Samacheer Kalvi Class 11 Business Maths Solutions Chapter 4 Trigonometry Exercise 4.5

Question 1. The degree measure of \( \frac{\pi}{8} \) is:
(a) \( 20^\circ 60' \)
(b) \( 22^\circ 30' \)
(c) \( 22^\circ 60' \)
(d) \( 20^\circ 30' \)
Answer: (b) \( 22^\circ 30' \)
We know that one radian is equal to \( \frac{180^{\circ}}{\pi} \). To convert radians to degrees, we multiply the radian measure by this value. Here, we are converting \( \frac{\pi}{8} \) radians.
\( \frac{\pi}{8} = \frac{180^{\circ}}{\pi} \times \frac{\pi}{8} \) degrees
\( \implies \frac{\pi}{8} = \frac{180^{\circ}}{8} \)
\( \implies \frac{\pi}{8} = \frac{45^{\circ}}{2} \)
\( \implies \frac{\pi}{8} = 22.5^{\circ} \)
To express \( 0.5^{\circ} \) in minutes, we multiply by 60 because there are 60 minutes in one degree.
\( 0.5^{\circ} = 0.5 \times 60' = 30' \)
So, \( 22.5^{\circ} \) is equal to \( 22^{\circ} 30' \).
In simple words: To change \( \frac{\pi}{8} \) from radians to degrees, we multiply it by \( \frac{180^{\circ}}{\pi} \). This calculation gives us \( 22.5^{\circ} \), which can be written as \( 22^{\circ} \) and 30 minutes.

🎯 Exam Tip: Remember the conversion factor: 1 radian = \( \frac{180^{\circ}}{\pi} \) and 1 degree = \( \frac{\pi}{180} \) radians. Also, \( 1^{\circ} = 60' \) (minutes) and \( 1' = 60'' \) (seconds).

Question 2. The radian measure of \( 37^\circ 30' \) is:
(a) \( \frac{5 \pi}{24} \)
(b) \( \frac{3 \pi}{24} \)
(c) \( \frac{7 \pi}{24} \)
(d) \( \frac{9 \pi}{24} \)
Answer: (a) \( \frac{5 \pi}{24} \)
First, we need to convert the given angle into a single degree measure. We know that \( 30' \) is half a degree.
\( 37^{\circ} 30' = 37^{\circ} + (\frac{30}{60})^{\circ} \)
\( \implies 37^{\circ} 30' = 37^{\circ} + 0.5^{\circ} \)
\( \implies 37^{\circ} 30' = 37.5^{\circ} \)
Now, we can write \( 37.5^{\circ} \) as a fraction: \( \frac{75^{\circ}}{2} \).
To convert degrees to radians, we multiply the degree measure by \( \frac{\pi}{180^{\circ}} \).
\( 37.5^{\circ} = \frac{75}{2} \times \frac{\pi}{180} \) radians
\( \implies 37.5^{\circ} = \frac{75 \pi}{360} \)
\( \implies 37.5^{\circ} = \frac{5 \pi}{24} \) radians, after simplifying the fraction by dividing the numerator and denominator by 15.
In simple words: First, change \( 37^{\circ} \) and 30 minutes into just degrees, which is \( 37.5^{\circ} \). Then, multiply this number by \( \frac{\pi}{180} \) to get the answer in radians, which simplifies to \( \frac{5 \pi}{24} \).

🎯 Exam Tip: Always convert minutes to degrees before converting the entire angle to radians or vice-versa, and simplify fractions to their lowest terms.

Question 3. If \( \tan \theta = \frac{1}{\sqrt{5}} \) and \( \theta \) lies in the first quadrant then \( \cos \theta \) is:
(a) \( \frac{1}{\sqrt{6}} \)
(b) \( \frac{-1}{\sqrt{6}} \)
(c) \( \frac{\sqrt{5}}{\sqrt{6}} \)
(d) \( \frac{-\sqrt{5}}{\sqrt{6}} \)
Answer: (c) \( \frac{\sqrt{5}}{\sqrt{6}} \)
We are given \( \tan \theta = \frac{1}{\sqrt{5}} \). In a right-angled triangle, \( \tan \theta \) is the ratio of the opposite side to the adjacent side. Let the opposite side be 1 unit and the adjacent side be \( \sqrt{5} \) units.

