cosec}^2\theta \)) is crucial for proofs and problem-solving.

(iv) When we see at a higher level, from the horizontal line, angle formed is _____
(A) angle of elevation.
(B) angle of depression.
(C) 0
(D) straight angle.
Answer: (A) angle of elevation.
In simple words: When an observer looks upwards from a horizontal line to an object above, the angle formed is called the angle of elevation.

🎯 Exam Tip: Clearly differentiate between angle of elevation (looking up) and angle of depression (looking down) in problems related to heights and distances.

Question 2. If sin θ = \( \frac{11}{61} \), find the value of cos θ using trigonometric identity.
Answer:
Solution:
sin θ = \( \frac{11}{61} \) ... [Given]
We know that,
\( \sin^2\theta + \cos^2\theta = 1 \)
\( \implies \left(\frac{11}{61}\right)^2 + \cos^2\theta = 1 \)
\( \implies \frac{121}{3721} + \cos^2\theta = 1 \)
\( \implies \cos^2\theta = 1 - \frac{121}{3721} \)
\( \implies \cos^2\theta = \frac{3721 - 121}{3721} \)
\( \implies \cos^2\theta = \frac{3600}{3721} \)
\( \implies \cos\theta = \frac{60}{61} \)
... [Taking square root of both sides]
In simple words: To find cos θ from sin θ, we use the fundamental identity \( \sin^2\theta + \cos^2\theta = 1 \), substitute the given sine value, and then solve for cos θ by taking the square root.

🎯 Exam Tip: Always state the trigonometric identity used. Be careful with calculations, especially when dealing with fractions and squares, and remember to take the square root for the final answer.

Question 3. If tan θ = 2, find the values of other trigonometric ratios.
Answer:
Solution:
tan θ = 2 ...[Given]
We know that,
\( 1 + \tan^2\theta = \sec^2\theta \)
\( \implies 1 + (2)^2 = \sec^2\theta \)
\( \implies 1 + 4 = \sec^2\theta \)
\( \implies \sec^2\theta = 5 \)
\( \implies \sec\theta = \sqrt{5} \) ...[Taking square root of both sides]
\( \implies \cos\theta = \frac{1}{\sec\theta} = \frac{1}{\sqrt{5}} \)
We know that, \( \sin^2\theta + \cos^2\theta = 1 \)
\( \implies \sin^2\theta + \left(\frac{1}{\sqrt{5}}\right)^2 = 1 \)
\( \implies \sin^2\theta + \frac{1}{5} = 1 \)
\( \implies \sin^2\theta = 1 - \frac{1}{5} \)
\( \implies \sin^2\theta = \frac{5-1}{5} \)
\( \implies \sin^2\theta = \frac{4}{5} \)
\( \implies \sin\theta = \frac{2}{\sqrt{5}} \) ...[Taking square root of both sides]
\( \implies \operatorname{cosec}\theta = \frac{1}{\sin\theta} = \frac{1}{\frac{2}{\sqrt{5}}} = \frac{\sqrt{5}}{2} \)
\( \implies \cot\theta = \frac{1}{\tan\theta} = \frac{1}{2} \)
Thus,
\( \sin\theta = \frac{2}{\sqrt{5}} \), \( \cos\theta = \frac{1}{\sqrt{5}} \), \( \cot\theta = \frac{1}{2} \), \( \sec\theta = \sqrt{5} \), \( \operatorname{cosec}\theta = \frac{\sqrt{5}}{2} \)
In simple words: Given tan θ, we first find sec θ using the identity \( 1 + \tan^2\theta = \sec^2\theta \). From sec θ, we find cos θ (its reciprocal). Then, using \( \sin^2\theta + \cos^2\theta = 1 \), we find sin θ, and finally, cosec θ and cot θ from their respective reciprocal relations.

🎯 Exam Tip: Systematically use the Pythagorean identities and reciprocal identities. Ensure all six trigonometric ratios are found and presented clearly.

