OP Malhotra Class 10 Maths Solutions Chapter 16 Trigonometrical Identities and Tables Exercise 16 (A)

S Chand Class 10 ICSE Maths Solutions Chapter 16 Trigonometrical Identities and Tables Ex 16(a)

Question 1. \( \frac{1-\cos ^2 \theta}{\sin ^2 \theta} = 1 \)
Answer:
L.H.S. \( = \frac{1-\cos ^2 \theta}{\sin ^2 \theta} \)
We know the identity \( 1-\cos ^2 \theta = \sin ^2 \theta \).
So, L.H.S. \( = \frac{\sin ^2 \theta}{\sin ^2 \theta} \)
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side of the equation. We use a basic rule of trigonometry that says "one minus cosine squared theta" is the same as "sine squared theta." Then, we have sine squared theta divided by sine squared theta, which simply equals one. This matches the right side of the equation.

🎯 Exam Tip: When proving identities, always state the trigonometric identities you are using in each step for clarity and to score full marks.

Question 2. \( \frac{1-\sin ^2 \theta}{\cos ^2 \theta} = 1 \)
Answer:
L.H.S. \( = \frac{1-\sin ^2 \theta}{\cos ^2 \theta} \)
We know the identity \( 1-\sin ^2 \theta = \cos ^2 \theta \). This is a fundamental identity derived from \( \sin^2 \theta + \cos^2 \theta = 1 \).
So, L.H.S. \( = \frac{\cos ^2 \theta}{\cos ^2 \theta} \)
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: For the left side, we replace "one minus sine squared theta" with "cosine squared theta" because they are equal. Then, cosine squared theta divided by itself is one, which is what the right side of the equation is.

🎯 Exam Tip: Remember the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) and its rearrangements (\( 1-\sin^2 \theta = \cos^2 \theta \) and \( 1-\cos^2 \theta = \sin^2 \theta \)) as they are very commonly used.

Question 3. \( \sin A \cdot \cot A = \cos A \)
Answer:
L.H.S. \( = \sin A \cdot \cot A \)
We know the identity \( \cot A = \frac{\cos A}{\sin A} \). This is the definition of cotangent.
So, L.H.S. \( = \sin A \cdot \frac{\cos A}{\sin A} \)
We can cancel \( \sin A \) from the numerator and denominator.
L.H.S. \( = \cos A \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. We change "cot A" to "cos A divided by sin A". Then, the "sin A" terms cancel each other out, leaving only "cos A", which is the right side of the equation.

🎯 Exam Tip: When dealing with trigonometric functions like cotangent, tangent, secant, and cosecant, converting them into sine and cosine is often the first step to simplify expressions.

Question 4. \( \frac{1}{\cos ^2 \theta}-\tan ^2 \theta=1 \)
Answer:
L.H.S. \( = \frac{1}{\cos ^2 \theta}-\tan ^2 \theta \)
We know the identity \( \frac{1}{\cos ^2 \theta} = \sec ^2 \theta \). Secant is the reciprocal of cosine.
So, L.H.S. \( = \sec ^2 \theta - \tan ^2 \theta \)
We also know the Pythagorean identity \( \sec ^2 \theta - \tan ^2 \theta = 1 \).
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: The left side starts with one over cosine squared theta, which is the same as secant squared theta. So, the equation becomes secant squared theta minus tangent squared theta. We know that this always equals one, which is the right side.

🎯 Exam Tip: Memorize the three main Pythagorean identities: \( \sin^2 \theta + \cos^2 \theta = 1 \), \( 1 + \tan^2 \theta = \sec^2 \theta \), and \( 1 + \cot^2 \theta = \cosec^2 \theta \).

Question 5. \( \tan ^2 A \cos ^2 A = 1 - \cos ^2 A \)
Answer:
L.H.S. \( = \tan ^2 A \cos ^2 A \)
We know the identity \( \tan ^2 A = \frac{\sin ^2 A}{\cos ^2 A} \). Tangent is sine divided by cosine.
So, L.H.S. \( = \frac{\sin ^2 A}{\cos ^2 A} \cdot \cos ^2 A \)
We can cancel \( \cos ^2 A \) from the numerator and denominator.
L.H.S. \( = \sin ^2 A \)
We also know the identity \( \sin ^2 A = 1 - \cos ^2 A \). This identity comes from \( \sin^2 A + \cos^2 A = 1 \).
So, L.H.S. \( = 1 - \cos ^2 A \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: On the left, we change tangent squared A to sine squared A over cosine squared A. The cosine squared A terms then cancel out, leaving sine squared A. We know that sine squared A is also equal to one minus cosine squared A, which is the right side of the equation.

🎯 Exam Tip: Always work from one side (usually the more complex side) to the other until both sides match. Avoid manipulating both sides simultaneously unless it's a very simple step.

Question 6. \( \tan \theta = \frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}} \)
Answer:
R.H.S. \( = \frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}} \)
We know the identity \( 1-\sin ^2 \theta = \cos ^2 \theta \). This is a rearranged Pythagorean identity.
So, R.H.S. \( = \frac{\sin \theta}{\sqrt{\cos ^2 \theta}} \)
The square root of \( \cos ^2 \theta \) is \( \cos \theta \).
R.H.S. \( = \frac{\sin \theta}{\cos \theta} \)
We know the identity \( \frac{\sin \theta}{\cos \theta} = \tan \theta \). This is the definition of tangent.
R.H.S. \( = \tan \theta \)
This equals the L.H.S.
Therefore, the identity is proven.
In simple words: We start with the right side. Inside the square root, "one minus sine squared theta" becomes "cosine squared theta". The square root of cosine squared theta is just cosine theta. So, we get sine theta divided by cosine theta, which we know is tangent theta, the left side.