Using the Pythagorean theorem, the hypotenuse (AC) can be calculated:
\( AC = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} \)
\( \implies AC = \sqrt{(1)^2 + (\sqrt{5})^2} \)
\( \implies AC = \sqrt{1 + 5} \)
\( \implies AC = \sqrt{6} \)
Now, \( \cos \theta \) is the ratio of the adjacent side to the hypotenuse.
\( \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \)
\( \implies \cos \theta = \frac{\sqrt{5}}{\sqrt{6}} \)
Since \( \theta \) lies in the first quadrant, \( \cos \theta \) will be positive.
In simple words: We know that \( \tan \theta \) is the ratio of the opposite side to the adjacent side in a right triangle. Using this, we find the length of the third side (hypotenuse) with the Pythagorean theorem. Then, \( \cos \theta \) is simply the adjacent side divided by the hypotenuse. Since \( \theta \) is in the first quarter of the circle, its cosine value is positive.

🎯 Exam Tip: Always draw a right-angled triangle to visualize the sides when given a trigonometric ratio, and use the Pythagorean theorem to find the missing side. Pay attention to the quadrant to determine the sign of the final trigonometric value.

Question 4. The value of \( \sin 15^{\circ} \) is:
(a) \( \frac{\sqrt{3}+1}{2 \sqrt{2}} \)
(b) \( \frac{\sqrt{3}-1}{2 \sqrt{2}} \)
(c) \( \frac{\sqrt{3}}{\sqrt{2}} \)
(d) \( \frac{\sqrt{3}}{2 \sqrt{2}} \)
Answer: (b) \( \frac{\sqrt{3}-1}{2 \sqrt{2}} \)
We can find the value of \( \sin 15^{\circ} \) by expressing \( 15^{\circ} \) as the difference of two standard angles whose sine and cosine values are known, such as \( 45^{\circ} - 30^{\circ} \).
Using the trigonometric identity for \( \sin(A - B) \):
\( \sin(A - B) = \sin A \cos B - \cos A \sin B \)
Here, \( A = 45^{\circ} \) and \( B = 30^{\circ} \).
\( \sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) \)
\( \implies \sin 15^{\circ} = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \)
Now, substitute the known values of these standard angles:
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
\( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \implies \sin 15^{\circ} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} \)
\( \implies \sin 15^{\circ} = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \)
\( \implies \sin 15^{\circ} = \frac{\sqrt{3}-1}{2\sqrt{2}} \).
In simple words: To find \( \sin 15^{\circ} \), we can use the formula for \( \sin(A - B) \) by thinking of \( 15^{\circ} \) as \( 45^{\circ} - 30^{\circ} \). Then, we put in the common values for sine and cosine of \( 45^{\circ} \) and \( 30^{\circ} \) and do the simple subtraction to get the final answer.

🎯 Exam Tip: Memorize the trigonometric values for standard angles like \( 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ} \). Also, be familiar with sum and difference identities for sine and cosine to solve problems for angles like \( 15^{\circ}, 75^{\circ}, 105^{\circ} \).

Question 5. The value of \( \sin(-420^{\circ}) \)
(a) \( \frac{\sqrt{3}}{2} \)
(b) \( -\frac{\sqrt{3}}{2} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{-1}{2} \)
Answer: (b) \( -\frac{\sqrt{3}}{2} \)
We use the property that \( \sin(-\theta) = -\sin \theta \).
So, \( \sin(-420^{\circ}) = -\sin(420^{\circ}) \).
Now, we need to find the value of \( \sin(420^{\circ}) \). We can express \( 420^{\circ} \) as \( 360^{\circ} + 60^{\circ} \).
The sine function has a period of \( 360^{\circ} \), which means \( \sin(360^{\circ} + \theta) = \sin \theta \).
So, \( \sin(420^{\circ}) = \sin(360^{\circ} + 60^{\circ}) = \sin 60^{\circ} \).
We know that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \).
Therefore, \( -\sin(420^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2} \).
In simple words: First, we use the rule that sine of a negative angle is negative sine of the positive angle. Then, we simplify \( \sin(420^{\circ}) \) by noticing that \( 420^{\circ} \) is one full circle plus \( 60^{\circ} \), so it's the same as \( \sin 60^{\circ} \). Finally, we put the minus sign back to get the answer.