Question 4. If sec θ = \( \frac{13}{12} \), find the values of other trigonometric ratios.
Answer:
Solution:
sec θ = \( \frac{13}{12} \) ... [Given]
We know that,
\( 1 + \tan^2\theta = \sec^2\theta \)
\( \implies 1 + \tan^2\theta = \left(\frac{13}{12}\right)^2 \)
\( \implies 1 + \tan^2\theta = \frac{169}{144} \)
\( \implies \tan^2\theta = \frac{169}{144} - 1 \)
\( \implies \tan^2\theta = \frac{169 - 144}{144} \)
\( \implies \tan^2\theta = \frac{25}{144} \)
\( \implies \tan\theta = \frac{5}{12} \) ...[Taking square root of both sides]
\( \implies \cot\theta = \frac{1}{\tan\theta} = \frac{1}{\frac{5}{12}} = \frac{12}{5} \)
\( \implies \cos\theta = \frac{1}{\sec\theta} = \frac{1}{\frac{13}{12}} = \frac{12}{13} \)
We know that, \( \sin^2\theta + \cos^2\theta = 1 \)
\( \implies \sin^2\theta + \left(\frac{12}{13}\right)^2 = 1 \)
\( \implies \sin^2\theta + \frac{144}{169} = 1 \)
\( \implies \sin^2\theta = 1 - \frac{144}{169} \)
\( \implies \sin^2\theta = \frac{169 - 144}{169} \)
\( \implies \sin^2\theta = \frac{25}{169} \)
\( \implies \sin\theta = \frac{5}{13} \) ...[Taking square root of both sides]
\( \implies \operatorname{cosec}\theta = \frac{1}{\sin\theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5} \)
Thus,
\( \sin\theta = \frac{5}{13} \), \( \cos\theta = \frac{12}{13} \), \( \tan\theta = \frac{5}{12} \), \( \cot\theta = \frac{12}{5} \), \( \operatorname{cosec}\theta = \frac{13}{5} \)
In simple words: Starting with sec θ, we derive tan θ using the Pythagorean identity \( 1 + \tan^2\theta = \sec^2\theta \). Then, we find cos θ as the reciprocal of sec θ. Finally, sin θ is found using \( \sin^2\theta + \cos^2\theta = 1 \), and its reciprocal cosec θ, along with cot θ (reciprocal of tan θ), are determined.

🎯 Exam Tip: Organize your steps clearly. Start with the given ratio, use the most direct identity to find a related ratio, and then use other identities to find the remaining ratios. Double-check all fraction arithmetic.

Question 5. Prove the following:
i. sec θ (1 – sin θ) (sec θ + tan θ) = 1
Answer:
Proof:
L.H.S. = sec θ (1 – sin θ) (sec θ + tan θ)
\( = \frac{1}{\cos\theta} (1 - \sin\theta) \left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right) \)
\( = \frac{1 - \sin\theta}{\cos\theta} \left(\frac{1 + \sin\theta}{\cos\theta}\right) \)
\( = \frac{(1 - \sin\theta)(1 + \sin\theta)}{\cos^2\theta} \)
\( = \frac{1 - \sin^2\theta}{\cos^2\theta} \)
\( = \frac{\cos^2\theta}{\cos^2\theta} \) [... \( \sin^2\theta + \cos^2\theta = 1 \implies 1 - \sin^2\theta = \cos^2\theta \)]
\( = 1 \)
\( = \text{R.H.S.} \)
\( \implies \sec\theta (1 - \sin\theta) (\sec\theta + \tan\theta) = 1 \)
In simple words: To prove this identity, express all terms in sine and cosine. Combine fractions and use the algebraic identity \((a-b)(a+b) = a^2-b^2\), along with the Pythagorean identity \( 1 - \sin^2\theta = \cos^2\theta \), to simplify the expression to 1.

