OP Malhotra Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Exercise 4 (C)

Question 1. If \( \cot \theta = \frac { 4 }{ 3 } \), find the values of other t-ratios of \( \theta \).
Answer: Given that \( \cot \theta = \frac { 4 }{ 3 } \).
First, we find \( \tan \theta \). Since \( \tan \theta = \frac{1}{\cot \theta} \), we have:
\( \tan \theta = \frac { 1 }{ \frac { 4 }{ 3 } } = \frac { 3 }{ 4 } \)
Next, we use the identity \( \csc^2 \theta = 1 + \cot^2 \theta \) to find \( \csc \theta \):
\( \csc^2 \theta = 1 + \left( \frac { 4 }{ 3 } \right)^2 \)
\( \csc^2 \theta = 1 + \frac { 16 }{ 9 } \)
\( \csc^2 \theta = \frac { 9 + 16 }{ 9 } = \frac { 25 }{ 9 } \)
\( \csc \theta = \pm \sqrt{ \frac { 25 }{ 9 } } = \pm \frac { 5 }{ 3 } \)
Now, we can find \( \sin \theta \) using \( \sin \theta = \frac{1}{\csc \theta} \):
\( \sin \theta = \pm \frac { 3 }{ 5 } \)
To find \( \cos \theta \), we use the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), so \( \cos \theta = \cot \theta \cdot \sin \theta \):
\( \cos \theta = \frac { 4 }{ 3 } \cdot \left( \pm \frac { 3 }{ 5 } \right) = \pm \frac { 4 }{ 5 } \)
Finally, we find \( \sec \theta \) using \( \sec \theta = \frac{1}{\cos \theta} \):
\( \sec \theta = \frac { 1 }{ \pm \frac { 4 }{ 5 } } = \pm \frac { 5 }{ 4 } \)
In simple words: We are given one trigonometric ratio. We use basic relationships and trigonometric identities to find the other ratios step by step. Remember that square roots can give both positive and negative results.

🎯 Exam Tip: Always remember to include the \( \pm \) sign for trigonometric ratios when the quadrant of the angle is not specified, as the square root can be positive or negative.

Question 2. If \( \cos A = 0.6 \), find the values of \( 5 \sin A – 3 \tan A \).
Answer: Given \( \cos A = 0.6 = \frac { 6 }{ 10 } = \frac { 3 }{ 5 } \).
First, we find \( \sin A \) using the identity \( \sin^2 A + \cos^2 A = 1 \):
\( \sin A = \pm \sqrt{1 - \cos^2 A} \)
\( \sin A = \pm \sqrt{1 - \left( \frac { 3 }{ 5 } \right)^2} \)
\( \sin A = \pm \sqrt{1 - \frac { 9 }{ 25 } } \)
\( \sin A = \pm \sqrt{ \frac { 25 - 9 }{ 25 } } = \pm \sqrt{ \frac { 16 }{ 25 } } = \pm \frac { 4 }{ 5 } \)
Next, we find \( \tan A \) using \( \tan A = \frac{\sin A}{\cos A} \):
\( \tan A = \frac { \pm \frac { 4 }{ 5 } }{ \frac { 3 }{ 5 } } = \pm \frac { 4 }{ 3 } \)
We need to consider two cases:
**Case I: When \( \sin A = \frac { 4 }{ 5 } \) and \( \tan A = \frac { 4 }{ 3 } \)**
Substitute these values into the expression \( 5 \sin A - 3 \tan A \):
\( 5 \left( \frac { 4 }{ 5 } \right) - 3 \left( \frac { 4 }{ 3 } \right) \)
\( = 4 - 4 = 0 \)
**Case II: When \( \sin A = - \frac { 4 }{ 5 } \) and \( \tan A = - \frac { 4 }{ 3 } \)**
Substitute these values into the expression \( 5 \sin A - 3 \tan A \):
\( 5 \left( - \frac { 4 }{ 5 } \right) - 3 \left( - \frac { 4 }{ 3 } \right) \)
\( = -4 + 4 = 0 \)
In both possible cases, the value of the expression is 0. This means the expression simplifies to 0 regardless of the quadrant where A lies, as long as cos A is positive (which makes it Quadrant I or IV).
In simple words: First, we find the values of sine and tangent using the given cosine value. Since sine and tangent can be positive or negative, we check both situations. In this problem, both situations lead to the same final answer for the expression.

