RBSE Solutions Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Exercise 14.3

Question 1. Prove that \( \cos \theta \tan \theta = \sin \theta \).
Answer: We start with the left-hand side (L.H.S.) of the equation and transform it to match the right-hand side (R.H.S.).
L.H.S. \( = \cos \theta \tan \theta \)
We know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substitute this into the expression:
\( = \cos \theta \times \frac{\sin \theta}{\cos \theta} \)
The \( \cos \theta \) terms in the numerator and denominator cancel each other out.
\( = \sin \theta \)
This matches the R.H.S., so the identity is proven.
In simple words: We changed tan into sin over cos, which let us cancel out the cos, leaving just sin. This shows both sides are the same.

🎯 Exam Tip: When proving trigonometric identities, always start with the more complex side (usually the L.H.S.) and simplify it step-by-step until it matches the simpler side (R.H.S.).

Question 2. Prove that \( (1 - \sin^2 \theta) \tan^2 \theta = \sin^2 \theta \).
Answer: We begin with the left-hand side (L.H.S.) of the given identity.
L.H.S. \( = (1 - \sin^2 \theta) \tan^2 \theta \)
From the fundamental trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we know that \( 1 - \sin^2 \theta = \cos^2 \theta \). Substitute this into the expression.
\( = (\cos^2 \theta) \tan^2 \theta \)
Next, replace \( \tan^2 \theta \) with \( \frac{\sin^2 \theta}{\cos^2 \theta} \).
\( = \cos^2 \theta \times \frac{\sin^2 \theta}{\cos^2 \theta} \)
The \( \cos^2 \theta \) terms cancel each other out.
\( = \sin^2 \theta \)
This result is equal to the right-hand side (R.H.S.), thus proving the identity.
In simple words: We used the rule that \( 1 - \sin^2 \theta \) is the same as \( \cos^2 \theta \). Then, we changed \( \tan^2 \theta \) into \( \sin^2 \theta \) over \( \cos^2 \theta \). This made the \( \cos^2 \theta \) parts cancel, leaving only \( \sin^2 \theta \).

🎯 Exam Tip: Remember the basic identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). These are very useful for simplifying expressions.

Question 3. Prove that \( \frac{\cos^2 \theta}{\sin \theta} + \sin \theta = \operatorname{cosec} \theta \).
Answer: We will simplify the left-hand side (L.H.S.) to reach the right-hand side (R.H.S.).
L.H.S. \( = \frac{\cos^2 \theta}{\sin \theta} + \sin \theta \)
To add these terms, find a common denominator, which is \( \sin \theta \). Multiply the second term by \( \frac{\sin \theta}{\sin \theta} \).
\( = \frac{\cos^2 \theta}{\sin \theta} + \frac{\sin \theta \cdot \sin \theta}{\sin \theta} \)
\( = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta} \)
Using the fundamental trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \).
\( = \frac{1}{\sin \theta} \)
We know that \( \frac{1}{\sin \theta} = \operatorname{cosec} \theta \).
\( = \operatorname{cosec} \theta \)
This is equal to the R.H.S., proving the identity. The reciprocal identity helps link sine to cosecant.
In simple words: We combined the two parts by giving them the same bottom number. Then, \( \cos^2 \theta + \sin^2 \theta \) became 1. Since \( 1 / \sin \theta \) is \( \operatorname{cosec} \theta \), the proof was complete.

🎯 Exam Tip: When adding fractions with trigonometric terms, always find a common denominator first. Also, remember that \( \operatorname{cosec} \theta \) is the reciprocal of \( \sin \theta \).

Question 4. Prove that \( (\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2 = 2 \).
Answer: We expand the left-hand side (L.H.S.) of the equation.
L.H.S. \( = (\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2 \)
Expand the squares using the formula \( (a+b)^2 = a^2 + b^2 + 2ab \) and \( (a-b)^2 = a^2 + b^2 - 2ab \).
\( = (\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta) + (\sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta) \)
Group similar terms together. The \( 2 \sin \theta \cos \theta \) and \( -2 \sin \theta \cos \theta \) terms cancel each other out.
\( = \sin^2 \theta + \cos^2 \theta + \sin^2 \theta + \cos^2 \theta \)
Now, use the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( = ( \sin^2 \theta + \cos^2 \theta ) + ( \sin^2 \theta + \cos^2 \theta ) \)
\( = 1 + 1 \)
\( = 2 \)
This result is equal to the right-hand side (R.H.S.), proving the identity. This shows how algebraic expansion simplifies trigonometric expressions.
In simple words: We opened up both brackets. The middle terms \( (2 \sin \theta \cos \theta) \) cancelled each other out. Then we used the rule that \( \sin^2 \theta + \cos^2 \theta \) equals 1, twice. So, it became \( 1 + 1 \), which is 2.