🎯 Exam Tip: Be careful with square roots: \( \sqrt{x^2} = |x| \). In trigonometric identities, \( \sqrt{\cos^2 \theta} \) is usually taken as \( \cos \theta \) assuming the angle is in a quadrant where \( \cos \theta \) is positive, or the context is for general identity proof.

Question 7. \( \frac{1+\cos \theta}{\sin ^2 \theta} = \frac{1}{1-\cos \theta} \)
Answer:
L.H.S. \( = \frac{1+\cos \theta}{\sin ^2 \theta} \)
We know the identity \( \sin ^2 \theta = 1 - \cos ^2 \theta \). This helps convert the expression to terms of cosine.
So, L.H.S. \( = \frac{1+\cos \theta}{1-\cos ^2 \theta} \)
We can factor the denominator using the difference of squares formula: \( a^2 - b^2 = (a+b)(a-b) \).
Here, \( 1-\cos ^2 \theta = (1+\cos \theta)(1-\cos \theta) \).
So, L.H.S. \( = \frac{1+\cos \theta}{(1+\cos \theta)(1-\cos \theta)} \)
We can cancel \( (1+\cos \theta) \) from the numerator and denominator.
L.H.S. \( = \frac{1}{1-\cos \theta} \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: For the left side, we change sine squared theta to one minus cosine squared theta. We then break down the bottom part using a special algebra rule, which allows us to cancel out "one plus cosine theta" from both the top and the bottom, leaving just one over "one minus cosine theta", which is the right side.

🎯 Exam Tip: The difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \) is extremely useful in simplifying trigonometric fractions when you have \( 1-\sin^2 \theta \) or \( 1-\cos^2 \theta \).

Question 8. \( \cot ^2 \theta (1-\cos ^2 \theta) = \cos ^2 \theta \)
Answer:
L.H.S. \( = \cot ^2 \theta (1-\cos ^2 \theta) \)
First, we use the identity \( 1-\cos ^2 \theta = \sin ^2 \theta \). This helps simplify the second term.
So, L.H.S. \( = \cot ^2 \theta \cdot \sin ^2 \theta \)
Next, we use the identity \( \cot ^2 \theta = \frac{\cos ^2 \theta}{\sin ^2 \theta} \). Cotangent is cosine over sine.
So, L.H.S. \( = \frac{\cos ^2 \theta}{\sin ^2 \theta} \cdot \sin ^2 \theta \)
We can cancel \( \sin ^2 \theta \) from the numerator and denominator.
L.H.S. \( = \cos ^2 \theta \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. "One minus cosine squared theta" becomes "sine squared theta". Then, "cotangent squared theta" becomes "cosine squared theta over sine squared theta". The sine squared theta terms cancel out, leaving only cosine squared theta, which is the right side.

🎯 Exam Tip: When terms are multiplied, converting all functions to sine and cosine is often effective, as it allows for cancellations and simplifications.

Question 9. \( \tan ^2 \theta (1 – \sin ^2 \theta) = \sin ^2 \theta \)
Answer:
L.H.S. \( = \tan ^2 \theta (1 – \sin ^2 \theta) \)
First, we use the identity \( 1-\sin ^2 \theta = \cos ^2 \theta \). This is a direct rearrangement of the Pythagorean identity.
So, L.H.S. \( = \tan ^2 \theta \cdot \cos ^2 \theta \)
Next, we use the identity \( \tan ^2 \theta = \frac{\sin ^2 \theta}{\cos ^2 \theta} \). Tangent is sine over cosine.
So, L.H.S. \( = \frac{\sin ^2 \theta}{\cos ^2 \theta} \cdot \cos ^2 \theta \)
We can cancel \( \cos ^2 \theta \) from the numerator and denominator.
L.H.S. \( = \sin ^2 \theta \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We take the left side. "One minus sine squared theta" is changed to "cosine squared theta". Then, "tangent squared theta" is written as "sine squared theta over cosine squared theta". The cosine squared theta terms cancel, leaving sine squared theta, which is the right side of the equation.

🎯 Exam Tip: Look for opportunities to simplify expressions by using Pythagorean identities or by converting tangent and cotangent into sine and cosine ratios.

Question 10. \( (1 – \sin ^2 \theta) \sec ^2 \theta = 1 \)
Answer:
L.H.S. \( = (1 – \sin ^2 \theta) \sec ^2 \theta \)
First, we use the identity \( 1-\sin ^2 \theta = \cos ^2 \theta \). This is a fundamental Pythagorean identity.
So, L.H.S. \( = \cos ^2 \theta \cdot \sec ^2 \theta \)
Next, we use the identity \( \sec ^2 \theta = \frac{1}{\cos ^2 \theta} \). Secant is the reciprocal of cosine.
So, L.H.S. \( = \cos ^2 \theta \cdot \frac{1}{\cos ^2 \theta} \)
We can cancel \( \cos ^2 \theta \) from the numerator and denominator.
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: Starting with the left side, we change "one minus sine squared theta" to "cosine squared theta". Then, "secant squared theta" is written as "one over cosine squared theta". The cosine squared theta terms cancel, resulting in one, which is the right side.

🎯 Exam Tip: When you see reciprocal functions (like secant and cosecant), consider converting them to their basic sine or cosine forms to reveal cancellations.