🎯 Exam Tip: When dealing with angles larger than \( 360^{\circ} \) or negative angles, always reduce them to an equivalent angle between \( 0^{\circ} \) and \( 360^{\circ} \) (or \( 0^{\circ} \) and \( 90^{\circ} \) using quadrant rules) using the periodicity and odd/even function properties of trigonometric functions.

Question 6. The value of \( \cos(-480^{\circ}) \) is:
(a) \( \sqrt{3} \)
(b) \( -\frac{\sqrt{3}}{2} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{-1}{2} \)
Answer: (d) \( \frac{-1}{2} \)
We use the property that \( \cos(-\theta) = \cos \theta \) because cosine is an even function.
So, \( \cos(-480^{\circ}) = \cos(480^{\circ}) \).
Now, we need to find the value of \( \cos(480^{\circ}) \). We can express \( 480^{\circ} \) as \( 360^{\circ} + 120^{\circ} \).
The cosine function has a period of \( 360^{\circ} \), which means \( \cos(360^{\circ} + \theta) = \cos \theta \).
So, \( \cos(480^{\circ}) = \cos(360^{\circ} + 120^{\circ}) = \cos 120^{\circ} \).
To find \( \cos 120^{\circ} \), we can use the property \( \cos(180^{\circ} - \theta) = -\cos \theta \).
\( \cos 120^{\circ} = \cos(180^{\circ} - 60^{\circ}) \)
\( \implies \cos 120^{\circ} = -\cos 60^{\circ} \)
We know that \( \cos 60^{\circ} = \frac{1}{2} \).
Therefore, \( \cos 120^{\circ} = -\frac{1}{2} \).
In simple words: First, we know that the cosine of a negative angle is the same as the cosine of the positive angle. So \( \cos(-480^{\circ}) \) becomes \( \cos(480^{\circ}) \). Then, we break \( 480^{\circ} \) into one full circle plus \( 120^{\circ} \), making it \( \cos 120^{\circ} \). Finally, we find \( \cos 120^{\circ} \) by using \( 180^{\circ} - 60^{\circ} \), which gives us \( -\frac{1}{2} \).

🎯 Exam Tip: Remember the even/odd properties of trigonometric functions (\(\cos(-\theta) = \cos \theta\), \(\sin(-\theta) = -\sin \theta\)) and their periodicity. This helps reduce angles to a simpler form within the first quadrant for easier calculation.

Question 7. The value of \( \sin 28^{\circ} \cos 17^{\circ} + \cos 28^{\circ} \sin 17^{\circ} \)
(a) \( \frac{1}{\sqrt{2}} \)
(b) 1
(c) \( \frac{-1}{\sqrt{2}} \)
(d) 0
Answer: (a) \( \frac{1}{\sqrt{2}} \)
The given expression is \( \sin 28^{\circ} \cos 17^{\circ} + \cos 28^{\circ} \sin 17^{\circ} \).
This expression matches the trigonometric identity for the sine of the sum of two angles, which is:
\( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
In this case, \( A = 28^{\circ} \) and \( B = 17^{\circ} \).
So, we can rewrite the expression as:
\( \sin(28^{\circ} + 17^{\circ}) \)
\( \implies \sin(45^{\circ}) \)
We know that the value of \( \sin 45^{\circ} \) is \( \frac{1}{\sqrt{2}} \).
Therefore, the value of the expression is \( \frac{1}{\sqrt{2}} \).
In simple words: The problem uses a special formula for sine of two added angles. We just need to add the two angles, \( 28^{\circ} \) and \( 17^{\circ} \), which gives \( 45^{\circ} \). The sine of \( 45^{\circ} \) is a known value, \( \frac{1}{\sqrt{2}} \).

🎯 Exam Tip: Recognize standard trigonometric identities like sum and difference formulas instantly. This saves time and prevents errors in complex expressions. For example, \( \sin(A+B) \), \( \cos(A+B) \), \( \tan(A+B) \).