🎯 Exam Tip: Always look for opportunities to apply the algebraic identities for squares, like \( (a+b)^2 \) and \( (a-b)^2 \), as they often lead to simplification using fundamental trigonometric identities.

Question 5. Prove that \( \operatorname{cosec}^6 \theta - \cot^6 \theta = 1 + 3 \operatorname{cosec}^2 \theta \cot^2 \theta \).
Answer: We will simplify the left-hand side (L.H.S.) to match the right-hand side (R.H.S.).
L.H.S. \( = \operatorname{cosec}^6 \theta - \cot^6 \theta \)
Rewrite this as a difference of cubes, \( (a^3 - b^3) \), where \( a = \operatorname{cosec}^2 \theta \) and \( b = \cot^2 \theta \).
\( = (\operatorname{cosec}^2 \theta)^3 - (\cot^2 \theta)^3 \)
Apply the difference of cubes formula: \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \).
\( = (\operatorname{cosec}^2 \theta - \cot^2 \theta) [(\operatorname{cosec}^2 \theta)^2 + (\cot^2 \theta)^2 + \operatorname{cosec}^2 \theta \cot^2 \theta] \)
From the identity \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \), we know that \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \). Substitute this into the expression.
\( = 1 \cdot [(\operatorname{cosec}^2 \theta)^2 + (\cot^2 \theta)^2 + \operatorname{cosec}^2 \theta \cot^2 \theta] \)
Now, we use the algebraic identity \( a^2 + b^2 = (a-b)^2 + 2ab \). Here, \( a = \operatorname{cosec}^2 \theta \) and \( b = \cot^2 \theta \).
\( = (\operatorname{cosec}^2 \theta - \cot^2 \theta)^2 + 2 \operatorname{cosec}^2 \theta \cot^2 \theta + \operatorname{cosec}^2 \theta \cot^2 \theta \)
Again, use \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \).
\( = (1)^2 + 3 \operatorname{cosec}^2 \theta \cot^2 \theta \)
\( = 1 + 3 \operatorname{cosec}^2 \theta \cot^2 \theta \)
This matches the R.H.S., thus proving the identity. Using algebraic formulas along with trigonometric identities is key here.
In simple words: We saw the problem as "something cubed minus something cubed". We used an algebra formula to break it down. Then we used the trigonometric rule that \( \operatorname{cosec}^2 \theta - \cot^2 \theta \) is 1. We used another algebra trick to simplify the squared terms, which helped us get the final answer.

🎯 Exam Tip: When dealing with higher powers like 6, try to rewrite them as cubes of squares (e.g., \( a^6 = (a^2)^3 \)) and apply algebraic identities like difference of cubes, which often simplify using fundamental trigonometric identities.

Question 7. Prove that \( \frac{\cos\theta}{1-\tan\theta} + \frac{\sin\theta}{1-\cot\theta} = \sin\theta + \cos\theta \).
Answer: We simplify the left-hand side (L.H.S.) of the identity.
L.H.S. \( = \frac{\cos\theta}{1-\tan\theta} + \frac{\sin\theta}{1-\cot\theta} \)
Convert \( \tan \theta \) and \( \cot \theta \) into \( \sin \theta \) and \( \cos \theta \). We know \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).
\( = \frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}} + \frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}} \)
Simplify the denominators by finding a common denominator for each.
\( = \frac{\cos\theta}{\frac{\cos\theta - \sin\theta}{\cos\theta}} + \frac{\sin\theta}{\frac{\sin\theta - \cos\theta}{\sin\theta}} \)
Invert and multiply.
\( = \frac{\cos^2\theta}{\cos\theta - \sin\theta} + \frac{\sin^2\theta}{\sin\theta - \cos\theta} \)
To combine these fractions, make the denominators the same. Notice that \( \sin\theta - \cos\theta = -(\cos\theta - \sin\theta) \).
\( = \frac{\cos^2\theta}{\cos\theta - \sin\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta} \)
Now, combine the numerators over the common denominator.
\( = \frac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta} \)
Use the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \), where \( a = \cos\theta \) and \( b = \sin\theta \).
\( = \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{\cos\theta - \sin\theta} \)
Cancel out the common term \( (\cos\theta - \sin\theta) \).
\( = \cos\theta + \sin\theta \)
This matches the R.H.S., proving the identity. Expressing everything in terms of sine and cosine is a common and effective strategy.
In simple words: First, we changed tan and cot into sin and cos fractions. Then we made the bottoms of the big fractions simple. After that, we noticed one bottom part was the negative of the other, so we changed a plus sign to a minus. Then we used the \( a^2-b^2 \) rule to simplify the top part, and the bottom part cancelled out, leaving \( \sin\theta + \cos\theta \).

