Samacheer Kalvi Class 10 Maths Solutions Chapter 6 Trigonometry Exercise 6.1

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

 

Question 1. Prove the following identities.
(i) cot \( \theta \) + tan \( \theta \) = sec \( \theta \) cosec \( \theta \)
(ii) \( \tan^4 \theta \) + \( \tan^2 \theta \) = \( \sec^4 \theta - \sec^2 \theta \)
Answer:
(i) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \cot \theta + \tan \theta \)
Now, we convert cotangent and tangent into sine and cosine:
\( = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \)
We find a common denominator for these fractions, which is \( \sin \theta \cos \theta \):
\( = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \)
Using the fundamental trigonometric identity, \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( = \frac{1}{\sin \theta \cos \theta} \)
This expression can be separated into two parts:
\( = \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} \)
We know that \( \frac{1}{\cos \theta} = \sec \theta \) and \( \frac{1}{\sin \theta} = \operatorname{cosec} \theta \):
\( = \sec \theta \cdot \operatorname{cosec} \theta \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven.
(ii) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \tan^4 \theta + \tan^2 \theta \)
We can factor out \( \tan^2 \theta \) from both terms:
\( = \tan^2 \theta (\tan^2 \theta + 1) \)
Using the identity \( \tan^2 \theta + 1 = \sec^2 \theta \):
\( = \tan^2 \theta \sec^2 \theta \)
Now, we look at the Right Hand Side (R.H.S.):
R.H.S = \( \sec^4 \theta - \sec^2 \theta \)
We can factor out \( \sec^2 \theta \) from both terms:
\( = \sec^2 \theta (\sec^2 \theta - 1) \)
Using the identity \( \sec^2 \theta - 1 = \tan^2 \theta \):
\( = \sec^2 \theta \tan^2 \theta \)
Since L.H.S = R.H.S, the identity is proven. Factoring common terms often helps simplify expressions, making them easier to prove.
In simple words: For part (i), we changed everything to sine and cosine, combined them, and used a basic identity to get secant and cosecant. For part (ii), we factored common terms and used a key identity for tangent and secant to make both sides match.

🎯 Exam Tip: When proving identities, if both sides look complex, try simplifying one side (usually L.H.S) until it matches the other, or simplify both sides independently to a common expression.

 

Question 2. Prove the following identities.
(i) \( \frac{1-\tan^2 \theta}{\cot^2 \theta-1} = \tan^2 \theta \)
(ii) \( \frac{\cos \theta}{1+\sin \theta} = \sec \theta - \tan \theta \)
Answer:
(i) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \frac{1-\tan^2 \theta}{\cot^2 \theta-1} \)
First, we change \( \cot^2 \theta \) to \( \frac{1}{\tan^2 \theta} \) in the denominator:
\( = \frac{1-\tan^2 \theta}{\frac{1}{\tan^2 \theta}-1} \)
Next, we simplify the denominator by finding a common denominator:
\( = \frac{1-\tan^2 \theta}{\frac{1-\tan^2 \theta}{\tan^2 \theta}} \)
Now, we multiply by the reciprocal of the denominator:
\( = (1-\tan^2 \theta) \times \frac{\tan^2 \theta}{1-\tan^2 \theta} \)
We can cancel out the \( (1-\tan^2 \theta) \) terms:
\( = \tan^2 \theta \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. Converting all terms to a common base like tangent or sine/cosine helps simplify complex fractions.
(ii) We start with the Right Hand Side (R.H.S.):
R.H.S = \( \sec \theta - \tan \theta \)
We convert secant and tangent into sine and cosine:
\( = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} \)
Since they have the same denominator, we can combine them:
\( = \frac{1-\sin \theta}{\cos \theta} \)
Now, we multiply the numerator and denominator by the conjugate of \( (1-\sin \theta) \), which is \( (1+\sin \theta) \):
\( = \frac{1-\sin \theta}{\cos \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \)
This simplifies the numerator using the difference of squares formula \( (a-b)(a+b)=a^2-b^2 \):
\( = \frac{1^2-\sin^2 \theta}{\cos \theta (1+\sin \theta)} \)
Using the identity \( 1-\sin^2 \theta = \cos^2 \theta \):
\( = \frac{\cos^2 \theta}{\cos \theta (1+\sin \theta)} \)
We can cancel out one \( \cos \theta \) term from the numerator and denominator:
\( = \frac{\cos \theta}{1+\sin \theta} \)
This is the Left Hand Side (L.H.S). Thus, the identity is proven. Multiplying by the conjugate of the denominator is a common technique to simplify trigonometric fractions.
In simple words: For part (i), we changed cotangent to tangent, simplified the fraction, and cancelled terms. For part (ii), we converted secant and tangent to sine and cosine, combined them, and then multiplied by the conjugate to reach the other side.