🎯 Exam Tip: When proving identities, convert all terms to sine and cosine first. Look for opportunities to apply algebraic identities and Pythagorean identities for simplification.

ii. (sec θ + tan θ) (1 – sin θ) = cos θ
Answer:
L.H.S. = (sec θ + tan θ) (1 – sin θ)
\( = \left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right) (1 - \sin\theta) \)
\( = \left(\frac{1 + \sin\theta}{\cos\theta}\right) (1 - \sin\theta) \)
\( = \frac{(1 + \sin\theta)(1 - \sin\theta)}{\cos\theta} \)
\( = \frac{1 - \sin^2\theta}{\cos\theta} \)
\( = \frac{\cos^2\theta}{\cos\theta} \) [... \( \sin^2\theta + \cos^2\theta = 1 \implies 1 - \sin^2\theta = \cos^2\theta \)]
\( = \cos\theta \)
\( = \text{R.H.S.} \)
\( \implies (\sec\theta + \tan\theta) (1 - \sin\theta) = \cos\theta \)
In simple words: Start by rewriting sec θ and tan θ in terms of sine and cosine. Combine the terms in the first parenthesis and then multiply by \((1 - \sin\theta)\). Use the difference of squares identity and the Pythagorean identity to simplify the numerator, which ultimately leads to cos θ.

🎯 Exam Tip: Always work from the more complex side (usually L.H.S.) to the simpler side (R.H.S.). Be meticulous with algebraic manipulations and trigonometric substitutions.

iii. sec² θ + cosec² θ = sec² θ × cosec² θ
Answer:
L.H.S. = sec² θ + cosec² θ
\( = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} \)
\( = \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta \cdot \sin^2\theta} \)
\( = \frac{1}{\cos^2\theta \cdot \sin^2\theta} \) [... \( \sin^2\theta + \cos^2\theta = 1 \)]
\( = \frac{1}{\cos^2\theta} \times \frac{1}{\sin^2\theta} \)
\( = \sec^2\theta \times \operatorname{cosec}^2\theta \)
\( = \text{R.H.S.} \)
\( \implies \sec^2\theta + \operatorname{cosec}^2\theta = \sec^2\theta \times \operatorname{cosec}^2\theta \)
In simple words: Convert sec² θ and cosec² θ into their sine and cosine forms. Add the fractions by finding a common denominator. Use the identity \( \sin^2\theta + \cos^2\theta = 1 \) in the numerator, then separate the terms to show the product of sec² θ and cosec² θ.

🎯 Exam Tip: When sums or differences of trigonometric functions are involved, converting to sine and cosine and finding a common denominator is often a good first step. The identity \( \sin^2\theta + \cos^2\theta = 1 \) is frequently used for simplification.

iv. cot² θ – tan² θ = cosec² θ – sec² θ
Answer:
L.H.S. = cot² θ – tan² θ
\( = (\operatorname{cosec}^2\theta - 1) - (\sec^2\theta - 1) \)
[: \( \tan^2\theta = \sec^2\theta - 1 \), \( \cot^2\theta = \operatorname{cosec}^2\theta - 1 \)]
\( = \operatorname{cosec}^2\theta - 1 - \sec^2\theta + 1 \)
\( = \operatorname{cosec}^2\theta - \sec^2\theta \)
\( = \text{R.H.S.} \)
\( \implies \cot^2\theta - \tan^2\theta = \operatorname{cosec}^2\theta - \sec^2\theta \)
In simple words: This proof uses the alternative forms of the Pythagorean identities: replace cot² θ with \( (\operatorname{cosec}^2\theta - 1) \) and tan² θ with \( (\sec^2\theta - 1) \). The terms \( -1 \) and \( +1 \) cancel out, directly yielding the right-hand side.

🎯 Exam Tip: Be familiar with all forms of the Pythagorean identities. Sometimes, substituting for tan² θ or cot² θ (e.g., \( \tan^2\theta = \sec^2\theta - 1 \)) can be more direct than converting to sine and cosine.

v. tan⁴ θ + tan² θ = sec⁴ θ – sec² θ
Answer:
L.H.S. = tan⁴ θ + tan² θ
\( = \tan^2\theta (\tan^2\theta + 1) \)
\( = \tan^2\theta \cdot \sec^2\theta \) [... \( 1 + \tan^2\theta = \sec^2\theta \)]
\( = (\sec^2\theta - 1) \sec^2\theta \) [... \( \tan^2\theta = \sec^2\theta - 1 \)]
\( = \sec^4\theta - \sec^2\theta \)
\( = \text{R.H.S.} \)
\( \implies \tan^4\theta + \tan^2\theta = \sec^4\theta - \sec^2\theta \)
In simple words: Factor out tan² θ from the left side. Then, use the identity \( 1 + \tan^2\theta = \sec^2\theta \) to replace \( (\tan^2\theta + 1) \) with sec² θ. Finally, replace the remaining tan² θ with \( (\sec^2\theta - 1) \) and distribute to get the desired result.