🎯 Exam Tip: When fractions involve \( (1-\tan\theta) \) or \( (1-\cot\theta) \), it's often helpful to convert \( \tan\theta \) and \( \cot\theta \) to \( \sin\theta \) and \( \cos\theta \) and then simplify the complex fractions.

Question 9. Prove that \( \frac{\sin\theta}{1-\cos\theta} = \frac{1+\cos\theta}{\sin\theta} \).
Answer: We will start with the right-hand side (R.H.S.) and convert it to the left-hand side (L.H.S.). This is sometimes easier when you have a sum in the numerator.
R.H.S. \( = \frac{1+\cos\theta}{\sin\theta} \)
To get a \( (1-\cos\theta) \) term in the denominator, multiply the numerator and denominator by \( (1-\cos\theta) \). This is a common technique known as multiplying by the conjugate.
\( = \frac{1+\cos\theta}{\sin\theta} \times \frac{1-\cos\theta}{1-\cos\theta} \)
In the numerator, use the difference of squares formula: \( (1+\cos\theta)(1-\cos\theta) = 1^2 - \cos^2\theta \).
\( = \frac{1-\cos^2\theta}{\sin\theta(1-\cos\theta)} \)
From the fundamental identity \( \sin^2\theta + \cos^2\theta = 1 \), we know that \( 1 - \cos^2\theta = \sin^2\theta \).
\( = \frac{\sin^2\theta}{\sin\theta(1-\cos\theta)} \)
Cancel one \( \sin\theta \) from the numerator and denominator.
\( = \frac{\sin\theta}{1-\cos\theta} \)
This matches the L.H.S., thus proving the identity. Multiplying by the conjugate is a powerful method for simplification.
In simple words: We took the right side and multiplied the top and bottom by \( (1-\cos\theta) \). This helped us use the \( a^2-b^2 \) rule on the top part to get \( \sin^2\theta \). Then, one \( \sin\theta \) cancelled out, leaving us with the left side of the problem.

🎯 Exam Tip: If you see \( (1 \pm \sin\theta) \) or \( (1 \pm \cos\theta) \) in an identity, consider multiplying by its conjugate \( (1 \mp \sin\theta) \) or \( (1 \mp \cos\theta) \). This often creates \( (1-\sin^2\theta) \) or \( (1-\cos^2\theta) \), which simplifies to \( \cos^2\theta \) or \( \sin^2\theta \) respectively.

Question 10. Prove that \( \frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = 2 \operatorname{cosec}\theta \).
Answer: We simplify the left-hand side (L.H.S.) of the identity.
L.H.S. \( = \frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} \)
To add these two fractions, find a common denominator, which is \( \sin\theta(1+\cos\theta) \).
\( = \frac{\sin\theta \cdot \sin\theta}{(1+\cos\theta)\sin\theta} + \frac{(1+\cos\theta)(1+\cos\theta)}{\sin\theta(1+\cos\theta)} \)
\( = \frac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)} \)
Expand \( (1+\cos\theta)^2 = 1^2 + 2(1)(\cos\theta) + \cos^2\theta = 1 + 2\cos\theta + \cos^2\theta \).
\( = \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1+\cos\theta)} \)
Rearrange the terms in the numerator and use the identity \( \sin^2\theta + \cos^2\theta = 1 \).
\( = \frac{(\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta}{\sin\theta(1+\cos\theta)} \)
\( = \frac{1 + 1 + 2\cos\theta}{\sin\theta(1+\cos\theta)} \)
\( = \frac{2 + 2\cos\theta}{\sin\theta(1+\cos\theta)} \)
Factor out 2 from the numerator.
\( = \frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} \)
Cancel out the common term \( (1+\cos\theta) \) from the numerator and denominator.
\( = \frac{2}{\sin\theta} \)
Since \( \frac{1}{\sin\theta} = \operatorname{cosec}\theta \).
\( = 2 \operatorname{cosec}\theta \)
This matches the R.H.S., thus proving the identity. Combining fractions and then applying fundamental identities is a standard approach.
In simple words: We added the two fractions by finding a common bottom. Then we expanded the top part, used the rule that \( \sin^2\theta + \cos^2\theta \) is 1, and simplified. We could then take out '2' from the top and cancel the \( (1+\cos\theta) \) part, leaving \( 2 / \sin\theta \), which is \( 2 \operatorname{cosec}\theta \).