🎯 Exam Tip: When dealing with fractions involving \( 1 \pm \sin \theta \) or \( 1 \pm \cos \theta \), multiplying by the conjugate is often the quickest way to use the \( \sin^2 \theta + \cos^2 \theta = 1 \) identity.

 

Question 3. Prove the following identities.
(i) \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} = \sec \theta + \tan \theta \)
(ii) \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} + \sqrt{\frac{1-\sin \theta}{1+\sin \theta}} = 2 \sec \theta \)
Answer:
(i) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \)
To remove the square root and simplify the denominator, we multiply the numerator and denominator inside the square root by the conjugate of \( (1-\sin \theta) \), which is \( (1+\sin \theta) \):
\( = \sqrt{\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}} \)
This simplifies to:
\( = \sqrt{\frac{(1+\sin \theta)^2}{1^2-\sin^2 \theta}} \)
Using the identity \( 1-\sin^2 \theta = \cos^2 \theta \):
\( = \sqrt{\frac{(1+\sin \theta)^2}{\cos^2 \theta}} \)
Now, we take the square root of both the numerator and the denominator:
\( = \frac{1+\sin \theta}{\cos \theta} \)
We can split this into two fractions:
\( = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} \)
Using the definitions \( \frac{1}{\cos \theta} = \sec \theta \) and \( \frac{\sin \theta}{\cos \theta} = \tan \theta \):
\( = \sec \theta + \tan \theta \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. Rationalizing the denominator inside a square root is a useful step for simplifying expressions involving square roots of fractions.
(ii) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} + \sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \)
From part (i), we know that \( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} = \sec \theta + \tan \theta \). We will use this for the first term.
For the second term, we follow similar steps as in part (i) by multiplying by its conjugate:
\( \sqrt{\frac{1-\sin \theta}{1+\sin \theta}} = \sqrt{\frac{(1-\sin \theta)(1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}} \)
\( = \sqrt{\frac{(1-\sin \theta)^2}{1-\sin^2 \theta}} \)
\( = \sqrt{\frac{(1-\sin \theta)^2}{\cos^2 \theta}} \)
\( = \frac{1-\sin \theta}{\cos \theta} \)
\( = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \sec \theta - \tan \theta \)
Now, we substitute these simplified expressions back into the L.H.S:
L.H.S \( = (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) \)
We can see that \( \tan \theta \) and \( -\tan \theta \) cancel each other out:
\( = \sec \theta + \sec \theta \)
\( = 2 \sec \theta \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. Combining results from previous parts can often simplify proving more complex identities.
In simple words: For part (i), we multiplied the top and bottom inside the square root by \( (1+\sin \theta) \) to simplify it into secant and tangent. For part (ii), we did the same for both terms and found that when added, the tangent parts cancelled out, leaving only 2 secant.

🎯 Exam Tip: Notice how proving part (i) directly helped solve part (ii). Always look for connections between sub-parts of a question to save time and effort.

 