🎯 Exam Tip: When dealing with higher powers of trigonometric functions, look for common factors and apply identities strategically to reduce powers or change functions. Sometimes, working with secant and tangent is more direct than sine and cosine.

vi. \( \frac{1}{1-\sin\theta} + \frac{1}{1+\sin\theta} = 2\sec^2\theta \)
Answer:
L.H.S. = \( \frac{1}{1-\sin\theta} + \frac{1}{1+\sin\theta} \)
\( = \frac{(1+\sin\theta) + (1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)} \)
\( = \frac{1+\sin\theta+1-\sin\theta}{(1-\sin\theta)(1+\sin\theta)} \)
\( = \frac{2}{1-\sin^2\theta} \)
\( = \frac{2}{\cos^2\theta} \) [... \( 1 - \sin^2\theta = \cos^2\theta \)]
\( = 2 \times \frac{1}{\cos^2\theta} \)
\( = 2\sec^2\theta \)
\( = \text{R.H.S.} \)
\( \implies \frac{1}{1-\sin\theta} + \frac{1}{1+\sin\theta} = 2\sec^2\theta \)
In simple words: Combine the two fractions on the left-hand side by finding a common denominator, which is \((1-\sin\theta)(1+\sin\theta)\). Simplify the numerator, then use the algebraic identity \((a-b)(a+b)=a^2-b^2\) and the Pythagorean identity \(1-\sin^2\theta = \cos^2\theta\) to simplify the expression to \(2\sec^2\theta\).

🎯 Exam Tip: When fractions with denominators like \((1 \pm \sin\theta)\) or \((1 \pm \cos\theta)\) are present, finding a common denominator often leads to \((1 - \sin^2\theta)\) or \((1 - \cos^2\theta)\), which simplifies to \( \cos^2\theta \) or \( \sin^2\theta \) respectively.

vii. sec⁶ x – tan⁶ x = 1 + 3 sec² x tan² x
Answer:
L.H.S. = sec⁶ x – tan⁶ x
\( = (\sec^2 x)^3 - (\tan^2 x)^3 \)
\( = (\sec^2 x - \tan^2 x) ((\sec^2 x)^2 + \sec^2 x \tan^2 x + (\tan^2 x)^2) \) [... \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \)]
We know \( \sec^2 x - \tan^2 x = 1 \)
\( = (1) ((\sec^2 x)^2 + \sec^2 x \tan^2 x + (\tan^2 x)^2) \)
\( = (\sec^2 x)^2 + \sec^2 x \tan^2 x + (\tan^2 x)^2 \)
We know \( \sec^2 x = 1 + \tan^2 x \)
So, \( (\sec^2 x)^2 = (1 + \tan^2 x)^2 = 1 + 2\tan^2 x + \tan^4 x \)
And \( (\tan^2 x)^2 = (\sec^2 x - 1)^2 = \sec^4 x - 2\sec^2 x + 1 \)
Let's use the identity: \( a^3 - b^3 = (a-b)((a+b)^2 - 3ab) \) if it simplifies.
\( (\sec^2 x)^3 - (\tan^2 x)^3 = (\sec^2 x - \tan^2 x) ((\sec^2 x - \tan^2 x)^2 + 3\sec^2 x \tan^2 x) \)
Since \( \sec^2 x - \tan^2 x = 1 \)
\( = (1) ((1)^2 + 3\sec^2 x \tan^2 x) \)
\( = 1 + 3\sec^2 x \tan^2 x \)
\( = \text{R.H.S.} \)
\( \implies \sec^6 x - \tan^6 x = 1 + 3\sec^2 x \tan^2 x \)
In simple words: Treat sec⁶ x and tan⁶ x as cubes of sec² x and tan² x, i.e., \((sec^2x)^3 - (tan^2x)^3\). Use the algebraic identity for \(a^3 - b^3\), specifically \( (a-b)( (a-b)^2 + 3ab ) \). Knowing that \( \sec^2x - \tan^2x = 1 \), substitute this directly into the expression to simplify it to \( 1 + 3\sec^2x \tan^2x \).