Question 11. \( (1 – \cos ^2 \theta) \cosec ^2 \theta = 1 \)
Answer:
L.H.S. \( = (1 – \cos ^2 \theta) \cosec ^2 \theta \)
First, we use the identity \( 1-\cos ^2 \theta = \sin ^2 \theta \). This is a fundamental trigonometric identity.
So, L.H.S. \( = \sin ^2 \theta \cdot \cosec ^2 \theta \)
Next, we use the identity \( \cosec ^2 \theta = \frac{1}{\sin ^2 \theta} \). Cosecant is the reciprocal of sine.
So, L.H.S. \( = \sin ^2 \theta \cdot \frac{1}{\sin ^2 \theta} \)
We can cancel \( \sin ^2 \theta \) from the numerator and denominator.
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: On the left side, "one minus cosine squared theta" is replaced by "sine squared theta". Then, "cosecant squared theta" is replaced by "one over sine squared theta". The sine squared theta terms cancel out, leaving one, which is the right side of the equation.

🎯 Exam Tip: Identifying reciprocal pairs like \( \sin \theta \) and \( \cosec \theta \) (where \( \sin \theta \cdot \cosec \theta = 1 \)) can quickly simplify complex expressions.

Question 12. \( \sin ^2 \theta + \frac{1}{1+\tan ^2 \theta} = 1 \)
Answer:
L.H.S. \( = \sin ^2 \theta + \frac{1}{1+\tan ^2 \theta} \)
We know the identity \( 1+\tan ^2 \theta = \sec ^2 \theta \). This is one of the Pythagorean identities.
So, L.H.S. \( = \sin ^2 \theta + \frac{1}{\sec ^2 \theta} \)
We know the identity \( \frac{1}{\sec ^2 \theta} = \cos ^2 \theta \). Cosine is the reciprocal of secant.
So, L.H.S. \( = \sin ^2 \theta + \cos ^2 \theta \)
We know the fundamental Pythagorean identity \( \sin ^2 \theta + \cos ^2 \theta = 1 \).
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: For the left side, we replace "one plus tangent squared theta" with "secant squared theta". Then, "one over secant squared theta" becomes "cosine squared theta". Finally, sine squared theta plus cosine squared theta always equals one, which is the right side.

🎯 Exam Tip: When simplifying, always try to bring expressions back to sine and cosine or to one of the fundamental identities. This provides a clear path to the solution.

Question 13. \( \cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}=1 \)
Answer:
L.H.S. \( = \cos ^2 \theta+\frac{1}{1+\cot ^2 \theta} \)
We know the identity \( 1+\cot ^2 \theta = \cosec ^2 \theta \). This is another Pythagorean identity.
So, L.H.S. \( = \cos ^2 \theta + \frac{1}{\cosec ^2 \theta} \)
We know the identity \( \frac{1}{\cosec ^2 \theta} = \sin ^2 \theta \). Sine is the reciprocal of cosecant.
So, L.H.S. \( = \cos ^2 \theta + \sin ^2 \theta \)
We know the fundamental Pythagorean identity \( \cos ^2 \theta + \sin ^2 \theta = 1 \).
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. "One plus cotangent squared theta" is changed to "cosecant squared theta". Then, "one over cosecant squared theta" is the same as "sine squared theta". Adding cosine squared theta and sine squared theta always gives one, which is the right side.

🎯 Exam Tip: Recognizing the sum of \( \sin^2 \theta \) and \( \cos^2 \theta \) as 1 is a crucial simplification step in many proofs.

Question 14. \( \frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta}=1 \)
Answer:
L.H.S. \( = \frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta} \)
We know the fundamental Pythagorean identity \( \sin ^2 \theta+\cos ^2 \theta = 1 \). This simplifies the numerator.
We also know the Pythagorean identity \( \sec ^2 \theta-\tan ^2 \theta = 1 \). This simplifies the denominator.
So, L.H.S. \( = \frac{1}{1} \)
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: The top part of the fraction, sine squared plus cosine squared, always equals one. The bottom part, secant squared minus tangent squared, also always equals one. So, the fraction becomes one divided by one, which is simply one.

🎯 Exam Tip: This question directly tests your knowledge of two key Pythagorean identities. Using them promptly simplifies the proof significantly.

Question 15. \( \left(\frac{\cos ^2 A}{\sin ^2 A}+1\right) \tan ^2 A = \frac{1}{\cos ^2 A} \)
Answer:
L.H.S. \( = \left(\frac{\cos ^2 A}{\sin ^2 A}+1\right) \tan ^2 A \)
We know that \( \frac{\cos ^2 A}{\sin ^2 A} = \cot ^2 A \). This helps convert the first term inside the bracket.
So, L.H.S. \( = (\cot ^2 A+1) \tan ^2 A \)
We know the identity \( \cot ^2 A+1 = \cosec ^2 A \). This simplifies the bracket.
So, L.H.S. \( = \cosec ^2 A \cdot \tan ^2 A \)
Next, we convert \( \cosec ^2 A \) to \( \frac{1}{\sin ^2 A} \) and \( \tan ^2 A \) to \( \frac{\sin ^2 A}{\cos ^2 A} \). It helps to express everything in terms of sine and cosine.
So, L.H.S. \( = \frac{1}{\sin ^2 A} \cdot \frac{\sin ^2 A}{\cos ^2 A} \)
We can cancel \( \sin ^2 A \) from the numerator and denominator.
L.H.S. \( = \frac{1}{\cos ^2 A} \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We begin with the left side. "Cosine squared A over sine squared A" is cotangent squared A. So the bracket becomes "cotangent squared A plus one", which simplifies to cosecant squared A. Then, we write cosecant squared A as one over sine squared A, and tangent squared A as sine squared A over cosine squared A. The sine squared A terms cancel, leaving one over cosine squared A, which is the right side.

🎯 Exam Tip: When faced with multiple trigonometric functions, converting them to sine and cosine is a reliable strategy for simplification, especially when addition or subtraction is involved.