🎯 Exam Tip: When calculating trigonometric ratios, always remember that square roots result in two possible signs (\( \pm \)). Consider all possible cases based on the signs of the ratios, especially when the quadrant is not specified.

Question 3. If \( \sin A = \frac { 3 }{ 5 } \), prove that \( \tan A + \frac { 1 }{ \cos A } = 2 \) or \( - 2 \).
Answer: Given \( \sin A = \frac { 3 }{ 5 } \).
First, find \( \cos A \) using the identity \( \sin^2 A + \cos^2 A = 1 \):
\( \cos A = \pm \sqrt{1 - \sin^2 A} \)
\( \cos A = \pm \sqrt{1 - \left( \frac { 3 }{ 5 } \right)^2} \)
\( \cos A = \pm \sqrt{1 - \frac { 9 }{ 25 } } \)
\( \cos A = \pm \sqrt{ \frac { 25 - 9 }{ 25 } } = \pm \sqrt{ \frac { 16 }{ 25 } } = \pm \frac { 4 }{ 5 } \)
Next, find \( \tan A \) using \( \tan A = \frac{\sin A}{\cos A} \):
\( \tan A = \frac { \frac { 3 }{ 5 } }{ \pm \frac { 4 }{ 5 } } = \pm \frac { 3 }{ 4 } \)
We will consider two cases for the values of \( \cos A \) and \( \tan A \):
**Case I: When \( \tan A = \frac { 3 }{ 4 } \) and \( \cos A = \frac { 4 }{ 5 } \)**
Substitute these values into the expression \( \tan A + \frac { 1 }{ \cos A } \):
\( \frac { 3 }{ 4 } + \frac { 1 }{ \frac { 4 }{ 5 } } \)
\( = \frac { 3 }{ 4 } + \frac { 5 }{ 4 } \)
\( = \frac { 3 + 5 }{ 4 } = \frac { 8 }{ 4 } = 2 \)
**Case II: When \( \tan A = - \frac { 3 }{ 4 } \) and \( \cos A = - \frac { 4 }{ 5 } \)**
Substitute these values into the expression \( \tan A + \frac { 1 }{ \cos A } \):
\( - \frac { 3 }{ 4 } + \frac { 1 }{ - \frac { 4 }{ 5 } } \)
\( = - \frac { 3 }{ 4 } - \frac { 5 }{ 4 } \)
\( = \frac { -3 - 5 }{ 4 } = \frac { -8 }{ 4 } = -2 \)
Thus, it is proven that \( \tan A + \frac { 1 }{ \cos A } = 2 \) or \( - 2 \). This shows that the value depends on the quadrant A lies in, giving two possible results.
In simple words: We find cosine and tangent from the given sine value. Because cosine can be positive or negative, tangent can also be positive or negative. We then calculate the expression for both possibilities, showing that the answer can be either 2 or -2.

🎯 Exam Tip: For "prove that" questions, clearly show each step of your calculation. When dealing with square roots, remember to account for both positive and negative values, and evaluate the expression for each case to demonstrate all possible outcomes.