Question 8. The value of \( \sin 15^{\circ} \cos 15^{\circ} \) is:
(a) 1
(b) \( \frac{1}{2} \)
(c) \( \frac{\sqrt{3}}{2} \)
(d) \( \frac{1}{4} \)
Answer: (d) \( \frac{1}{4} \)
The given expression is \( \sin 15^{\circ} \cos 15^{\circ} \).
We can use the double angle identity for sine, which is \( \sin 2A = 2 \sin A \cos A \).
To make our expression fit this identity, we can multiply and divide by 2:
\( \sin 15^{\circ} \cos 15^{\circ} = \frac{1}{2} (2 \sin 15^{\circ} \cos 15^{\circ}) \)
Now, the term inside the parenthesis is \( \sin(2 \times 15^{\circ}) \), which is \( \sin 30^{\circ} \).
So, the expression becomes:
\( \frac{1}{2} (\sin 30^{\circ}) \)
We know that the value of \( \sin 30^{\circ} \) is \( \frac{1}{2} \).
\( \implies \frac{1}{2} \times \frac{1}{2} \)
\( \implies \frac{1}{4} \).
In simple words: We can solve this by using the double angle formula for sine, which is \( 2 \sin A \cos A = \sin 2A \). We adjust the given expression by adding a \( \frac{1}{2} \) factor. This turns the problem into finding \( \frac{1}{2} \) of \( \sin(2 \times 15^{\circ}) \), which is \( \frac{1}{2} \) of \( \sin 30^{\circ} \), giving us \( \frac{1}{4} \).

🎯 Exam Tip: Look for opportunities to use double angle formulas (like \( \sin 2A \), \( \cos 2A \), \( \tan 2A \)) to simplify expressions involving products of sine and cosine of the same angle. Sometimes, multiplying and dividing by 2 can help transform the expression.

Question 9. The value of \( \sec A \sin(270^{\circ} + A) \) is:
(a) -1
(b) \( \cos^2 A \)
(c) \( \sec^2 A \)
(d) 1
Answer: (a) -1
The given expression is \( \sec A \sin(270^{\circ} + A) \).
First, let's simplify \( \sin(270^{\circ} + A) \).
When an angle is \( (270^{\circ} + A) \), it falls into the fourth quadrant. In the fourth quadrant, the sine function is negative.
Also, when the angle involves \( 270^{\circ} \), the trigonometric function changes from sine to cosine.
So, \( \sin(270^{\circ} + A) = -\cos A \). This is because \( \sin(270^{\circ} + A) \) is in the 4th quadrant where sin is negative, and \( 270^{\circ} \) converts sine to cosine.
Now, substitute this back into the original expression:
\( \sec A \times (-\cos A) \)
We know that \( \sec A = \frac{1}{\cos A} \).
So, \( \frac{1}{\cos A} \times (-\cos A) \)
\( \implies -1 \).
In simple words: We first simplify \( \sin(270^{\circ} + A) \). Since \( 270^{\circ} + A \) is in the fourth section of a circle, sine becomes negative cosine. Then, we put this back into the problem. Because \( \sec A \) is \( \frac{1}{\cos A} \), multiplying it by \( -\cos A \) simply results in -1.

🎯 Exam Tip: Master the angle reduction formulas and quadrant rules. Specifically, remember that for angles like \( 90^{\circ} \pm \theta \), \( 270^{\circ} \pm \theta \), the function changes (e.g., sin to cos, tan to cot, sec to cosec), and for \( 180^{\circ} \pm \theta \), \( 360^{\circ} \pm \theta \), the function remains the same. Always consider the quadrant of the original angle to determine the sign.

Question 10. If \( \sin A + \cos A = 1 \) then \( \sin 2A \) is equal to:
(a) 1
(b) 2
(c) 0
(d) \( \frac{1}{2} \)
Answer: (c) 0
We are given the equation \( \sin A + \cos A = 1 \).
To find \( \sin 2A \), we can square both sides of the given equation:
\( (\sin A + \cos A)^2 = 1^2 \)
Expand the left side using the formula \( (a+b)^2 = a^2 + b^2 + 2ab \):
\( \sin^2 A + \cos^2 A + 2 \sin A \cos A = 1 \)
We know the fundamental trigonometric identity \( \sin^2 A + \cos^2 A = 1 \).
Substitute this into the equation:
\( 1 + 2 \sin A \cos A = 1 \)
Now, we also know the double angle identity for sine: \( 2 \sin A \cos A = \sin 2A \).
Substitute this into the equation:
\( 1 + \sin 2A = 1 \)
Subtract 1 from both sides:
\( \sin 2A = 1 - 1 \)
\( \sin 2A = 0 \).
In simple words: We start with the given equation \( \sin A + \cos A = 1 \). By squaring both sides, we use the identity \( \sin^2 A + \cos^2 A = 1 \) and the double angle formula \( 2 \sin A \cos A = \sin 2A \). This simplifies the equation to \( 1 + \sin 2A = 1 \), which clearly means \( \sin 2A \) must be 0.