Question 4. Prove the following identities.
(i) \( \sec^6 \theta = \tan^6 \theta + 3 \tan^2 \theta \sec^2 \theta + 1 \)
(ii) \( (\sin \theta + \sec \theta)^2 + (\cos \theta + \operatorname{cosec} \theta)^2 = 1 + (\sec \theta + \operatorname{cosec} \theta)^2 \)
Answer:
(i) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \sec^6 \theta \)
We can rewrite \( \sec^6 \theta \) as \( (\sec^2 \theta)^3 \):
\( = (\sec^2 \theta)^3 \)
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we substitute it into the expression:
\( = (1 + \tan^2 \theta)^3 \)
Now, we expand this using the algebraic formula \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \), where \( a=1 \) and \( b=\tan^2 \theta \):
\( = 1^3 + (\tan^2 \theta)^3 + 3(1)(\tan^2 \theta)(1 + \tan^2 \theta) \)
Simplifying the terms:
\( = 1 + \tan^6 \theta + 3 \tan^2 \theta (1 + \tan^2 \theta) \)
We again use the identity \( 1 + \tan^2 \theta = \sec^2 \theta \) in the last term:
\( = 1 + \tan^6 \theta + 3 \tan^2 \theta \sec^2 \theta \)
Rearranging the terms to match the R.H.S:
\( = \tan^6 \theta + 3 \tan^2 \theta \sec^2 \theta + 1 \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. The binomial expansion formula for cubes is very useful when dealing with higher powers of trigonometric functions.
(ii) We start with the Left Hand Side (L.H.S.):
L.H.S = \( (\sin \theta + \sec \theta)^2 + (\cos \theta + \operatorname{cosec} \theta)^2 \)
First, we expand both squared terms using \( (a+b)^2 = a^2+b^2+2ab \):
\( = (\sin^2 \theta + \sec^2 \theta + 2 \sin \theta \sec \theta) + (\cos^2 \theta + \operatorname{cosec}^2 \theta + 2 \cos \theta \operatorname{cosec} \theta) \)
Now, we group the \( \sin^2 \theta \) and \( \cos^2 \theta \) terms, and rewrite \( \sec \theta \) and \( \operatorname{cosec} \theta \) in terms of sine and cosine:
\( = (\sin^2 \theta + \cos^2 \theta) + \sec^2 \theta + \operatorname{cosec}^2 \theta + 2 \sin \theta \left(\frac{1}{\cos \theta}\right) + 2 \cos \theta \left(\frac{1}{\sin \theta}\right) \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = 1 + \sec^2 \theta + \operatorname{cosec}^2 \theta + 2 \frac{\sin \theta}{\cos \theta} + 2 \frac{\cos \theta}{\sin \theta} \)
We find a common denominator for the last two terms:
\( = 1 + \sec^2 \theta + \operatorname{cosec}^2 \theta + 2 \left(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\right) \)
Again, using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = 1 + \sec^2 \theta + \operatorname{cosec}^2 \theta + 2 \left(\frac{1}{\sin \theta \cos \theta}\right) \)
We know that \( \frac{1}{\sin \theta \cos \theta} = \frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} = \operatorname{cosec} \theta \sec \theta \):
\( = 1 + \sec^2 \theta + \operatorname{cosec}^2 \theta + 2 \sec \theta \operatorname{cosec} \theta \)
The terms \( \sec^2 \theta + \operatorname{cosec}^2 \theta + 2 \sec \theta \operatorname{cosec} \theta \) form a perfect square, \( (\sec \theta + \operatorname{cosec} \theta)^2 \):
\( = 1 + (\sec \theta + \operatorname{cosec} \theta)^2 \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. Always expand squared terms first and then use fundamental identities to simplify the expression.
In simple words: For part (i), we wrote \( \sec^6 \theta \) as \( (\sec^2 \theta)^3 \), changed \( \sec^2 \theta \) to \( (1+\tan^2 \theta) \), and then expanded it using an algebraic cube formula to match the other side. For part (ii), we expanded both squared terms, grouped the sine and cosine squared parts to make 1, and then simplified the remaining terms to match the required square expression.

🎯 Exam Tip: For higher powers, consider using algebraic expansion formulas like \( (a+b)^3 \) or \( (a+b)^2 \) after expressing trigonometric terms using fundamental identities.

 

Question 5. Prove the following identities.
(i) \( \sec^4 \theta (1 - \sin^4 \theta) - 2 \tan^2 \theta = 1 \)
(ii) \( \frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1} \)
Answer:
(i) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \sec^4 \theta (1 - \sin^4 \theta) - 2 \tan^2 \theta \)
First, we rewrite \( \sec^4 \theta \) as \( \frac{1}{\cos^4 \theta} \) and \( \tan^2 \theta \) as \( \frac{\sin^2 \theta}{\cos^2 \theta} \). Also, factor \( (1 - \sin^4 \theta) \) using the difference of squares formula, \( (a^2-b^2)=(a-b)(a+b) \), where \( a=1 \) and \( b=\sin^2 \theta \):
\( = \frac{1}{\cos^4 \theta} (1 - \sin^2 \theta)(1 + \sin^2 \theta) - 2 \frac{\sin^2 \theta}{\cos^2 \theta} \)
Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \):
\( = \frac{1}{\cos^4 \theta} (\cos^2 \theta)(1 + \sin^2 \theta) - 2 \frac{\sin^2 \theta}{\cos^2 \theta} \)
Cancel \( \cos^2 \theta \) from the first term:
\( = \frac{1}{\cos^2 \theta} (1 + \sin^2 \theta) - 2 \frac{\sin^2 \theta}{\cos^2 \theta} \)
Now, we combine the fractions since they have a common denominator \( \cos^2 \theta \):
\( = \frac{1(1 + \sin^2 \theta) - 2 \sin^2 \theta}{\cos^2 \theta} \)
Simplify the numerator:
\( = \frac{1 + \sin^2 \theta - 2 \sin^2 \theta}{\cos^2 \theta} \)
\( = \frac{1 - \sin^2 \theta}{\cos^2 \theta} \)
Again, using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \):
\( = \frac{\cos^2 \theta}{\cos^2 \theta} \)
\( = 1 \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. Remember that \( (a^2-b^2) = (a-b)(a+b) \) is a very powerful factorization tool in trigonometry.
(ii) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta} \)
We convert \( \cot \theta \) into \( \frac{\cos \theta}{\sin \theta} \):
\( = \frac{\frac{\cos \theta}{\sin \theta}-\cos \theta}{\frac{\cos \theta}{\sin \theta}+\cos \theta} \)
Now, we can factor out \( \cos \theta \) from both the numerator and the denominator:
\( = \frac{\cos \theta (\frac{1}{\sin \theta}-1)}{\cos \theta (\frac{1}{\sin \theta}+1)} \)
We cancel out the common factor \( \cos \theta \):
\( = \frac{\frac{1}{\sin \theta}-1}{\frac{1}{\sin \theta}+1} \)
Using the definition \( \frac{1}{\sin \theta} = \operatorname{cosec} \theta \):
\( = \frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1} \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. When proving identities, look for common factors that can be cancelled to simplify the expression.
In simple words: For part (i), we expanded \( (1-\sin^4 \theta) \) and changed secant to cosine to simplify the expression to 1. For part (ii), we changed cotangent to cosine/sine, factored out cosine, and then replaced 1/sine with cosecant to get the right side.