🎯 Exam Tip: For higher powers like 6, consider algebraic factorizations like \( a^3 - b^3 \). The key is to recognize that \( \sec^2x - \tan^2x = 1 \), which greatly simplifies the expression when used correctly in the algebraic identity.

viii. \( \frac{\tan\theta}{\sec\theta+1} = \frac{\sec\theta-1}{\tan\theta} \)
Answer:
L.H.S. = \( \frac{\tan\theta}{\sec\theta+1} \)
Multiply numerator and denominator by \( (\sec\theta - 1) \):
\( = \frac{\tan\theta}{\sec\theta+1} \times \frac{\sec\theta-1}{\sec\theta-1} \) [... On rationalising the denominator]
\( = \frac{\tan\theta(\sec\theta-1)}{\sec^2\theta-1} \)
\( = \frac{\tan\theta(\sec\theta-1)}{\tan^2\theta} \) [... \( 1 + \tan^2\theta = \sec^2\theta \implies \sec^2\theta - 1 = \tan^2\theta \)]
\( = \frac{\sec\theta-1}{\tan\theta} \)
\( = \text{R.H.S.} \)
\( \implies \frac{\tan\theta}{\sec\theta+1} = \frac{\sec\theta-1}{\tan\theta} \)
In simple words: To prove this identity, start with the left-hand side and multiply the numerator and denominator by \( (\sec\theta - 1) \). This creates a difference of squares in the denominator \( (\sec^2\theta - 1) \), which simplifies to tan² θ. Cancel out a tan θ term to reach the right-hand side.

🎯 Exam Tip: Rationalizing the denominator (or numerator) by multiplying by its conjugate is a common technique in trigonometric proofs, especially when terms like \( (\sec\theta \pm 1) \) or \( (1 \pm \sin\theta) \) are involved.

ix. \( \frac{\tan^3\theta-1}{\tan\theta-1} = \sec^2\theta + \tan\theta \)
Answer:
L.H.S. = \( \frac{\tan^3\theta-1}{\tan\theta-1} \)
This is in the form \( \frac{a^3-b^3}{a-b} \), where \( a = \tan\theta \) and \( b = 1 \).
Using the algebraic identity \( a^3-b^3 = (a-b)(a^2+ab+b^2) \):
\( = \frac{(\tan\theta-1)(\tan^2\theta + \tan\theta \cdot 1 + 1^2)}{\tan\theta-1} \)
\( = \tan^2\theta + \tan\theta + 1 \)
Rearranging: \( = (1 + \tan^2\theta) + \tan\theta \)
\( = \sec^2\theta + \tan\theta \) [... \( 1 + \tan^2\theta = \sec^2\theta \)]
\( = \text{R.H.S.} \)
\( \implies \frac{\tan^3\theta-1}{\tan\theta-1} = \sec^2\theta + \tan\theta \)
In simple words: Recognize the numerator as a difference of cubes (\( \tan^3\theta - 1^3 \)) and factor it using the algebraic formula \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \). The \(( \tan\theta - 1 )\) terms cancel, leaving \( \tan^2\theta + \tan\theta + 1 \). Finally, apply the Pythagorean identity \( 1 + \tan^2\theta = \sec^2\theta \) to complete the proof.

🎯 Exam Tip: Be mindful of algebraic factorizations (difference of squares, sum/difference of cubes) as they are frequently combined with trigonometric identities to simplify expressions.

x. \( \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} = \frac{1}{\sec\theta-\tan\theta} \)
Answer:
We know that,
\( \sin^2\theta + \cos^2\theta = 1 \)
\( \implies 1 - \sin^2\theta = \cos^2\theta \)
\( \implies (1 - \sin\theta)(1 + \sin\theta) = \cos\theta \cdot \cos\theta \)
By theorem on equal ratios, if \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{a+c}{b+d} \) is equal to them.
So, \( \frac{1+\sin\theta}{\cos\theta} = \frac{\cos\theta}{1-\sin\theta} \)
\( \implies \frac{1+\sin\theta}{\cos\theta} = \frac{\cos\theta}{1-\sin\theta} = \frac{1+\sin\theta+\cos\theta}{\cos\theta+1-\sin\theta} \) ... (i)
Consider,
\( \frac{1}{\sec\theta-\tan\theta} \)
\( = \frac{1}{\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}} \)
\( = \frac{1}{\frac{1-\sin\theta}{\cos\theta}} \)
\( = \frac{\cos\theta}{1-\sin\theta} \)
From (i) and (ii), we get
\( \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} = \frac{1}{\sec\theta-\tan\theta} \)