🎯 Exam Tip: When adding or subtracting trigonometric fractions, always aim for a common denominator. Look for binomial squares like \( (1+\cos\theta)^2 \) that can be expanded and then simplified using \( \sin^2\theta + \cos^2\theta = 1 \).

Question 11. Prove that \( \frac{(\sin\theta+\cos\theta-1)(\sin\theta+\cos\theta+1)}{\sin\theta \cos\theta} = 2 \).
Answer: We simplify the left-hand side (L.H.S.) of the expression.
L.H.S. \( = \frac{(\sin\theta+\cos\theta-1)(\sin\theta+\cos\theta+1)}{\sin\theta \cos\theta} \)
Notice the numerator is in the form \( (A-B)(A+B) \), where \( A = (\sin\theta+\cos\theta) \) and \( B = 1 \).
Apply the difference of squares formula: \( (A-B)(A+B) = A^2 - B^2 \).
\( = \frac{(\sin\theta+\cos\theta)^2 - 1^2}{\sin\theta \cos\theta} \)
Expand \( (\sin\theta+\cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta \).
\( = \frac{\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta - 1}{\sin\theta \cos\theta} \)
Use the fundamental trigonometric identity \( \sin^2\theta + \cos^2\theta = 1 \).
\( = \frac{1 + 2\sin\theta\cos\theta - 1}{\sin\theta \cos\theta} \)
The \( +1 \) and \( -1 \) in the numerator cancel each other out.
\( = \frac{2\sin\theta\cos\theta}{\sin\theta \cos\theta} \)
Cancel out the common term \( \sin\theta \cos\theta \) from the numerator and denominator.
\( = 2 \)
This matches the R.H.S., thus proving the identity. This problem is a good example of how algebraic identities simplify complex trigonometric expressions.
In simple words: We saw the top part was like \( (A-B)(A+B) \), so we changed it to \( A^2-B^2 \). After expanding \( (\sin\theta+\cos\theta)^2 \), we used the rule that \( \sin^2\theta + \cos^2\theta \) is 1. The '1's cancelled out, and the \( \sin\theta \cos\theta \) parts also cancelled, leaving just 2.

🎯 Exam Tip: Recognize the difference of squares pattern \( (A-B)(A+B) \) in complex expressions. It simplifies quickly, often allowing the use of \( \sin^2\theta + \cos^2\theta = 1 \).

Question 12. Prove that \( \sqrt{\frac{\sec\theta+1}{\sec\theta-1}} = \cot\theta + \operatorname{cosec}\theta \).
Answer: We start by simplifying the left-hand side (L.H.S.).
L.H.S. \( = \sqrt{\frac{\sec\theta+1}{\sec\theta-1}} \)
Multiply the numerator and denominator inside the square root by the conjugate of the denominator, \( (\sec\theta+1) \), to rationalize it.
\( = \sqrt{\frac{\sec\theta+1}{\sec\theta-1} \times \frac{\sec\theta+1}{\sec\theta+1}} \)
This results in a perfect square in the numerator and a difference of squares in the denominator.
\( = \sqrt{\frac{(\sec\theta+1)^2}{\sec^2\theta-1^2}} \)
From the identity \( 1 + \tan^2\theta = \sec^2\theta \), we know that \( \sec^2\theta - 1 = \tan^2\theta \).
\( = \sqrt{\frac{(\sec\theta+1)^2}{\tan^2\theta}} \)
Take the square root of the numerator and denominator.
\( = \frac{\sec\theta+1}{\tan\theta} \)
Separate the fraction into two terms.
\( = \frac{\sec\theta}{\tan\theta} + \frac{1}{\tan\theta} \)
Convert \( \sec\theta \) and \( \tan\theta \) into \( \sin\theta \) and \( \cos\theta \). We know \( \sec\theta = \frac{1}{\cos\theta} \) and \( \tan\theta = \frac{\sin\theta}{\cos\theta} \). Also, \( \frac{1}{\tan\theta} = \cot\theta \).
\( = \frac{1/\cos\theta}{\sin\theta/\cos\theta} + \cot\theta \)
Simplify the first term by inverting and multiplying.
\( = \frac{1}{\cos\theta} \times \frac{\cos\theta}{\sin\theta} + \cot\theta \)
The \( \cos\theta \) terms cancel.
\( = \frac{1}{\sin\theta} + \cot\theta \)
We know that \( \frac{1}{\sin\theta} = \operatorname{cosec}\theta \).
\( = \operatorname{cosec}\theta + \cot\theta \)
This matches the R.H.S., proving the identity. Rationalizing the denominator inside the square root is a key first step.
In simple words: We multiplied the top and bottom inside the square root by \( \sec\theta+1 \). This made the top a perfect square and the bottom \( \tan^2\theta \). After taking the square root, we split the fraction into two parts. We then changed sec and tan into sin and cos, which helped us get cot and cosec.