Question 4. If \( \tan \theta = \frac{1}{\sqrt{7}} \), find the value of \( \frac{{\operatorname{cosec}}^2 \theta-\sec ^2 \theta}{{\operatorname{cosec}}^2 \theta+\sec ^2 \theta} \).
Answer: Given \( \tan \theta = \frac{1}{\sqrt{7}} \).
From this, we can find \( \cot \theta = \frac{1}{\tan \theta} = \sqrt{7} \).
Now, we use the identities \( \operatorname{cosec}^2 \theta = 1 + \cot^2 \theta \) and \( \sec^2 \theta = 1 + \tan^2 \theta \) to simplify the expression.
Substitute these into the given expression:
\( \frac{{\operatorname{cosec}}^2 \theta-\sec ^2 \theta}{{\operatorname{cosec}}^2 \theta+\sec ^2 \theta} = \frac{(1 + \cot^2 \theta) - (1 + \tan^2 \theta)}{(1 + \cot^2 \theta) + (1 + \tan^2 \theta)} \)
Simplify the expression:
\( = \frac{1 + \cot^2 \theta - 1 - \tan^2 \theta}{1 + \cot^2 \theta + 1 + \tan^2 \theta} \)
\( = \frac{\cot^2 \theta - \tan^2 \theta}{2 + \cot^2 \theta + \tan^2 \theta} \)
Now, substitute the values of \( \tan \theta \) and \( \cot \theta \):
\( \cot^2 \theta = (\sqrt{7})^2 = 7 \)
\( \tan^2 \theta = \left( \frac{1}{\sqrt{7}} \right)^2 = \frac{1}{7} \)
Substitute these squared values into the simplified expression:
\( = \frac{7 - \frac{1}{7}}{2 + 7 + \frac{1}{7}} \)
\( = \frac{\frac{49 - 1}{7}}{\frac{14 + 49 + 1}{7}} \)
\( = \frac{\frac{48}{7}}{\frac{64}{7}} \)
\( = \frac{48}{7} \times \frac{7}{64} \)
\( = \frac{48}{64} = \frac{3 \times 16}{4 \times 16} = \frac{3}{4} \)
In simple words: We are given the value of tangent. We use basic trigonometric identities to change the expression into a simpler form using only tangent and cotangent. Then, we plug in the given value and calculate the final answer. This method avoids calculating secant and cosecant values separately.

🎯 Exam Tip: When an expression involves powers of trigonometric ratios, it is often easier to convert the expression into terms of known ratios (like tangent and cotangent) using fundamental identities before substituting numerical values. This can simplify calculations and reduce errors.

Question 5. If \( \sin \theta = \frac { 21 }{ 29 } \), prove that \( \sec \theta + \tan \theta = 2\frac { 1 }{ 2 } \) if \( \theta \) lies between \( 0 \) and \( \frac { \pi }{ 2 } \). What will be the value of the expression when \( \theta \) lies (i) between \( \frac { \pi }{ 2 } \) and \( \pi \) and (ii) between \( \pi \) and \( \frac { 3\pi }{ 2 } \)?
Answer: Given \( \sin \theta = \frac { 21 }{ 29 } \).
First, find \( \cos \theta \) using \( \cos^2 \theta = 1 - \sin^2 \theta \).
Since \( \theta \) lies between \( 0 \) and \( \frac { \pi }{ 2 } \) (first quadrant), \( \cos \theta \) is positive.
\( \cos \theta = \sqrt{1 - \left( \frac { 21 }{ 29 } \right)^2} \)
\( \cos \theta = \sqrt{1 - \frac { 441 }{ 841 } } \)
\( \cos \theta = \sqrt{ \frac { 841 - 441 }{ 841 } } = \sqrt{ \frac { 400 }{ 841 } } = \frac { 20 }{ 29 } \)
Next, find \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\( \tan \theta = \frac { \frac { 21 }{ 29 } }{ \frac { 20 }{ 29 } } = \frac { 21 }{ 20 } \)
Now, calculate \( \sec \theta + \tan \theta \). Since \( \sec \theta = \frac{1}{\cos \theta} \), \( \sec \theta = \frac { 29 }{ 20 } \).
\( \sec \theta + \tan \theta = \frac { 29 }{ 20 } + \frac { 21 }{ 20 } \)
\( = \frac { 29 + 21 }{ 20 } = \frac { 50 }{ 20 } = \frac { 5 }{ 2 } = 2\frac { 1 }{ 2 } \)
This proves the first part.
Now, let's find the value of the expression for other quadrants:
(i) **When \( \theta \) lies between \( \frac { \pi }{ 2 } \) and \( \pi \) (second quadrant):**
In the second quadrant, \( \sin \theta \) is positive, but \( \cos \theta \) and \( \tan \theta \) are negative.
So, \( \cos \theta = - \frac { 20 }{ 29 } \) and \( \tan \theta = - \frac { 21 }{ 20 } \).
Therefore, \( \sec \theta = \frac{1}{\cos \theta} = - \frac { 29 }{ 20 } \).
\( \sec \theta + \tan \theta = - \frac { 29 }{ 20 } + \left( - \frac { 21 }{ 20 } \right) \)
\( = - \frac { 29 + 21 }{ 20 } = - \frac { 50 }{ 20 } = - \frac { 5 }{ 2 } = -2\frac { 1 }{ 2 } \)
(ii) **When \( \theta \) lies between \( \pi \) and \( \frac { 3\pi }{ 2 } \) (third quadrant):**
In the third quadrant, \( \sin \theta \) is negative. However, the problem states \( \sin \theta = \frac { 21 }{ 29 } \), which is positive.
Since \( \sin \theta > 0 \) is given, the condition that \( \theta \) lies in the third quadrant is not possible. For the third quadrant, the sine value must be negative.
Thus, this case (ii) is impossible with the given \( \sin \theta \) value.
In simple words: First, we find cosine and tangent using the given sine value, assuming the angle is in the first quadrant. Then we add secant and tangent to prove the first part. After that, we check what happens if the angle is in other quadrants. For the second quadrant, cosine and tangent are negative, so the sum changes its sign. For the third quadrant, sine should be negative, but our given sine is positive, so the angle cannot be in the third quadrant.