🎯 Exam Tip: When an expression involves the sum of sine and cosine and you need to find a double angle, squaring both sides of the original equation is often the key. This approach effectively brings in both \( \sin^2 A + \cos^2 A \) and \( 2 \sin A \cos A \).

Question 11. The value of \( \cos^2 45^{\circ} - \sin^2 45^{\circ} \) is:
(a) \( \frac{\sqrt{3}}{2} \)
(b) \( \frac{1}{2} \)
(c) 0
(d) \( \frac{1}{\sqrt{2}} \)
Answer: (c) 0
The given expression is \( \cos^2 45^{\circ} - \sin^2 45^{\circ} \).
We can use the double angle identity for cosine, which states:
\( \cos 2A = \cos^2 A - \sin^2 A \)
In this problem, \( A = 45^{\circ} \).
So, we can rewrite the expression as:
\( \cos(2 \times 45^{\circ}) \)
\( \implies \cos(90^{\circ}) \)
We know that the value of \( \cos 90^{\circ} \) is 0.
Therefore, \( \cos^2 45^{\circ} - \sin^2 45^{\circ} = 0 \).
Alternatively, we can directly substitute the values:
\( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( (\frac{1}{\sqrt{2}})^2 - (\frac{1}{\sqrt{2}})^2 \)
\( \implies \frac{1}{2} - \frac{1}{2} \)
\( \implies 0 \).
In simple words: The problem is asking for the value of \( \cos^2 45^{\circ} - \sin^2 45^{\circ} \). This matches the formula for \( \cos 2A \). So, we can just calculate \( \cos(2 \times 45^{\circ}) \), which is \( \cos 90^{\circ} \). The value of \( \cos 90^{\circ} \) is 0.

🎯 Exam Tip: Always look for ways to apply trigonometric identities. In cases like \( \cos^2 A - \sin^2 A \), recognizing it as \( \cos 2A \) is quicker and less prone to calculation errors than substituting individual values, especially if the angle is not standard.

Question 12. The value of \( 1 - 2 \sin^2 45^{\circ} \) is:
(a) 1
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{4} \)
(d) 0
Answer: (d) 0
The given expression is \( 1 - 2 \sin^2 45^{\circ} \).
We can use another form of the double angle identity for cosine, which states:
\( \cos 2A = 1 - 2 \sin^2 A \)
In this problem, \( A = 45^{\circ} \).
So, we can rewrite the expression as:
\( \cos(2 \times 45^{\circ}) \)
\( \implies \cos(90^{\circ}) \)
We know that the value of \( \cos 90^{\circ} \) is 0.
Therefore, \( 1 - 2 \sin^2 45^{\circ} = 0 \).
Alternatively, we can directly substitute the values:
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( 1 - 2 \times (\frac{1}{\sqrt{2}})^2 \)
\( \implies 1 - 2 \times \frac{1}{2} \)
\( \implies 1 - 1 \)
\( \implies 0 \).
In simple words: The expression \( 1 - 2 \sin^2 45^{\circ} \) is the same as the formula for \( \cos 2A \). So, we just need to calculate \( \cos(2 \times 45^{\circ}) \), which means \( \cos 90^{\circ} \). The value of \( \cos 90^{\circ} \) is 0.

🎯 Exam Tip: Be familiar with all three forms of the double angle identity for cosine: \( \cos 2A = \cos^2 A - \sin^2 A \), \( \cos 2A = 2 \cos^2 A - 1 \), and \( \cos 2A = 1 - 2 \sin^2 A \). Choose the form that directly matches the given expression to simplify calculations.