🎯 Exam Tip: When fractions have common terms in the numerator and denominator (like \( \cos \theta \) in part ii), always factor them out first. This simplification often reveals the next steps clearly.

 

Question 6. Prove the following identities.
(i) \( \frac{\sin A-\sin B}{\cos A+\cos B} + \frac{\cos A-\cos B}{\sin A+\sin B} = 0 \)
(ii) \( \frac{\sin^3 A+\cos^3 A}{\sin A+\cos A} + \frac{\sin^3 A-\cos^3 A}{\sin A-\cos A} = 2 \)
Answer:
(i) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \frac{\sin A-\sin B}{\cos A+\cos B} + \frac{\cos A-\cos B}{\sin A+\sin B} \)
To add these two fractions, we find a common denominator, which is \( (\cos A+\cos B)(\sin A+\sin B) \):
\( = \frac{(\sin A-\sin B)(\sin A+\sin B) + (\cos A-\cos B)(\cos A+\cos B)}{(\cos A+\cos B)(\sin A+\sin B)} \)
Now, we expand the numerators using the difference of squares formula, \( (x-y)(x+y) = x^2-y^2 \):
\( = \frac{(\sin^2 A-\sin^2 B) + (\cos^2 A-\cos^2 B)}{(\cos A+\cos B)(\sin A+\sin B)} \)
We rearrange the terms in the numerator to group \( \sin^2 \) and \( \cos^2 \) terms:
\( = \frac{\sin^2 A+\cos^2 A - (\sin^2 B+\cos^2 B)}{(\cos A+\cos B)(\sin A+\sin B)} \)
Using the fundamental identity \( \sin^2 x + \cos^2 x = 1 \):
\( = \frac{1 - 1}{(\cos A+\cos B)(\sin A+\sin B)} \)
The numerator becomes 0:
\( = \frac{0}{(\cos A+\cos B)(\sin A+\sin B)} \)
\( = 0 \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. The difference of squares formula, \( (a-b)(a+b) = a^2-b^2 \), is extremely useful for simplifying sums and differences of trigonometric terms.
(ii) We start with the Left Hand Side (L.H.S.):
L.H.S = \( \frac{\sin^3 A+\cos^3 A}{\sin A+\cos A} + \frac{\sin^3 A-\cos^3 A}{\sin A-\cos A} \)
We use the algebraic sum and difference of cubes formulas:
\( a^3+b^3 = (a+b)(a^2-ab+b^2) \)
\( a^3-b^3 = (a-b)(a^2+ab+b^2) \)
Applying the first formula to the first term (with \( a=\sin A, b=\cos A \)):
\( \frac{(\sin A+\cos A)(\sin^2 A-\sin A \cos A+\cos^2 A)}{\sin A+\cos A} \)
We cancel out \( (\sin A+\cos A) \) and use \( \sin^2 A+\cos^2 A=1 \):
\( = 1-\sin A \cos A \)
Applying the second formula to the second term (with \( a=\sin A, b=\cos A \)):
\( \frac{(\sin A-\cos A)(\sin^2 A+\sin A \cos A+\cos^2 A)}{\sin A-\cos A} \)
We cancel out \( (\sin A-\cos A) \) and use \( \sin^2 A+\cos^2 A=1 \):
\( = 1+\sin A \cos A \)
Now, we add the simplified terms:
L.H.S \( = (1-\sin A \cos A) + (1+\sin A \cos A) \)
The \( -\sin A \cos A \) and \( +\sin A \cos A \) terms cancel each other:
\( = 1+1 \)
\( = 2 \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. Knowing algebraic factorization formulas like sum and difference of cubes can greatly simplify complex trigonometric expressions.
In simple words: For part (i), we found a common denominator and used the difference of squares rule for the numerators. The terms cancelled out to give 0. For part (ii), we used the sum and difference of cubes formulas to simplify each fraction. After cancelling terms, we added the two simplified expressions to get 2.