Question 16. \( \sin ^4 \theta + \sin ^2 \theta \cos ^2 \theta = \sin ^2 \theta \)
Answer:
L.H.S. \( = \sin ^4 \theta + \sin ^2 \theta \cos ^2 \theta \)
We can factor out \( \sin ^2 \theta \) from both terms. This is a common first step when terms share a common factor.
So, L.H.S. \( = \sin ^2 \theta (\sin ^2 \theta + \cos ^2 \theta) \)
We know the fundamental Pythagorean identity \( \sin ^2 \theta + \cos ^2 \theta = 1 \).
So, L.H.S. \( = \sin ^2 \theta \cdot 1 \)
L.H.S. \( = \sin ^2 \theta \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. We can take out "sine squared theta" as a common part. Inside the bracket, "sine squared theta plus cosine squared theta" becomes one. So, we are left with sine squared theta multiplied by one, which is just sine squared theta. This is the same as the right side.

🎯 Exam Tip: Always look for common factors in expressions. Factoring can simplify a problem by reducing it to a known identity, like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 17. \( \sin ^4 \theta + 2 \sin ^2 \theta \cos ^2 \theta + \cos ^4 \theta = 1 \)
Answer:
L.H.S. \( = \sin ^4 \theta + 2 \sin ^2 \theta \cos ^2 \theta + \cos ^4 \theta \)
We can rewrite this expression to match the algebraic identity \( a^2 + 2ab + b^2 = (a+b)^2 \).
Here, let \( a = \sin ^2 \theta \) and \( b = \cos ^2 \theta \).
So, L.H.S. \( = (\sin ^2 \theta)^2 + 2 (\sin ^2 \theta)(\cos ^2 \theta) + (\cos ^2 \theta)^2 \)
L.H.S. \( = (\sin ^2 \theta + \cos ^2 \theta)^2 \)
We know the fundamental Pythagorean identity \( \sin ^2 \theta + \cos ^2 \theta = 1 \). This simplifies the bracket.
So, L.H.S. \( = (1)^2 \)
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We look at the left side and see it looks like an algebra pattern: "a squared plus two ab plus b squared". If we let "a" be sine squared theta and "b" be cosine squared theta, the whole expression becomes "(sine squared theta plus cosine squared theta) squared". Since sine squared theta plus cosine squared theta is one, we end up with one squared, which is one.

🎯 Exam Tip: Keep an eye out for algebraic identities like \( (a+b)^2 \) or \( (a-b)^2 \) hidden within trigonometric expressions, as they can greatly simplify the proof.

Question 18. \( \sin ^4 A \cosec ^2 A + \cos ^4 A \sec ^2 A = 1 \)
Answer:
L.H.S. \( = \sin ^4 A \cosec ^2 A + \cos ^4 A \sec ^2 A \)
We know that \( \cosec ^2 A = \frac{1}{\sin ^2 A} \) and \( \sec ^2 A = \frac{1}{\cos ^2 A} \). These reciprocal identities help simplify.
So, L.H.S. \( = \sin ^4 A \cdot \frac{1}{\sin ^2 A} + \cos ^4 A \cdot \frac{1}{\cos ^2 A} \)
We can cancel \( \sin ^2 A \) from the first term and \( \cos ^2 A \) from the second term.
L.H.S. \( = \sin ^2 A + \cos ^2 A \)
We know the fundamental Pythagorean identity \( \sin ^2 A + \cos ^2 A = 1 \).
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: For the left side, we change cosecant squared A to one over sine squared A, and secant squared A to one over cosine squared A. This allows us to cancel out some sine squared A and cosine squared A terms, leaving sine squared A plus cosine squared A. This sum is always one, which matches the right side.

🎯 Exam Tip: When you see powers of trigonometric functions multiplied by their reciprocals, anticipate direct cancellations that simplify the expression.

Question 19. \( \sin ^2 A \cot ^2 A + \cos ^2 A \tan ^2 A = 1 \)
Answer:
L.H.S. \( = \sin ^2 A \cot ^2 A + \cos ^2 A \tan ^2 A \)
We know that \( \cot ^2 A = \frac{\cos ^2 A}{\sin ^2 A} \) and \( \tan ^2 A = \frac{\sin ^2 A}{\cos ^2 A} \). Converting to sine and cosine is a general strategy.
So, L.H.S. \( = \sin ^2 A \cdot \frac{\cos ^2 A}{\sin ^2 A} + \cos ^2 A \cdot \frac{\sin ^2 A}{\cos ^2 A} \)
In the first term, we can cancel \( \sin ^2 A \). In the second term, we can cancel \( \cos ^2 A \).
L.H.S. \( = \cos ^2 A + \sin ^2 A \)
We know the fundamental Pythagorean identity \( \cos ^2 A + \sin ^2 A = 1 \).
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We take the left side and change cotangent squared A to cosine squared A over sine squared A, and tangent squared A to sine squared A over cosine squared A. After canceling common terms in each part, we are left with cosine squared A plus sine squared A. This sum is always one, matching the right side.

🎯 Exam Tip: For expressions involving tangent and cotangent, converting them to their sine/cosine ratios is often the most direct route to simplification, especially if other terms are already in sine/cosine.