Question 13. The value of \( 4 \cos^3 40^{\circ} - 3 \cos 40^{\circ} \) is
(a) \( \frac{\sqrt{3}}{2} \)
(b) \( -\frac{1}{2} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{\sqrt{2}} \)
Answer: (b) \( -\frac{1}{2} \)
The given expression is \( 4 \cos^3 40^{\circ} - 3 \cos 40^{\circ} \).
This expression matches the triple angle identity for cosine, which states:
\( \cos 3A = 4 \cos^3 A - 3 \cos A \)
In this problem, \( A = 40^{\circ} \).
So, we can rewrite the expression as:
\( \cos(3 \times 40^{\circ}) \)
\( \implies \cos(120^{\circ}) \)
To find \( \cos 120^{\circ} \), we can use the property \( \cos(180^{\circ} - \theta) = -\cos \theta \).
\( \cos 120^{\circ} = \cos(180^{\circ} - 60^{\circ}) \)
\( \implies \cos 120^{\circ} = -\cos 60^{\circ} \)
We know that the value of \( \cos 60^{\circ} \) is \( \frac{1}{2} \).
Therefore, \( \cos 120^{\circ} = -\frac{1}{2} \).
In simple words: The expression looks exactly like the formula for \( \cos 3A \). So, we replace A with \( 40^{\circ} \) to get \( \cos(3 \times 40^{\circ}) \), which is \( \cos 120^{\circ} \). Then, we find \( \cos 120^{\circ} \) by thinking of it as \( \cos(180^{\circ} - 60^{\circ}) \), which results in \( -\cos 60^{\circ} \), or \( -\frac{1}{2} \).

🎯 Exam Tip: Familiarize yourself with triple angle identities (e.g., \( \sin 3A \), \( \cos 3A \), \( \tan 3A \)). Recognizing these patterns in expressions allows for quick simplification and accurate calculation.

Question 14. The value of \( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \) is:
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{\sqrt{3}} \)
(c) \( \frac{\sqrt{3}}{2} \)
(d) \( \sqrt{3} \)
Answer: (c) \( \frac{\sqrt{3}}{2} \)
The given expression is \( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \).
This expression matches a form of the double angle identity for sine, which states:
\( \sin 2A = \frac{2 \tan A}{1+\tan^2 A} \)
In this problem, \( A = 30^{\circ} \).
So, we can rewrite the expression as:
\( \sin(2 \times 30^{\circ}) \)
\( \implies \sin(60^{\circ}) \)
We know that the value of \( \sin 60^{\circ} \) is \( \frac{\sqrt{3}}{2} \).
Therefore, \( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} = \frac{\sqrt{3}}{2} \). This is a helpful identity that connects tangent with sine for double angles.
In simple words: This problem uses a special formula that relates the tangent of an angle to the sine of twice that angle. By applying the formula \( \sin 2A = \frac{2 \tan A}{1+\tan^2 A} \), and knowing that A is \( 30^{\circ} \), we calculate \( \sin(2 \times 30^{\circ}) \), which is \( \sin 60^{\circ} \). The value of \( \sin 60^{\circ} \) is \( \frac{\sqrt{3}}{2} \).

🎯 Exam Tip: Remember the double angle identities in terms of tangent: \( \sin 2A = \frac{2 \tan A}{1+\tan^2 A} \), \( \cos 2A = \frac{1-\tan^2 A}{1+\tan^2 A} \), and \( \tan 2A = \frac{2 \tan A}{1-\tan^2 A} \). These are very useful for simplifying expressions or solving equations when tangent is involved.

Question 15. If \( \sin A = \frac{1}{2} \) then \( 4 \cos^3 A - 3 \cos A \) is:
(a) 1
(b) 0
(c) \( \frac{\sqrt{3}}{2} \)
(d) \( \frac{1}{\sqrt{2}} \)
Answer: (b) 0
We are given \( \sin A = \frac{1}{2} \).
From this, we know that \( A = 30^{\circ} \) (since \( 30^{\circ} \) is the angle whose sine is \( \frac{1}{2} \) and A is typically considered in the first quadrant unless specified otherwise).
The expression we need to evaluate is \( 4 \cos^3 A - 3 \cos A \).
This expression is the triple angle identity for cosine:
\( \cos 3A = 4 \cos^3 A - 3 \cos A \)
Substitute \( A = 30^{\circ} \) into the identity:
\( \cos(3 \times 30^{\circ}) \)
\( \implies \cos(90^{\circ}) \)
We know that the value of \( \cos 90^{\circ} \) is 0.
Therefore, \( 4 \cos^3 A - 3 \cos A = 0 \).
In simple words: First, we use \( \sin A = \frac{1}{2} \) to find that A is \( 30^{\circ} \). Then, we recognize that the expression \( 4 \cos^3 A - 3 \cos A \) is the formula for \( \cos 3A \). So, we replace A with \( 30^{\circ} \) to get \( \cos(3 \times 30^{\circ}) \), which simplifies to \( \cos 90^{\circ} \). The value of \( \cos 90^{\circ} \) is 0.