🎯 Exam Tip: Memorizing common algebraic factorization formulas like \( a^2-b^2 \), \( a^3+b^3 \), and \( a^3-b^3 \) is essential for efficiently proving trigonometric identities.

 

Question 7.
(i) If sin \( \theta \) + cos \( \theta \) = \( \sqrt{3} \), then prove that tan \( \theta \) + cot \( \theta \) = 1.
(ii) If \( \sqrt{3} \) sin \( \theta \) - cos \( \theta \) = 0, then show that tan 3\( \theta \) = \( \frac{3 \tan \theta-\tan^{3} \theta}{1-3 \tan^{2} \theta} \)
Answer:
(i) We are given: \( \sin \theta + \cos \theta = \sqrt{3} \)
To work with this, we square both sides of the equation:
\( (\sin \theta + \cos \theta)^2 = (\sqrt{3})^2 \)
Expand the left side using \( (a+b)^2 = a^2+b^2+2ab \):
\( \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3 \)
Using the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( 1 + 2 \sin \theta \cos \theta = 3 \)
Subtract 1 from both sides:
\( 2 \sin \theta \cos \theta = 3 - 1 \)
\( 2 \sin \theta \cos \theta = 2 \)
Divide by 2:
\( \sin \theta \cos \theta = 1 \)
Now, we need to prove that \( \tan \theta + \cot \theta = 1 \). We start with the L.H.S.:
L.H.S = \( \tan \theta + \cot \theta \)
Convert tangent and cotangent to sine and cosine:
\( = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
Find a common denominator, \( \sin \theta \cos \theta \):
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \frac{1}{\sin \theta \cos \theta} \)
Substitute the value \( \sin \theta \cos \theta = 1 \) that we found earlier:
\( = \frac{1}{1} \)
\( = 1 \)
This is the R.H.S. Thus, the identity is proven. Squaring both sides of an equation is a common strategy to introduce terms like \( \sin^2 \theta + \cos^2 \theta \) which simplify to 1.
(ii) We are given: \( \sqrt{3} \sin \theta - \cos \theta = 0 \)
Rearrange the equation to find a value for \( \tan \theta \):
\( \sqrt{3} \sin \theta = \cos \theta \)
Divide both sides by \( \cos \theta \):
\( \sqrt{3} \frac{\sin \theta}{\cos \theta} = 1 \)
\( \sqrt{3} \tan \theta = 1 \)
\( \tan \theta = \frac{1}{\sqrt{3}} \)
This means \( \theta = 30^\circ \).
Now, we evaluate the Left Hand Side (L.H.S) of the identity to show: \( \tan 3\theta \)
Substitute \( \theta = 30^\circ \):
\( = \tan (3 \times 30^\circ) = \tan 90^\circ \)
We know that \( \tan 90^\circ \) is undefined.
Next, we evaluate the Right Hand Side (R.H.S) of the identity: \( \frac{3 \tan \theta-\tan^{3} \theta}{1-3 \tan^{2} \theta} \)
Substitute \( \tan \theta = \frac{1}{\sqrt{3}} \):
\[ = \frac{3 \left(\frac{1}{\sqrt{3}}\right) - \left(\frac{1}{\sqrt{3}}\right)^3}{1 - 3 \left(\frac{1}{\sqrt{3}}\right)^2} \]
Calculate the powers:
\( \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \)
\( \left(\frac{1}{\sqrt{3}}\right)^3 = \frac{1}{3\sqrt{3}} \)
Substitute these back into the R.H.S expression:
\[ = \frac{\frac{3}{\sqrt{3}} - \frac{1}{3\sqrt{3}}}{1 - 3 \left(\frac{1}{3}\right)} \]
Simplify the numerator and denominator:
\[ = \frac{\frac{9}{3\sqrt{3}} - \frac{1}{3\sqrt{3}}}{1 - 1} \]
\[ = \frac{\frac{8}{3\sqrt{3}}}{0} \]
This expression is also undefined.
Since both L.H.S and R.H.S are undefined for \( \theta = 30^\circ \), the identity holds true for this value. This problem demonstrates a special case where both sides of an identity are undefined, yet the equality still holds.
In simple words: For part (i), we squared the given equation to find that sine multiplied by cosine is 1. Then we showed that tangent plus cotangent is also 1 using this fact. For part (ii), we found that \( \tan \theta = 1/\sqrt{3} \), which means \( \theta = 30^\circ \). When we put \( 30^\circ \) into both sides of the identity, both sides became undefined, showing they are equal.

🎯 Exam Tip: When a question asks to "show that" an identity is true, finding a specific value for the angle (if possible) and checking both sides can be a useful verification method, especially if the angle is a standard one.