Alternate method:
L.H.S. = \( \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} \)
Dividing numerator and denominator by cos θ:
\( = \frac{\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\cos\theta}+\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\cos\theta}-\frac{1}{\cos\theta}} \) [...Dividing numerator and denominator by cos θ]
\( = \frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta} \)
\( = \frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1} \)
Replace the '1' in the numerator with \( (\sec^2\theta - \tan^2\theta) \):
\( = \frac{\tan\theta+\sec\theta-(\sec^2\theta-\tan^2\theta)}{\tan\theta-\sec\theta+1} \) [... \( \sec^2\theta-\tan^2\theta = 1 \)]
\( = \frac{(\tan\theta+\sec\theta)-(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}{\tan\theta-\sec\theta+1} \)
Factor out \( (\tan\theta+\sec\theta) \) from the numerator:
\( = \frac{(\tan\theta+\sec\theta)[1-(\sec\theta-\tan\theta)]}{\tan\theta-\sec\theta+1} \)
\( = \frac{(\tan\theta+\sec\theta)(1-\sec\theta+\tan\theta)}{\tan\theta-\sec\theta+1} \)
\( = \tan\theta+\sec\theta \)
Now for the R.H.S.:
R.H.S. = \( \frac{1}{\sec\theta-\tan\theta} \)
Multiply numerator and denominator by \( (\sec\theta+\tan\theta) \):
\( = \frac{1}{\sec\theta-\tan\theta} \times \frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta} \)
\( = \frac{\sec\theta+\tan\theta}{\sec^2\theta-\tan^2\theta} \)
\( = \frac{\sec\theta+\tan\theta}{1} \) [... \( \sec^2\theta-\tan^2\theta = 1 \)]
\( = \sec\theta+\tan\theta \)
Since L.H.S. = R.H.S., the identity is proven.
In simple words: This complex identity can be proven by either using properties of equal ratios or by a more common method: divide the numerator and denominator of the L.H.S. by cos θ. Then, in the numerator, replace '1' with \( (\sec^2\theta - \tan^2\theta) \), factor using difference of squares, and simplify. The result will match the simplified R.H.S. after rationalizing it.

🎯 Exam Tip: For complex fractions, dividing by cos θ (or sin θ) to convert to tan/sec (or cot/cosec) can be very effective. Look for ways to use \( \sec^2\theta - \tan^2\theta = 1 \) strategically, either by substitution or by rationalization.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज ABC दिखाता है जहाँ कोण C 30 डिग्री है और भुजा BC 48 मीटर है। बिंदु A शीर्ष पर है और B आधार पर है, जो इमारत की ऊंचाई AB का प्रतिनिधित्व करता है।

Question 6. A boy standing at a distance of 48 metres from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
Answer:
Solution:
Let AB represent the height of the building and point C represent the position of the boy.
Angle of elevation = ∠ACB = 30°
BC = 48 m
In right angled \( \triangle \)ABC,
\( \tan 30^\circ = \frac{AB}{BC} \) ... [By definition]
\( \implies \frac{1}{\sqrt{3}} = \frac{AB}{48} \)
\( \implies AB = \frac{48}{\sqrt{3}} \)
To rationalise the denominator:
\( \implies AB = \frac{48}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \) [... On rationalising the denominator]
\( \implies AB = \frac{48\sqrt{3}}{3} \)
\( \implies AB = 16\sqrt{3} \) m
The height of the building is \( 16\sqrt{3} \) m.
In simple words: We have a right-angled triangle formed by the building, the ground, and the line of sight. Given the distance from the building (BC) and the angle of elevation (∠ACB), we use the tangent function (\( \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} \)) to find the height of the building (AB). The denominator is rationalized for a clean final answer.