Question 20. \( \tan \theta + \cot \theta = \sec \theta \cosec \theta \)
Answer:
L.H.S. \( = \tan \theta + \cot \theta \)
We convert \( \tan \theta \) to \( \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta \) to \( \frac{\cos \theta}{\sin \theta} \). This puts both terms in a common base.
So, L.H.S. \( = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
To add these fractions, we find a common denominator, which is \( \cos \theta \sin \theta \).
L.H.S. \( = \frac{\sin ^2 \theta + \cos ^2 \theta}{\cos \theta \sin \theta} \)
We know the fundamental Pythagorean identity \( \sin ^2 \theta + \cos ^2 \theta = 1 \).
So, L.H.S. \( = \frac{1}{\cos \theta \sin \theta} \)
We can split this into two fractions: \( \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} \).
We know that \( \frac{1}{\cos \theta} = \sec \theta \) and \( \frac{1}{\sin \theta} = \cosec \theta \). These are reciprocal identities.
So, L.H.S. \( = \sec \theta \cosec \theta \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side and change tangent to sine over cosine, and cotangent to cosine over sine. When we add these fractions, we get sine squared plus cosine squared on top, which is one. The bottom is cosine times sine. This can be split into one over cosine times one over sine, which is secant times cosecant, matching the right side.

🎯 Exam Tip: When adding or subtracting trigonometric fractions, always find a common denominator. This often leads to the appearance of \( \sin^2 \theta + \cos^2 \theta \) in the numerator.

Question 21. \( (\tan A + \cot A) \sin A \cos A = 1 \)
Answer:
L.H.S. \( = (\tan A + \cot A) \sin A \cos A \)
First, we convert \( \tan A \) to \( \frac{\sin A}{\cos A} \) and \( \cot A \) to \( \frac{\cos A}{\sin A} \). This is a standard approach for mixed functions.
So, L.H.S. \( = \left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right) \sin A \cos A \)
Inside the bracket, we find a common denominator for the fractions, which is \( \cos A \sin A \).
L.H.S. \( = \left(\frac{\sin ^2 A+\cos ^2 A}{\cos A \sin A}\right) \sin A \cos A \)
We know the fundamental Pythagorean identity \( \sin ^2 A+\cos ^2 A = 1 \).
So, L.H.S. \( = \left(\frac{1}{\cos A \sin A}\right) \sin A \cos A \)
We can cancel \( \cos A \sin A \) from the numerator and denominator.
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. Inside the bracket, we change tangent A and cotangent A to sine and cosine. Adding them gives "one over (cosine A times sine A)". When we multiply this by "sine A times cosine A", all terms cancel out, leaving just one, which is the right side.

🎯 Exam Tip: Expanding expressions by first converting to sine and cosine often reveals opportunities for simplification and cancellation, leading to the desired result.

Question 22. \( \frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)}=\cot \theta \)
Answer:
L.H.S. \( = \frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \)
Rearrange the terms in the numerator to group \( (1-\sin ^2 \theta) \). This helps to use an identity.
L.H.S. \( = \frac{(1-\sin ^2 \theta)+\cos \theta}{\sin \theta(1+\cos \theta)} \)
We know the identity \( 1-\sin ^2 \theta = \cos ^2 \theta \). This is a direct rearrangement of a Pythagorean identity.
So, L.H.S. \( = \frac{\cos ^2 \theta+\cos \theta}{\sin \theta(1+\cos \theta)} \)
Factor out \( \cos \theta \) from the numerator. This prepares for cancellation.
L.H.S. \( = \frac{\cos \theta( \cos \theta+1)}{\sin \theta(1+\cos \theta)} \)
We can cancel \( (1+\cos \theta) \) (which is the same as \( (\cos \theta+1) \)) from the numerator and denominator.
L.H.S. \( = \frac{\cos \theta}{\sin \theta} \)
We know the identity \( \frac{\cos \theta}{\sin \theta} = \cot \theta \). This is the definition of cotangent.
L.H.S. \( = \cot \theta \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: For the left side, we swap "one minus sine squared theta" with "cosine squared theta" in the top part. Then, we can take out "cosine theta" as a common factor from the top. After that, "one plus cosine theta" terms cancel from top and bottom. We are left with cosine theta over sine theta, which is cotangent theta, the right side.

🎯 Exam Tip: When an expression involves \( 1-\sin^2 \theta \) or \( 1-\cos^2 \theta \), immediately substitute with \( \cos^2 \theta \) or \( \sin^2 \theta \) respectively. Also, look for common factors after applying identities.

Question 23. \( \frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta} = 2 \cosec ^2 \theta \)
Answer:
L.H.S. \( = \frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta} \)
To add these fractions, we find a common denominator, which is \( (1-\cos \theta)(1+\cos \theta) \).
L.H.S. \( = \frac{(1+\cos \theta) + (1-\cos \theta)}{(1-\cos \theta)(1+\cos \theta)} \)
In the numerator, \( \cos \theta \) and \( -\cos \theta \) cancel out, leaving \( 1+1=2 \).
L.H.S. \( = \frac{2}{(1-\cos \theta)(1+\cos \theta)} \)
In the denominator, we use the difference of squares formula: \( (a-b)(a+b) = a^2 - b^2 \).
So, \( (1-\cos \theta)(1+\cos \theta) = 1^2 - \cos ^2 \theta = 1 - \cos ^2 \theta \).
L.H.S. \( = \frac{2}{1-\cos ^2 \theta} \)
We know the identity \( 1-\cos ^2 \theta = \sin ^2 \theta \). This simplifies the denominator.
So, L.H.S. \( = \frac{2}{\sin ^2 \theta} \)
We can rewrite this as \( 2 \cdot \frac{1}{\sin ^2 \theta} \).
We know that \( \frac{1}{\sin ^2 \theta} = \cosec ^2 \theta \). Cosecant is the reciprocal of sine.
So, L.H.S. \( = 2 \cosec ^2 \theta \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We combine the two fractions on the left side. The top part becomes "one plus cosine theta plus one minus cosine theta", which simplifies to two. The bottom part becomes "one minus cosine squared theta". We then change "one minus cosine squared theta" to "sine squared theta". Finally, two divided by sine squared theta is the same as two times cosecant squared theta, which is the right side.