 

Question 8.
(i) If \( \frac{\cos \alpha}{\cos \beta}=m \) and \( \frac{\cos \alpha}{\sin \beta}=n \) then prove that \( (m^2 + n^2) \cos^2 \beta = n^2 \)
(ii) If cot \( \theta \) + tan \( \theta \) = x and sec \( \theta \) - cos \( \theta \) = y, then prove that \( (x^2y)^{2/3} - (xy^2)^{2/3} = 1 \)
Answer:
(i) We are given:
\( m = \frac{\cos \alpha}{\cos \beta} \)
\( n = \frac{\cos \alpha}{\sin \beta} \)
We need to prove \( (m^2 + n^2) \cos^2 \beta = n^2 \). We start with the Left Hand Side (L.H.S.):
L.H.S = \( (m^2 + n^2) \cos^2 \beta \)
Substitute the expressions for \( m \) and \( n \):
\[ = \left( \left(\frac{\cos \alpha}{\cos \beta}\right)^2 + \left(\frac{\cos \alpha}{\sin \beta}\right)^2 \right) \cos^2 \beta \]
Square the terms inside the parenthesis:
\[ = \left( \frac{\cos^2 \alpha}{\cos^2 \beta} + \frac{\cos^2 \alpha}{\sin^2 \beta} \right) \cos^2 \beta \]
Factor out \( \cos^2 \alpha \) from the terms inside the parenthesis:
\[ = \cos^2 \alpha \left( \frac{1}{\cos^2 \beta} + \frac{1}{\sin^2 \beta} \right) \cos^2 \beta \]
Find a common denominator for the terms inside the parenthesis, which is \( \cos^2 \beta \sin^2 \beta \):
\[ = \cos^2 \alpha \left( \frac{\sin^2 \beta + \cos^2 \beta}{\cos^2 \beta \sin^2 \beta} \right) \cos^2 \beta \]
Using the identity \( \sin^2 \beta + \cos^2 \beta = 1 \):
\[ = \cos^2 \alpha \left( \frac{1}{\cos^2 \beta \sin^2 \beta} \right) \cos^2 \beta \]
Cancel out \( \cos^2 \beta \):
\( = \frac{\cos^2 \alpha}{\sin^2 \beta} \)
We know that \( n = \frac{\cos \alpha}{\sin \beta} \), so \( n^2 = \left(\frac{\cos \alpha}{\sin \beta}\right)^2 = \frac{\cos^2 \alpha}{\sin^2 \beta} \).
Therefore, L.H.S \( = n^2 \). This is the Right Hand Side (R.H.S). Thus, the identity is proven. Substituting given expressions into the L.H.S and simplifying is a standard approach to proving identities.
(ii) We are given:
\( x = \cot \theta + \tan \theta \)
\( y = \sec \theta - \cos \theta \)
We will first simplify \( x \) and \( y \):
For \( x \):
\( x = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \)
\( = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \)
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( x = \frac{1}{\sin \theta \cos \theta} \)
For \( y \):
\( y = \frac{1}{\cos \theta} - \cos \theta \)
\( = \frac{1-\cos^2 \theta}{\cos \theta} \)
Using \( 1-\cos^2 \theta = \sin^2 \theta \):
\( y = \frac{\sin^2 \theta}{\cos \theta} \)
Now, we need to prove \( (x^2y)^{2/3} - (xy^2)^{2/3} = 1 \). We calculate \( x^2y \) and \( xy^2 \):
For \( x^2y \):
\( x^2y = \left(\frac{1}{\sin \theta \cos \theta}\right)^2 \left(\frac{\sin^2 \theta}{\cos \theta}\right) \)
\( = \frac{1}{\sin^2 \theta \cos^2 \theta} \cdot \frac{\sin^2 \theta}{\cos \theta} \)
Cancel \( \sin^2 \theta \):
\( = \frac{1}{\cos^3 \theta} \)
For \( xy^2 \):
\( xy^2 = \left(\frac{1}{\sin \theta \cos \theta}\right) \left(\frac{\sin^2 \theta}{\cos \theta}\right)^2 \)
\( = \frac{1}{\sin \theta \cos \theta} \cdot \frac{\sin^4 \theta}{\cos^2 \theta} \)
Cancel \( \sin \theta \):
\( = \frac{\sin^3 \theta}{\cos^3 \theta} \)
\( = \tan^3 \theta \)
Now, substitute these into the L.H.S of the identity we want to prove:
L.H.S \( = \left(\frac{1}{\cos^3 \theta}\right)^{2/3} - (\tan^3 \theta)^{2/3} \)
Apply the power of \( 2/3 \):
\( = \frac{1}{(\cos^3 \theta)^{2/3}} - (\tan^3 \theta)^{2/3} \)
\( = \frac{1}{\cos^2 \theta} - \tan^2 \theta \)
We know that \( \frac{1}{\cos^2 \theta} = \sec^2 \theta \):
\( = \sec^2 \theta - \tan^2 \theta \)
Using the identity \( \sec^2 \theta - \tan^2 \theta = 1 \):
\( = 1 \)
This is the R.H.S. Thus, the identity is proven. Always simplify the initial given expressions (x and y in this case) before substituting them into the main identity to be proved.
In simple words: For part (i), we put the given m and n into the left side of the equation. After simplifying by combining fractions and using \( \sin^2 \beta + \cos^2 \beta = 1 \), we found it equal to \( n^2 \). For part (ii), we first simplified x and y separately using sine and cosine. Then, we calculated \( x^2y \) and \( xy^2 \), put them into the main equation, and simplified further using basic trigonometric identities to get 1.