🎯 Exam Tip: Draw a clear diagram for problems involving heights and distances. Identify the right-angled triangle, the known sides/angles, and the unknown quantity. Choose the appropriate trigonometric ratio (sin, cos, or tan) based on the given information. Rationalize denominators if square roots remain.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक प्रकाशस्तंभ AB को दर्शाता है जिसकी ऊंचाई 100 मीटर है। एक प्रेक्षक बिंदु A से एक जहाज C को 30 डिग्री के अवसाद कोण (∠PAC) पर देख रहा है, जहाँ AP क्षैतिज रेखा है।

Question 7. From the top of the lighthouse, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the lighthouse is 100 metres, then find how far the ship is from the lighthouse.
Answer:
Solution:
Let AB represent the height of lighthouse and point C represent the position of the ship.
Angle of depression ∠PAC = 30°
AB = 100m.
Now, ray AP || seg BC
\( \implies \angle ACB = \angle PAC \) ... [Alternate angles]
\( \implies \angle ACB = 30^\circ \)
In right angled \( \triangle \)ABC,
\( \tan 30^\circ = \frac{AB}{BC} \) ... [By definition]
\( \implies \frac{1}{\sqrt{3}} = \frac{100}{BC} \)
\( \implies BC = 100\sqrt{3} \) m
The ship is \( 100\sqrt{3} \) m far from the lighthouse.
In simple words: The angle of depression from the lighthouse top to the ship is equal to the angle of elevation from the ship to the lighthouse top (alternate interior angles). Using the height of the lighthouse (AB) and this angle (∠ACB), we apply the tangent function (\( \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} \)) to find the distance of the ship from the lighthouse (BC).

🎯 Exam Tip: Correctly identify and use alternate angles when given an angle of depression. Always draw the horizontal line at the observer's eye level to define angles of elevation or depression. Select the appropriate trigonometric ratio to solve for the unknown distance.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो इमारतों AB और CD को दर्शाता है, जहाँ AB की ऊंचाई 12 मीटर है और सड़क BD की चौड़ाई 15 मीटर है। बिंदु A से दूसरी इमारत CD के शीर्ष C का उन्नयन कोण 30 डिग्री है, और AM एक क्षैतिज रेखा है जो CD पर लंबवत है।

Question 8. Two buildings are in front of each other on a road of width 15 metres. From the top of the first building, having a height of 12 metre, the angle of elevation of the top of the second building is 30°. What is the height of the second building?
Answer:
Solution:
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.
Draw seg AM \( \perp \) seg CD.
Angle of elevation = ∠CAM = 30°
AB = 12m
BD = 15m
In \( \square \)ABDM,
∠B = ∠D = 90°
∠M = 90° ...[seg AM \( \perp \) seg CD]
∠A = 90° ...[Remaining angle of \( \square \)ABDM]
\( \square \)ABDM is a rectangle ... [Each angle is 90°]
\( \implies AM = BD = 15 \) m
\( \implies DM = AB = 12 \) m } ... [Opposite sides of a rectangle]
In right angled \( \triangle \)AMC,
\( \tan 30^\circ = \frac{CM}{AM} \) ...[By definition]
\( \implies \frac{1}{\sqrt{3}} = \frac{CM}{15} \)
\( \implies CM = \frac{15}{\sqrt{3}} \)
To rationalise the denominator:
\( \implies CM = \frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \) [... On rationalising the denominator]
\( \implies CM = \frac{15\sqrt{3}}{3} \)
\( \implies CM = 5\sqrt{3} \) m
Now, CD = DM + CM ... [C-M-D]
\( = 12 + 5\sqrt{3} \)
\( = 12 + 5 \times 1.73 \)
\( = 12 + 8.65 = 20.65 \) m
The height of the second building is 20.65 m.
In simple words: First, create a rectangle from the base of the buildings and the line from the top of the first building to the second. This helps determine the horizontal distance (AM) and vertical height (DM) for the triangle formed. Use the angle of elevation and the horizontal distance to find the additional height (CM) of the second building using the tangent function. Add this to the height of the first building to get the total height of the second building.