🎯 Exam Tip: Always pay attention to the quadrant given for an angle, as it determines the sign of trigonometric ratios. If a given ratio conflicts with the quadrant, state that the condition is impossible. Clearly show how the signs affect the final expression's value.

Question 6. If \( \theta \) lies in the second quadrant and \( \tan \theta = \frac { -5 }{ 12 } \), find the value of \( \frac{2 \cos \theta}{1-\sin \theta} \).
Answer: Given that \( \theta \) lies in the second quadrant and \( \tan \theta = \frac { -5 }{ 12 } \).
In the second quadrant, \( \sin \theta > 0 \) and \( \cos \theta < 0 \).
First, find \( \sec \theta \) using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
\( \sec \theta = \pm \sqrt{1 + \left( \frac { -5 }{ 12 } \right)^2} \)
\( \sec \theta = \pm \sqrt{1 + \frac { 25 }{ 144 } } \)
\( \sec \theta = \pm \sqrt{ \frac { 144 + 25 }{ 144 } } = \pm \sqrt{ \frac { 169 }{ 144 } } = \pm \frac { 13 }{ 12 } \)
Since \( \theta \) is in the second quadrant, \( \sec \theta \) must be negative.
So, \( \sec \theta = - \frac { 13 }{ 12 } \).
From \( \sec \theta \), we can find \( \cos \theta = \frac{1}{\sec \theta} = \frac { 1 }{ - \frac { 13 }{ 12 } } = - \frac { 12 }{ 13 } \).
Now, find \( \sin \theta \) using \( \sin \theta = \tan \theta \cdot \cos \theta \):
\( \sin \theta = \left( \frac { -5 }{ 12 } \right) \cdot \left( - \frac { 12 }{ 13 } \right) = \frac { 5 }{ 13 } \)
Now, substitute the values of \( \sin \theta \) and \( \cos \theta \) into the expression \( \frac{2 \cos \theta}{1-\sin \theta} \):
\( = \frac{2 \left( - \frac { 12 }{ 13 } \right)}{1 - \frac { 5 }{ 13 }} \)
\( = \frac{- \frac { 24 }{ 13 }}{\frac { 13 - 5 }{ 13 }} \)
\( = \frac{- \frac { 24 }{ 13 }}{\frac { 8 }{ 13 }} \)
\( = - \frac { 24 }{ 13 } \times \frac { 13 }{ 8 } \)
\( = - \frac { 24 }{ 8 } = -3 \)
In simple words: We are given the tangent value and that the angle is in the second quadrant. In this quadrant, sine is positive and cosine is negative. We use these facts with trigonometric identities to find sine and cosine. Finally, we plug these values into the given expression to get the answer.

🎯 Exam Tip: When the quadrant is specified, use that information to correctly assign the signs (\( + \) or \( - \)) to the trigonometric ratios after taking square roots. A common mistake is to ignore the quadrant information and take only the positive root.