🎯 Exam Tip: When you see a square root involving \( (A \pm B) \), try multiplying by the conjugate \( (A \mp B) \) inside the root. This often creates perfect squares or differences of squares that simplify nicely.

Question 13. Prove that \( \sqrt{\frac{\operatorname{cosec}^2\theta-1}{\operatorname{cosec}\theta}} = \cos\theta \).
Answer: We begin by simplifying the left-hand side (L.H.S.) of the identity.
L.H.S. \( = \sqrt{\frac{\operatorname{cosec}^2\theta-1}{\operatorname{cosec}\theta}} \)
From the identity \( 1 + \cot^2\theta = \operatorname{cosec}^2\theta \), we know that \( \operatorname{cosec}^2\theta - 1 = \cot^2\theta \). Substitute this into the numerator.
\( = \sqrt{\frac{\cot^2\theta}{\operatorname{cosec}\theta}} \)
Take the square root of the numerator.
\( = \frac{\cot\theta}{\sqrt{\operatorname{cosec}\theta}} \)
This form seems incorrect based on the expected R.H.S. Let's re-examine the question's MathJax or if a different approach is needed. The question is ` \sqrt{\frac { \operatorname{cosec}^2\theta-1 }{ \operatorname{cosec}\theta }} = \cos\theta `. This implies the `\operatorname{cosec}\theta` in the denominator is *outside* the square root, or the problem statement is different from what the OCR picked up. Looking at the OCR image on page 8: the L.H.S. is written as `L.H.S. = \frac{\sqrt{\operatorname{cosec}^2\theta-1}}{\operatorname{cosec}\theta}`. The square root is only over `\operatorname{cosec}^2\theta-1`. I will proceed with this interpretation.
L.H.S. \( = \frac{\sqrt{\operatorname{cosec}^2\theta-1}}{\operatorname{cosec}\theta} \)
Using the identity \( \operatorname{cosec}^2\theta - 1 = \cot^2\theta \), substitute this into the square root.
\( = \frac{\sqrt{\cot^2\theta}}{\operatorname{cosec}\theta} \)
Take the square root of \( \cot^2\theta \).
\( = \frac{\cot\theta}{\operatorname{cosec}\theta} \)
Now, express \( \cot\theta \) and \( \operatorname{cosec}\theta \) in terms of \( \sin\theta \) and \( \cos\theta \). We know \( \cot\theta = \frac{\cos\theta}{\sin\theta} \) and \( \operatorname{cosec}\theta = \frac{1}{\sin\theta} \).
\( = \frac{\cos\theta/\sin\theta}{1/\sin\theta} \)
To simplify, multiply the numerator by the reciprocal of the denominator.
\( = \frac{\cos\theta}{\sin\theta} \times \frac{\sin\theta}{1} \)
The \( \sin\theta \) terms cancel each other out.
\( = \cos\theta \)
This matches the R.H.S., thus proving the identity. Carefully handling square roots and converting to sine/cosine are useful techniques.
In simple words: First, we changed \( \operatorname{cosec}^2\theta - 1 \) into \( \cot^2\theta \) under the square root. Taking the square root, we got \( \cot\theta \) on top. Then we changed \( \cot\theta \) and \( \operatorname{cosec}\theta \) into fractions with \( \sin\theta \) and \( \cos\theta \). The \( \sin\theta \) parts cancelled out, leaving us with \( \cos\theta \).

🎯 Exam Tip: Always pay close attention to where the square root symbol ends. Misinterpreting its scope can lead to incorrect simplification. Also, remember to express cotangent and cosecant in terms of sine and cosine for easier manipulation.