🎯 Exam Tip: When given expressions for variables (like m, n, x, y), simplify these expressions first as much as possible before substituting them into the more complex identity to be proved. This makes the algebra more manageable.

 

Question 9.
(i) If sin \( \theta \) + cos \( \theta \) = p and sec \( \theta \) + cosec \( \theta \) = q, then prove that q \( (p^2 - 1) = 2p \)
(ii) If sin \( \theta \) \( (1 + \sin^2 \theta) = \cos^2 \theta \), then prove that \( \cos^6 \theta - 4 \cos^4 \theta + 8 \cos^2 \theta = 4 \)
Answer:
(i) We are given:
\( p = \sin \theta + \cos \theta \)
\( q = \sec \theta + \operatorname{cosec} \theta \)
First, we calculate \( p^2 \):
\( p^2 = (\sin \theta + \cos \theta)^2 \)
Expand this:
\( = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = 1 + 2 \sin \theta \cos \theta \)
Now, we find \( p^2 - 1 \):
\( p^2 - 1 = (1 + 2 \sin \theta \cos \theta) - 1 \)
\( p^2 - 1 = 2 \sin \theta \cos \theta \)
Next, we simplify \( q \) by converting secant and cosecant to sine and cosine:
\( q = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \)
Find a common denominator:
\( q = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \)
We need to prove \( q(p^2 - 1) = 2p \). Let's start with the L.H.S.:
L.H.S = \( q(p^2 - 1) \)
Substitute the expressions we found for \( q \) and \( (p^2 - 1) \):
\[ = \left(\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}\right) (2 \sin \theta \cos \theta) \]
The \( \sin \theta \cos \theta \) terms cancel out:
\( = (\sin \theta + \cos \theta) (2) \)
\( = 2 (\sin \theta + \cos \theta) \)
Since \( p = \sin \theta + \cos \theta \), we can substitute \( p \) back:
\( = 2p \)
This is the R.H.S. Thus, the identity is proven. Expressing all terms in terms of sine and cosine before substitution often makes algebraic simplification easier.
(ii) We are given: \( \sin \theta (1 + \sin^2 \theta) = \cos^2 \theta \)
We need to prove that \( \cos^6 \theta - 4 \cos^4 \theta + 8 \cos^2 \theta = 4 \).
First, rewrite \( \sin^2 \theta \) using \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( \sin \theta (1 + (1 - \cos^2 \theta)) = \cos^2 \theta \)
Simplify the term in the parenthesis:
\( \sin \theta (2 - \cos^2 \theta) = \cos^2 \theta \)
To eliminate \( \sin \theta \), we square both sides of the equation:
\( (\sin \theta (2 - \cos^2 \theta))^2 = (\cos^2 \theta)^2 \)
\( \sin^2 \theta (2 - \cos^2 \theta)^2 = \cos^4 \theta \)
Again, replace \( \sin^2 \theta \) with \( (1 - \cos^2 \theta) \):
\( (1 - \cos^2 \theta) (2 - \cos^2 \theta)^2 = \cos^4 \theta \)
Expand \( (2 - \cos^2 \theta)^2 \) using \( (a-b)^2 = a^2-2ab+b^2 \):
\( (1 - \cos^2 \theta) (4 - 4 \cos^2 \theta + \cos^4 \theta) = \cos^4 \theta \)
Now, multiply the two terms on the left side:
\( 1(4 - 4 \cos^2 \theta + \cos^4 \theta) - \cos^2 \theta (4 - 4 \cos^2 \theta + \cos^4 \theta) = \cos^4 \theta \)
\( 4 - 4 \cos^2 \theta + \cos^4 \theta - 4 \cos^2 \theta + 4 \cos^4 \theta - \cos^6 \theta = \cos^4 \theta \)
Combine like terms on the L.H.S.:
\( -\cos^6 \theta + (\cos^4 \theta + 4 \cos^4 \theta) + (-4 \cos^2 \theta - 4 \cos^2 \theta) + 4 = \cos^4 \theta \)
\( -\cos^6 \theta + 5 \cos^4 \theta - 8 \cos^2 \theta + 4 = \cos^4 \theta \)
Move \( \cos^4 \theta \) from the R.H.S to the L.H.S:
\( -\cos^6 \theta + 5 \cos^4 \theta - \cos^4 \theta - 8 \cos^2 \theta + 4 = 0 \)
\( -\cos^6 \theta + 4 \cos^4 \theta - 8 \cos^2 \theta + 4 = 0 \)
Multiply the entire equation by -1 to match the required form:
\( \cos^6 \theta - 4 \cos^4 \theta + 8 \cos^2 \theta - 4 = 0 \)
Finally, move the constant term to the R.H.S:
\( \cos^6 \theta - 4 \cos^4 \theta + 8 \cos^2 \theta = 4 \)
This matches the required identity. Hence it is proven. When dealing with powers, converting all trigonometric functions to a single base (like cosine in this case) often simplifies the algebra significantly.
In simple words: For part (i), we calculated \( p^2 \) and \( p^2-1 \), and simplified q into sine and cosine. Then, we substituted these into \( q(p^2-1) \) and found it equal to \( 2p \). For part (ii), we started by changing sine terms to cosine terms in the given equation. We then squared both sides and expanded carefully. After grouping similar terms, we rearranged them to get the final required identity.