Question 7. If \( \sin \theta \sec \theta = -1 \) and \( \theta \) lies in the second quadrant, find \( \sin \theta \) and \( \sec \theta \).
Answer: Given \( \sin \theta \sec \theta = -1 \).
We know that \( \sec \theta = \frac{1}{\cos \theta} \). Substitute this into the given equation:
\( \sin \theta \cdot \frac{1}{\cos \theta} = -1 \)
\( \frac{\sin \theta}{\cos \theta} = -1 \)
\( \implies \tan \theta = -1 \)
We are also given that \( \theta \) lies in the second quadrant.
In the second quadrant, \( \sin \theta \) is positive, \( \cos \theta \) is negative, and \( \tan \theta \) is negative. Our calculated \( \tan \theta = -1 \) matches this.
Now, find \( \sec \theta \) using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
Since \( \theta \) is in the second quadrant, \( \sec \theta \) must be negative.
\( \sec \theta = - \sqrt{1 + (-1)^2} \)
\( \sec \theta = - \sqrt{1 + 1} = - \sqrt{2} \)
Now, find \( \cos \theta = \frac{1}{\sec \theta} = \frac{1}{-\sqrt{2}} = - \frac{1}{\sqrt{2}} \).
Finally, find \( \sin \theta \) using \( \sin \theta = \tan \theta \cdot \cos \theta \):
\( \sin \theta = (-1) \cdot \left( - \frac{1}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} \)
So, \( \sin \theta = \frac{1}{\sqrt{2}} \) and \( \sec \theta = - \sqrt{2} \).
In simple words: We are given an equation and the quadrant of the angle. First, we use the identity for secant to find that tangent is -1. Since the angle is in the second quadrant, we know secant should be negative. We use this to find the exact value of secant, and then cosine and sine.

🎯 Exam Tip: When an equation involves products of trigonometric ratios, try to simplify it into a single ratio (like tangent, as done here). Always double-check your signs based on the given quadrant to ensure accuracy in your final answers.

Question 8. If \( A \) is in the fourth quadrant and \( \cos A = \frac { 5 }{ 13 } \), find the value of \( \frac{13 \sin A+5 \sec A}{5 \tan A+6 {\operatorname{cosec}} A} \).
Answer: Given that \( A \) is in the fourth quadrant and \( \cos A = \frac { 5 }{ 13 } \).
In the fourth quadrant, \( \cos A \) is positive, but \( \sin A \), \( \tan A \), and \( \operatorname{cosec} A \) are all negative.
First, find \( \sin A \) using \( \sin^2 A = 1 - \cos^2 A \). Since \( A \) is in the fourth quadrant, \( \sin A \) is negative.
\( \sin A = - \sqrt{1 - \left( \frac { 5 }{ 13 } \right)^2} \)
\( \sin A = - \sqrt{1 - \frac { 25 }{ 169 } } \)
\( \sin A = - \sqrt{ \frac { 169 - 25 }{ 169 } } = - \sqrt{ \frac { 144 }{ 169 } } = - \frac { 12 }{ 13 } \)
Next, find the other required ratios:
\( \sec A = \frac{1}{\cos A} = \frac { 1 }{ \frac { 5 }{ 13 } } = \frac { 13 }{ 5 } \)
\( \tan A = \frac{\sin A}{\cos A} = \frac { - \frac { 12 }{ 13 } }{ \frac { 5 }{ 13 } } = - \frac { 12 }{ 5 } \)
\( \operatorname{cosec} A = \frac{1}{\sin A} = \frac { 1 }{ - \frac { 12 }{ 13 } } = - \frac { 13 }{ 12 } \)
Now, substitute these values into the expression \( \frac{13 \sin A+5 \sec A}{5 \tan A+6 {\operatorname{cosec}} A} \):
Numerator: \( 13 \sin A + 5 \sec A = 13 \left( - \frac { 12 }{ 13 } \right) + 5 \left( \frac { 13 }{ 5 } \right) \)
\( = -12 + 13 = 1 \)
Denominator: \( 5 \tan A + 6 \operatorname{cosec} A = 5 \left( - \frac { 12 }{ 5 } \right) + 6 \left( - \frac { 13 }{ 12 } \right) \)
\( = -12 + \left( - \frac { 13 }{ 2 } \right) \)
\( = -12 - \frac { 13 }{ 2 } = \frac { -24 - 13 }{ 2 } = - \frac { 37 }{ 2 } \)
Finally, divide the numerator by the denominator:
\( \frac{1}{- \frac { 37 }{ 2 } } = - \frac { 2 }{ 37 } \)
In simple words: We are given the cosine value and the quadrant. We use this information to find sine, tangent, secant, and cosecant, making sure to use the correct positive or negative signs for each. Then, we put all these values into the big fraction and simplify it carefully to get the final answer.