🎯 Exam Tip: Sums or differences of fractions often involve the difference of squares in the denominator. Always simplify the numerator and apply identities in the denominator to reach the solution.

Question 24. \( \frac{1-\tan ^2 \theta}{\cot ^2 \theta-1}=\tan ^2 \theta \)
Answer:
L.H.S. \( = \frac{1-\tan ^2 \theta}{\cot ^2 \theta-1} \)
We convert \( \cot ^2 \theta \) to \( \frac{1}{\tan ^2 \theta} \). This will express the entire fraction in terms of tangent.
So, L.H.S. \( = \frac{1-\tan ^2 \theta}{\frac{1}{\tan ^2 \theta}-1} \)
Now, we simplify the denominator by finding a common denominator for the terms inside it.
L.H.S. \( = \frac{1-\tan ^2 \theta}{\frac{1-\tan ^2 \theta}{\tan ^2 \theta}} \)
When dividing by a fraction, we multiply by its reciprocal.
L.H.S. \( = (1-\tan ^2 \theta) \cdot \frac{\tan ^2 \theta}{1-\tan ^2 \theta} \)
We can cancel \( (1-\tan ^2 \theta) \) from the numerator and denominator.
L.H.S. \( = \tan ^2 \theta \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. We change "cotangent squared theta" in the bottom part to "one over tangent squared theta". Then, we combine the terms in the bottom to make a single fraction. Finally, we flip the bottom fraction and multiply, which allows us to cancel out the "one minus tangent squared theta" terms, leaving just tangent squared theta, which is the right side.

🎯 Exam Tip: When an expression contains both a function and its reciprocal, try to convert one into the other (e.g., cotangent to tangent) to simplify the entire expression to a single function.

Question 25. \( \frac{1}{\sec \theta+\tan \theta} = \frac{1-\sin \theta}{\cos \theta} \)
Answer:
L.H.S. \( = \frac{1}{\sec \theta+\tan \theta} \)
To simplify, we multiply the numerator and denominator by the conjugate of the denominator, which is \( (\sec \theta-\tan \theta) \). This is a common technique for denominators involving sums/differences.
L.H.S. \( = \frac{1}{\sec \theta+\tan \theta} \cdot \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta} \)
L.H.S. \( = \frac{\sec \theta-\tan \theta}{\sec ^2 \theta-\tan ^2 \theta} \)
We know the Pythagorean identity \( \sec ^2 \theta-\tan ^2 \theta = 1 \). This simplifies the denominator.
So, L.H.S. \( = \frac{\sec \theta-\tan \theta}{1} \)
L.H.S. \( = \sec \theta-\tan \theta \)
Now, we convert \( \sec \theta \) to \( \frac{1}{\cos \theta} \) and \( \tan \theta \) to \( \frac{\sin \theta}{\cos \theta} \). This puts the expression in terms of sine and cosine.
So, L.H.S. \( = \frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \)
Since they have a common denominator, we can combine the terms.
L.H.S. \( = \frac{1-\sin \theta}{\cos \theta} \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. We multiply the top and bottom by "secant theta minus tangent theta". The bottom part then becomes "secant squared theta minus tangent squared theta", which is equal to one. So, the left side simplifies to "secant theta minus tangent theta". Next, we change secant to one over cosine and tangent to sine over cosine. Since they now have the same bottom, we can subtract the tops, getting "one minus sine theta over cosine theta", which is the right side.

🎯 Exam Tip: When the denominator involves a sum or difference of \( \sec \theta \) and \( \tan \theta \) or \( \cosec \theta \) and \( \cot \theta \), multiplying by the conjugate is a powerful technique because it leads to the Pythagorean identities \( \sec^2 \theta - \tan^2 \theta = 1 \) or \( \cosec^2 \theta - \cot^2 \theta = 1 \).

Question 26. \( (\cosec A – \sin A) (\sec A – \cos A) (\tan A + \cot A ) = 1 \)
Answer:
L.H.S. \( = (\cosec A – \sin A) (\sec A – \cos A) (\tan A + \cot A ) \)
We will simplify each bracket separately, converting everything to sine and cosine.
For the first bracket: \( \cosec A – \sin A = \frac{1}{\sin A} - \sin A = \frac{1-\sin ^2 A}{\sin A} = \frac{\cos ^2 A}{\sin A} \).
For the second bracket: \( \sec A – \cos A = \frac{1}{\cos A} - \cos A = \frac{1-\cos ^2 A}{\cos A} = \frac{\sin ^2 A}{\cos A} \).
For the third bracket: \( \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin ^2 A+\cos ^2 A}{\cos A \sin A} = \frac{1}{\cos A \sin A} \).
Now, we multiply these simplified expressions:
L.H.S. \( = \left(\frac{\cos ^2 A}{\sin A}\right) \cdot \left(\frac{\sin ^2 A}{\cos A}\right) \cdot \left(\frac{1}{\cos A \sin A}\right) \)
L.H.S. \( = \frac{\cos ^2 A \sin ^2 A}{\sin A \cos A \cos A \sin A} \)
L.H.S. \( = \frac{\cos ^2 A \sin ^2 A}{\cos ^2 A \sin ^2 A} \)
We can cancel \( \cos ^2 A \sin ^2 A \) from the numerator and denominator.
L.H.S. \( = 1 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We simplify each part of the left side. The first part becomes cosine squared A over sine A. The second part becomes sine squared A over cosine A. The third part becomes one over (cosine A times sine A). When we multiply these three results, all the sine and cosine terms in the top and bottom cancel out perfectly, leaving just one, which is the right side.