🎯 Exam Tip: When given a value for one trigonometric function, first find the angle (if it's a standard angle) and then substitute it into the expression. This often leads to a quick solution, especially when the expression is a recognizable identity.

Question 16. The value of \( \frac{3 \tan 10^{\circ}-\tan ^{3} 10^{\circ}}{1-3 \tan ^{2} 10^{\circ}} \) is:
(a) \( \frac{1}{\sqrt{3}} \)
(b) \( \frac{-\sqrt{3}}{2} \)
(c) \( \frac{\sqrt{3}}{2} \)
(d) \( \frac{1}{\sqrt{2}} \)
Answer: (a) \( \frac{1}{\sqrt{3}} \)
The given expression is \( \frac{3 \tan 10^{\circ}-\tan ^{3} 10^{\circ}}{1-3 \tan ^{2} 10^{\circ}} \).
This expression matches the triple angle identity for tangent, which states:
\( \tan 3A = \frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A} \)
In this problem, \( A = 10^{\circ} \).
So, we can rewrite the expression as:
\( \tan(3 \times 10^{\circ}) \)
\( \implies \tan(30^{\circ}) \)
We know that the value of \( \tan 30^{\circ} \) is \( \frac{1}{\sqrt{3}} \).
Therefore, \( \frac{3 \tan 10^{\circ}-\tan ^{3} 10^{\circ}}{1-3 \tan ^{2} 10^{\circ}} = \frac{1}{\sqrt{3}} \).
In simple words: We can solve this by recognizing that the given expression fits the formula for \( \tan 3A \). By replacing A with \( 10^{\circ} \), the problem becomes finding \( \tan(3 \times 10^{\circ}) \), which is \( \tan 30^{\circ} \). The value of \( \tan 30^{\circ} \) is a known constant, \( \frac{1}{\sqrt{3}} \).

🎯 Exam Tip: Memorizing triple angle identities for tangent can save a lot of time. If you don't recall it, you can derive it from \( \tan(A+B) \) where \( B=2A \), but this takes longer in an exam setting.

Question 17. The value of \( \operatorname{cosec}^{-1} (\frac{2}{\sqrt{3}}) \) is:
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{6} \)
Answer: (c) \( \frac{\pi}{3} \)
Let \( y = \operatorname{cosec}^{-1} (\frac{2}{\sqrt{3}}) \).
This means \( \operatorname{cosec} y = \frac{2}{\sqrt{3}} \).
We know that \( \operatorname{cosec} y = \frac{1}{\sin y} \).
So, \( \frac{1}{\sin y} = \frac{2}{\sqrt{3}} \).
Taking the reciprocal of both sides gives:
\( \sin y = \frac{\sqrt{3}}{2} \).
We need to find an angle \( y \) such that its sine is \( \frac{\sqrt{3}}{2} \). We know that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \).
In radians, \( 60^{\circ} = \frac{\pi}{3} \).
Therefore, \( y = \frac{\pi}{3} \).
In simple words: To find the value of \( \operatorname{cosec}^{-1} (\frac{2}{\sqrt{3}}) \), we can first think of it as finding an angle whose cosecant is \( \frac{2}{\sqrt{3}} \). Since cosecant is the inverse of sine, this means we are looking for an angle whose sine is \( \frac{\sqrt{3}}{2} \). That angle is \( 60^{\circ} \), which is \( \frac{\pi}{3} \) in radians.

🎯 Exam Tip: When dealing with inverse trigonometric functions of cosecant, secant, or cotangent, convert them into their reciprocal sine, cosine, or tangent forms. This often makes it easier to recognize standard angle values.