🎯 Exam Tip: Be very careful with the signs of the trigonometric ratios in the specified quadrant. A single sign error will lead to an incorrect final answer. Calculate the numerator and denominator separately for complex fractions to avoid confusion.

Question 9. Verify that \( \sin 60^{\circ} = \frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}} \).
Answer: We need to verify that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
First, let's find the value of the L.H.S.:
L.H.S. = \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
Now, let's evaluate the R.H.S.:
R.H.S. = \( \frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}} \)
We know that \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \). Substitute this value:
\( = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 + \left( \frac{1}{\sqrt{3}} \right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} \)
\( = \frac{2}{\sqrt{3}} \times \frac{3}{4} \)
\( = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} \)
To simplify \( \frac{3}{2\sqrt{3}} \), we can multiply the numerator and denominator by \( \sqrt{3} \):
\( = \frac{3\sqrt{3}}{2\sqrt{3}\sqrt{3}} = \frac{3\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2} \)
Since L.H.S. = \( \frac{\sqrt{3}}{2} \) and R.H.S. = \( \frac{\sqrt{3}}{2} \), we have L.H.S. = R.H.S.
Thus, the identity is verified. This is a common trigonometric identity, specifically the tangent half-angle formula for sine.
In simple words: We check if the left side of the equation is equal to the right side. We find the known value of sine 60 degrees. Then we calculate the right side by putting in the known value of tangent 30 degrees and simplify it. If both sides are the same, the statement is true.

🎯 Exam Tip: For "verify" or "prove" questions involving trigonometric identities, it's often best to work with one side (usually the more complex one) and simplify it until it matches the other side. Always show all steps clearly and use known standard values or identities.

Question 10. Verify that \( \cos 60^{\circ} = \frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30^{\circ}} \).
Answer: We need to verify that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).
First, let's find the value of the L.H.S.:
L.H.S. = \( \cos 60^{\circ} = \frac{1}{2} \)
Now, let's evaluate the R.H.S.:
R.H.S. = \( \frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30^{\circ}} \)
We know that \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \). Substitute this value:
\( = \frac{1 - \left( \frac{1}{\sqrt{3}} \right)^2}{1 + \left( \frac{1}{\sqrt{3}} \right)^2} \)
\( = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} \)
\( = \frac{\frac{3-1}{3}}{\frac{3+1}{3}} \)
\( = \frac{\frac{2}{3}}{\frac{4}{3}} \)
\( = \frac{2}{3} \times \frac{3}{4} \)
\( = \frac{2}{4} = \frac{1}{2} \)
Since L.H.S. = \( \frac{1}{2} \) and R.H.S. = \( \frac{1}{2} \), we have L.H.S. = R.H.S.
Thus, the identity is verified. This is another form of the tangent half-angle formula, specifically for cosine.
In simple words: We are checking if the left side of the equation equals the right side. We know the value of cosine 60 degrees. We then calculate the right side by putting in the known value of tangent 30 degrees and simplify the fraction. If both sides give the same answer, the statement is true.

🎯 Exam Tip: Always remember the standard trigonometric values for common angles (0°, 30°, 45°, 60°, 90°). These values are essential for verifying identities and solving many trigonometric problems. Show all algebraic steps when simplifying expressions.