🎯 Exam Tip: Clearly label your diagram and identify any rectangles or right-angled triangles formed. Remember that the horizontal distance between the buildings is the same as the base of the right-angled triangle used for the angle of elevation. Use the value of \( \sqrt{3} \) accurately if required for numerical answers.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक सीढ़ी AB को दर्शाता है जिसकी लंबाई 20 मीटर है और यह एक प्लेटफॉर्म AE से 70 डिग्री के कोण पर ऊपर की ओर उठी हुई है। प्लेटफॉर्म की ऊंचाई 2 मीटर है। बिंदु B सीढ़ी का शीर्ष है और C वह बिंदु है जहाँ से सीढ़ी जमीन से ऊंचाई BC तक पहुँचती है।

Question 9. A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)
Answer:
Solution:
Let AB represent the length of the ladder and AE represent the height of the platform.
Draw seg AC \( \perp \) seg BD.
Angle of elevation = ∠BAC = 70°
AB = 20 m
AE = 2m
In right angled \( \triangle \)ABC,
\( \sin 70^\circ = \frac{BC}{AB} \) .....[By definition]
\( \implies 0.94 = \frac{BC}{20} \)
\( \implies BC = 0.94 \times 20 \)
\( = 18.80 \) m
In \( \square \)ACDE,
∠E = ∠D = 90°
∠C = 90° ... [seg AC \( \perp \) seg BD]
∠A = 90° ... [Remaining angle of \( \square \)ACDE]
\( \square \)ACDE is a rectangle. ... [Each angle is 90°]
\( \implies CD = AE = 2 \) m ... [Opposite sides of a rectangle]
Now, BD = BC + CD ... [B - C - D]
\( = 18.80 + 2 \)
\( = 20.80 \) m
The maximum height from the ground upto which the ladder can reach is 20.80 metres.
In simple words: The ladder's extended length (AB) and elevation angle (∠BAC) form a right triangle above the platform. Use the sine function to find the height (BC) the ladder extends above the platform. Then, add this height to the platform's height (CD, which equals AE) to find the total maximum height from the ground.

🎯 Exam Tip: For problems involving platforms or raised observation points, remember to add the initial height of the platform to the calculated vertical distance from the trigonometric calculation to get the total height from the ground. Use the provided trigonometric values accurately.

Question 10. While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing, (sin 20° = 0.342)
Answer: Solution:
Let AC represent the initial height and point A represent the initial position of the plane.
Let point B represent the position where plane lands.
Angle of depression = \( \angle EAB = 20° \)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिखाया गया है, जिसमें बिंदु A पर हवाई जहाज की शुरुआती स्थिति है और C हवाई जहाज की शुरुआती ऊंचाई दर्शाती है। क्षैतिज रेखा AE है। कोण EAB 20° का अवसाद कोण है। बिंदु B वह जगह है जहाँ हवाई जहाज उतरता है, और ABC भी 20° है। Now, seg AE || seg BC
\( \therefore \angle ABC = \angle EAB \) ... [Alternate angles]
\( \therefore \angle ABC = 20° \) Speed of the plane = 200 km/hr
\( = 200 \times \frac{1000}{3600} \) m/sec
\( = \frac{500}{9} \) m/sec
\( \therefore \) Distance travelled in 54 sec = speed \( \times \) time
\( = \frac{500}{9} \times 54 \)
\( = 3000 \) m
\( \therefore AB = 3000 \) m In right angled \( \triangle ABC \),
\( \sin 20° = \frac{AC}{AB} \) ...[By definition]
\( \therefore 0.342 = \frac{AC}{3000} \)
\( \therefore AC = 0.342 \times 3000 \)
\( = 1026 \) m
\( \therefore \) The plane was at a height of 1026 m when it started landing.
In simple words: First, calculate the distance the plane traveled using its speed and time. Then, use the given angle of depression and the sine trigonometric ratio in the right-angled triangle formed by the plane's path to find its initial height.

🎯 Exam Tip: Remember to convert units (km/hr to m/s) and clearly label your diagram. Showing the trigonometric ratio formula and substituting values correctly are key steps for full marks.