🎯 Exam Tip: When an identity has multiple factors, simplify each factor separately first. Then multiply them together. This reduces complexity and makes cancellations easier to spot.

Question 27. \( \frac{1+\sin \theta}{1-\sin \theta} = (\sec \theta + \tan \theta)^2 \)
Answer:
L.H.S. \( = \frac{1+\sin \theta}{1-\sin \theta} \)
To simplify the fraction, multiply the numerator and denominator by \( (1+\sin \theta) \). This makes the denominator a difference of squares.
L.H.S. \( = \frac{1+\sin \theta}{1-\sin \theta} \cdot \frac{1+\sin \theta}{1+\sin \theta} \)
L.H.S. \( = \frac{(1+\sin \theta)^2}{1-\sin ^2 \theta} \)
We know the identity \( 1-\sin ^2 \theta = \cos ^2 \theta \). This simplifies the denominator.
So, L.H.S. \( = \frac{(1+\sin \theta)^2}{\cos ^2 \theta} \)
We can write this as a single squared term: \( \left(\frac{1+\sin \theta}{\cos \theta}\right)^2 \).
Now, split the fraction inside the square: \( \left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right)^2 \).
We know that \( \frac{1}{\cos \theta} = \sec \theta \) and \( \frac{\sin \theta}{\cos \theta} = \tan \theta \). These are reciprocal and ratio identities.
So, L.H.S. \( = (\sec \theta + \tan \theta)^2 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. We multiply the top and bottom by "one plus sine theta". The top becomes "one plus sine theta squared", and the bottom becomes "one minus sine squared theta", which is cosine squared theta. We can write this as a single fraction squared. Then, we split that fraction into "one over cosine theta" plus "sine theta over cosine theta". This changes to "secant theta plus tangent theta", all squared, which is the right side.

🎯 Exam Tip: Rationalizing the denominator (multiplying by the conjugate) is a powerful technique for expressions with \( 1 \pm \sin \theta \) or \( 1 \pm \cos \theta \), as it often leads to \( \sin^2 \theta \) or \( \cos^2 \theta \).

Question 28. \( \frac{1+\cos \theta}{1-\cos \theta} = (\cosec \theta + \cot \theta)^2 \)
Answer:
L.H.S. \( = \frac{1+\cos \theta}{1-\cos \theta} \)
To simplify the fraction, multiply the numerator and denominator by \( (1+\cos \theta) \). This helps create a difference of squares in the denominator.
L.H.S. \( = \frac{1+\cos \theta}{1-\cos \theta} \cdot \frac{1+\cos \theta}{1+\cos \theta} \)
L.H.S. \( = \frac{(1+\cos \theta)^2}{1-\cos ^2 \theta} \)
We know the identity \( 1-\cos ^2 \theta = \sin ^2 \theta \). This simplifies the denominator.
So, L.H.S. \( = \frac{(1+\cos \theta)^2}{\sin ^2 \theta} \)
We can write this as a single squared term: \( \left(\frac{1+\cos \theta}{\sin \theta}\right)^2 \).
Now, split the fraction inside the square: \( \left(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right)^2 \).
We know that \( \frac{1}{\sin \theta} = \cosec \theta \) and \( \frac{\cos \theta}{\sin \theta} = \cot \theta \). These are reciprocal and ratio identities.
So, L.H.S. \( = (\cosec \theta + \cot \theta)^2 \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We start with the left side. We multiply the top and bottom by "one plus cosine theta". The top becomes "one plus cosine theta squared", and the bottom becomes "one minus cosine squared theta", which is sine squared theta. This can be written as one fraction squared. Then, we split this fraction into "one over sine theta" plus "cosine theta over sine theta". This changes to "cosecant theta plus cotangent theta", all squared, which is the right side.

🎯 Exam Tip: This problem is similar to Q27. Always recognize the pattern of multiplying by the conjugate to simplify expressions of the form \( \frac{1 \pm \cos \theta}{1 \mp \cos \theta} \) or \( \frac{1 \pm \sin \theta}{1 \mp \sin \theta} \).

Question 29. \( \frac{\cot \theta+\cosec \theta-1}{\cot \theta-\cosec \theta+1} = \frac{1+\cos \theta}{\sin \theta} \)
Answer:
L.H.S. \( = \frac{\cot \theta+\cosec \theta-1}{\cot \theta-\cosec \theta+1} \)
We know the identity \( 1 = \cosec ^2 \theta - \cot ^2 \theta \). We substitute this into the numerator.
L.H.S. \( = \frac{\cot \theta+\cosec \theta-(\cosec ^2 \theta - \cot ^2 \theta)}{\cot \theta-\cosec \theta+1} \)
Factor the term \( (\cosec ^2 \theta - \cot ^2 \theta) \) using difference of squares: \( (a-b)(a+b) \).
L.H.S. \( = \frac{(\cot \theta+\cosec \theta)-(\cosec \theta - \cot \theta)(\cosec \theta + \cot \theta)}{\cot \theta-\cosec \theta+1} \)
Now, factor out \( (\cot \theta+\cosec \theta) \) from the numerator.
L.H.S. \( = \frac{(\cot \theta+\cosec \theta)[1-(\cosec \theta - \cot \theta)]}{\cot \theta-\cosec \theta+1} \)
Distribute the negative sign inside the square bracket.
L.H.S. \( = \frac{(\cot \theta+\cosec \theta)[1-\cosec \theta + \cot \theta]}{\cot \theta-\cosec \theta+1} \)
Notice that the term in the square bracket in the numerator is the same as the denominator.
So, we can cancel \( [1-\cosec \theta + \cot \theta] \) from numerator and denominator.
L.H.S. \( = \cot \theta+\cosec \theta \)
Now, convert to sine and cosine.
L.H.S. \( = \frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta} \)
Since they have a common denominator, combine them.
L.H.S. \( = \frac{1+\cos \theta}{\sin \theta} \)
This equals the R.H.S.
Therefore, the identity is proven.
In simple words: We take the left side. In the top part, we replace the number 1 with "cosecant squared theta minus cotangent squared theta". This allows us to factor the top part, taking out "cotangent theta plus cosecant theta". We find that the remaining part of the top matches the entire bottom part, so they cancel. This leaves us with just "cotangent theta plus cosecant theta". We then change these to cosine over sine and one over sine. Adding them gives "one plus cosine theta over sine theta", which is the right side.