Question 11. Prove that \( \sec 30^{\circ} \tan 60^{\circ} + \sin 45^{\circ} \operatorname{cosec} 45^{\circ} + \cos 30^{\circ} \cot 60^{\circ} = \frac { 7 }{ 2 } \).
Answer: We need to prove that the left-hand side (L.H.S.) equals \( \frac { 7 }{ 2 } \).
L.H.S. = \( \sec 30^{\circ} \tan 60^{\circ} + \sin 45^{\circ} \operatorname{cosec} 45^{\circ} + \cos 30^{\circ} \cot 60^{\circ} \)
First, recall the standard values of the trigonometric ratios:
\( \sec 30^{\circ} = \frac{2}{\sqrt{3}} \)
\( \tan 60^{\circ} = \sqrt{3} \)
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \operatorname{cosec} 45^{\circ} = \sqrt{2} \)
\( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
\( \cot 60^{\circ} = \frac{1}{\sqrt{3}} \)
Now, substitute these values into the L.H.S. expression:
\( = \left( \frac{2}{\sqrt{3}} \right) (\sqrt{3}) + \left( \frac{1}{\sqrt{2}} \right) (\sqrt{2}) + \left( \frac{\sqrt{3}}{2} \right) \left( \frac{1}{\sqrt{3}} \right) \)
\( = 2 + 1 + \frac{1}{2} \)
\( = 3 + \frac{1}{2} \)
\( = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} \)
Since L.H.S. = \( \frac{7}{2} \), which is equal to the R.H.S., the identity is proven.
In simple words: We need to show that a complex sum of trigonometric values equals 7/2. We write down the known values for each trigonometric function at the given angles. Then, we put these numbers into the expression and do the math to get the final answer, which should be 7/2.

🎯 Exam Tip: Memorizing the standard trigonometric values for angles like 30°, 45°, and 60° is crucial for solving such problems quickly and accurately. Always be careful with fraction and radical arithmetic.

Question 12. Prove that \( \frac { 4 }{ 3 } \cot^2 30^{\circ} + 3 \sin^2 60^{\circ} – 2 \operatorname{cosec}^2 60^{\circ} – \frac { 3 }{ 4 } \tan^2 30^{\circ} = 3\frac { 1 }{ 3 } \).
Answer: We need to prove that the left-hand side (L.H.S.) equals \( 3\frac { 1 }{ 3 } \).
L.H.S. = \( \frac { 4 }{ 3 } \cot^2 30^{\circ} + 3 \sin^2 60^{\circ} – 2 \operatorname{cosec}^2 60^{\circ} – \frac { 3 }{ 4 } \tan^2 30^{\circ} \)
First, recall the standard values of the trigonometric ratios:
\( \cot 30^{\circ} = \sqrt{3} \implies \cot^2 30^{\circ} = (\sqrt{3})^2 = 3 \)
\( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \implies \sin^2 60^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \)
\( \operatorname{cosec} 60^{\circ} = \frac{2}{\sqrt{3}} \implies \operatorname{cosec}^2 60^{\circ} = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \)
\( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \implies \tan^2 30^{\circ} = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \)
Now, substitute these squared values into the L.H.S. expression:
\( = \frac { 4 }{ 3 } (3) + 3 \left( \frac { 3 }{ 4 } \right) - 2 \left( \frac { 4 }{ 3 } \right) - \frac { 3 }{ 4 } \left( \frac { 1 }{ 3 } \right) \)
\( = 4 + \frac { 9 }{ 4 } - \frac { 8 }{ 3 } - \frac { 1 }{ 4 } \)
Group terms with common denominators or prepare for a common denominator:
\( = (4) + \left( \frac { 9 }{ 4 } - \frac { 1 }{ 4 } \right) - \frac { 8 }{ 3 } \)
\( = 4 + \frac { 8 }{ 4 } - \frac { 8 }{ 3 } \)
\( = 4 + 2 - \frac { 8 }{ 3 } \)
\( = 6 - \frac { 8 }{ 3 } \)
To subtract, find a common denominator (3):
\( = \frac { 18 }{ 3 } - \frac { 8 }{ 3 } = \frac { 18 - 8 }{ 3 } = \frac { 10 }{ 3 } \)
Convert the improper fraction to a mixed number:
\( = 3\frac { 1 }{ 3 } \)
Since L.H.S. = \( 3\frac { 1 }{ 3 } \), which is equal to the R.H.S., the identity is proven. This shows how combining several trigonometric terms with standard angles results in a simple fraction.
In simple words: We need to prove that a long sum and subtraction of squared trigonometric values equals 3 and 1/3. We find the known values for each squared trigonometric function. Then, we put these numbers into the equation and carefully perform the additions and subtractions of fractions to show that the final answer is 10/3, or 3 and 1/3.

🎯 Exam Tip: When dealing with multiple terms involving fractions, first calculate the squared values of the trigonometric ratios. Then, simplify the expression by combining fractions with common denominators or by finding a least common multiple for all denominators to perform the arithmetic efficiently.