🎯 Exam Tip: In complex proofs, always simplify each part of the given information (like p and q here) separately before combining them. This modular approach helps in managing complexity and reducing errors.

 

Question 10. If \( \frac{\cos \theta}{1+\sin \theta} = \frac{1}{a} \), then prove that \( \frac{a^2-1}{a^2+1} = \sin \theta \)
Answer:
We are given: \( \frac{\cos \theta}{1+\sin \theta} = \frac{1}{a} \)
From this, we can write \( a \) as:
\( a = \frac{1+\sin \theta}{\cos \theta} \)
Next, we need to find \( a^2 \). We square both sides of the expression for \( a \):
\( a^2 = \left(\frac{1+\sin \theta}{\cos \theta}\right)^2 \)
\( a^2 = \frac{(1+\sin \theta)^2}{\cos^2 \theta} \)
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we substitute this into the denominator:
\( a^2 = \frac{(1+\sin \theta)^2}{1 - \sin^2 \theta} \)
Factor the denominator using the difference of squares formula, \( (x^2-y^2)=(x-y)(x+y) \):
\( a^2 = \frac{(1+\sin \theta)^2}{(1-\sin \theta)(1+\sin \theta)} \)
Cancel one \( (1+\sin \theta) \) term from the numerator and denominator:
\( a^2 = \frac{1+\sin \theta}{1-\sin \theta} \)
Now, we work with the Left Hand Side (L.H.S.) of the expression we need to prove: \( \frac{a^2-1}{a^2+1} \)
Substitute the expression for \( a^2 \):
\[ = \frac{\frac{1+\sin \theta}{1-\sin \theta} - 1}{\frac{1+\sin \theta}{1-\sin \theta} + 1} \]
To simplify, we find a common denominator for the numerator and the denominator separately:
Numerator: \( \frac{(1+\sin \theta) - 1(1-\sin \theta)}{1-\sin \theta} = \frac{1+\sin \theta - 1+\sin \theta}{1-\sin \theta} = \frac{2 \sin \theta}{1-\sin \theta} \)
Denominator: \( \frac{(1+\sin \theta) + 1(1-\sin \theta)}{1-\sin \theta} = \frac{1+\sin \theta + 1-\sin \theta}{1-\sin \theta} = \frac{2}{1-\sin \theta} \)
Now, we divide the simplified numerator by the simplified denominator:
\[ = \frac{\frac{2 \sin \theta}{1-\sin \theta}}{\frac{2}{1-\sin \theta}} \]
We can cancel out the common factor \( \frac{2}{1-\sin \theta} \):
\( = \sin \theta \)
This is the Right Hand Side (R.H.S). Thus, the identity is proven. When proving identities involving fractions, replacing \( a^2 \) completely can help in simplifying complex fractional expressions.
In simple words: We first used the given equation to find what 'a' squared is in terms of sine and cosine. Then, we put this expression for \( a^2 \) into the left side of the equation we needed to prove. After carefully simplifying the fractions, we found that it became equal to \( \sin \theta \).

🎯 Exam Tip: When an unknown variable (like 'a') is introduced, always aim to express it and its powers (like \( a^2 \)) in terms of standard trigonometric functions first. This sets up the problem for easier algebraic manipulation.

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