🎯 Exam Tip: When \( +1 \) or \( -1 \) appear in complex trigonometric fractions, consider replacing them with appropriate Pythagorean identities (like \( \sec^2 \theta - \tan^2 \theta \) or \( \cosec^2 \theta - \cot^2 \theta \)) to enable factorization.

Question 30. If \( \tan \theta + \cot \theta = 2 \), Prove that \( \tan ^2 \theta + \cot ^2 \theta = 2 \).
Answer:
Given that \( \tan \theta + \cot \theta = 2 \).
We need to prove \( \tan ^2 \theta + \cot ^2 \theta = 2 \).
Start by squaring both sides of the given equation:
\( (\tan \theta + \cot \theta)^2 = 2^2 \)
Using the algebraic identity \( (a+b)^2 = a^2+b^2+2ab \):
\( \tan ^2 \theta + \cot ^2 \theta + 2 \tan \theta \cot \theta = 4 \)
We know the identity \( \tan \theta \cot \theta = 1 \). This is because tangent and cotangent are reciprocals.
Substitute \( \tan \theta \cot \theta = 1 \) into the equation:
\( \tan ^2 \theta + \cot ^2 \theta + 2(1) = 4 \)
\( \tan ^2 \theta + \cot ^2 \theta + 2 = 4 \)
Subtract 2 from both sides:
\( \tan ^2 \theta + \cot ^2 \theta = 4 - 2 \)
\( \tan ^2 \theta + \cot ^2 \theta = 2 \)
Hence Proved.
In simple words: We start with the given information: tangent theta plus cotangent theta equals 2. We square both sides of this equation. This gives us tangent squared theta plus cotangent squared theta plus two times (tangent theta times cotangent theta) equals 4. Since tangent theta times cotangent theta is always 1, the equation becomes tangent squared theta plus cotangent squared theta plus 2 equals 4. Subtracting 2 from both sides proves that tangent squared theta plus cotangent squared theta equals 2.

🎯 Exam Tip: When given an equation and asked to prove another involving higher powers, squaring or cubing the given equation is often the correct approach. Remember key reciprocal identities like \( \tan \theta \cot \theta = 1 \).

Question. If \( \sin \theta + \cos \theta = a \), \( \sin \theta – \cos \theta = b \), Prove that \( a^2 + b^2 = 2 \)
Answer:
Given: \( \sin \theta + \cos \theta = a \)
Given: \( \sin \theta - \cos \theta = b \)
We need to prove that \( a^2 + b^2 = 2 \).
First, calculate \( a^2 \):
\( a^2 = (\sin \theta + \cos \theta)^2 \)
Using the identity \( (x+y)^2 = x^2+y^2+2xy \):
\( a^2 = \sin ^2 \theta + \cos ^2 \theta + 2 \sin \theta \cos \theta \)
Next, calculate \( b^2 \):
\( b^2 = (\sin \theta - \cos \theta)^2 \)
Using the identity \( (x-y)^2 = x^2+y^2-2xy \):
\( b^2 = \sin ^2 \theta + \cos ^2 \theta - 2 \sin \theta \cos \theta \)
Now, add \( a^2 \) and \( b^2 \):
\( a^2 + b^2 = (\sin ^2 \theta + \cos ^2 \theta + 2 \sin \theta \cos \theta) + (\sin ^2 \theta + \cos ^2 \theta - 2 \sin \theta \cos \theta) \)
Combine like terms. The \( +2 \sin \theta \cos \theta \) and \( -2 \sin \theta \cos \theta \) terms cancel each other out.
\( a^2 + b^2 = \sin ^2 \theta + \cos ^2 \theta + \sin ^2 \theta + \cos ^2 \theta \)
Group the terms:
\( a^2 + b^2 = (\sin ^2 \theta + \cos ^2 \theta) + (\sin ^2 \theta + \cos ^2 \theta) \)
We know the fundamental Pythagorean identity \( \sin ^2 \theta + \cos ^2 \theta = 1 \).
So, \( a^2 + b^2 = 1 + 1 \)
\( a^2 + b^2 = 2 \)
Hence Proved.
In simple words: We are given what 'a' and 'b' represent. We find 'a squared' by squaring the expression for 'a', which gives us sine squared theta plus cosine squared theta plus two sine theta cosine theta. Similarly, we find 'b squared' by squaring the expression for 'b', which gives us sine squared theta plus cosine squared theta minus two sine theta cosine theta. When we add 'a squared' and 'b squared', the "two sine theta cosine theta" terms cancel out. We are left with two sets of "sine squared theta plus cosine squared theta". Since each set equals one, the total sum is 1 plus 1, which is 2.

🎯 Exam Tip: Remember the basic algebraic identities for squares: \( (x+y)^2 \) and \( (x-y)^2 \). When adding such squared terms, the \( 2xy \) terms often cancel, leading to a simplification involving \( x^2+y^2 \).