Question 18. \( \sec^{-1} (\frac{2}{3}) + \operatorname{cosec}^{-1} (\frac{2}{3}) = \)
(a) \( \frac{-\pi}{2} \)
(b) \( \frac{\pi}{2} \)
(c) \( \pi \)
(d) \( -\pi \)
Answer: (b) \( \frac{\pi}{2} \)
This problem uses a fundamental property of inverse trigonometric functions. For any real number \( x \) where \( |x| \geq 1 \), the identity is:
\( \sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2} \)
In this question, \( x = \frac{2}{3} \). However, the domain for \( \sec^{-1} x \) and \( \operatorname{cosec}^{-1} x \) is \( |x| \ge 1 \). So, the input \( x = \frac{2}{3} \) is technically outside the standard domain for these functions.
However, if we assume the problem intends to test this identity in a context where the values are treated formally, or if it's a simplification in a broader algebraic context, the identity holds for values where both functions are defined. Given the MCQ options, it most likely expects the application of this identity. If this were a question requiring rigorous domain checks, it would be "undefined" or "no solution". Since we have options that are specific values, we apply the identity.
Applying the identity directly:
\( \sec^{-1} (\frac{2}{3}) + \operatorname{cosec}^{-1} (\frac{2}{3}) = \frac{\pi}{2} \).
In simple words: There is a special rule for inverse secant and inverse cosecant functions. When you add \( \sec^{-1} \) of a number to \( \operatorname{cosec}^{-1} \) of the same number, the answer is always \( \frac{\pi}{2} \), as long as the number is valid for both functions. So, for the number \( \frac{2}{3} \), the sum is \( \frac{\pi}{2} \).

🎯 Exam Tip: Memorize the complementary inverse trigonometric identities: \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \), \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \), and \( \sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2} \). These identities are frequently tested and can simplify complex problems significantly. Always be mindful of the domain constraints for inverse trigonometric functions.

Question 19. If \( \alpha \) and \( \beta \) be between 0 and \( \frac{\pi}{2} \) and if \( \cos(\alpha + \beta) = \frac{12}{13} \) and \( \sin(\alpha – \beta) = \frac{3}{5} \) then \( \sin 2\alpha \) is:
(a) \( \frac{16}{15} \)
(b) 0
(c) \( \frac{56}{65} \)
(d) \( \frac{64}{65} \)
Answer: (c) \( \frac{56}{65} \)
We are given that \( \alpha \) and \( \beta \) are between 0 and \( \frac{\pi}{2} \). This means both angles are in the first quadrant, so all trigonometric ratios for \( \alpha \), \( \beta \), \( \alpha + \beta \), and \( \alpha - \beta \) (if \( \alpha - \beta \) is also in the first quadrant or for which the sine and cosine values are positive) will be positive.
Given \( \cos(\alpha + \beta) = \frac{12}{13} \). Since \( \cos^2 x + \sin^2 x = 1 \), we can find \( \sin(\alpha + \beta) \):
\( \sin(\alpha + \beta) = \sqrt{1 - \cos^2(\alpha + \beta)} = \sqrt{1 - (\frac{12}{13})^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{169-144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \).

Given \( \sin(\alpha – \beta) = \frac{3}{5} \). Similarly, we find \( \cos(\alpha – \beta) \):
\( \cos(\alpha – \beta) = \sqrt{1 - \sin^2(\alpha – \beta)} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25-9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \).

We want to find \( \sin 2\alpha \). We can write \( 2\alpha \) as \( (\alpha + \beta) + (\alpha – \beta) \).
Using the identity \( \sin(X+Y) = \sin X \cos Y + \cos X \sin Y \), where \( X = (\alpha + \beta) \) and \( Y = (\alpha – \beta) \):
\( \sin 2\alpha = \sin[(\alpha + \beta) + (\alpha – \beta)] \)
\( \implies \sin 2\alpha = \sin(\alpha + \beta) \cos(\alpha – \beta) + \cos(\alpha + \beta) \sin(\alpha – \beta) \)
Substitute the values we found:
\( \sin 2\alpha = (\frac{5}{13}) \times (\frac{4}{5}) + (\frac{12}{13}) \times (\frac{3}{5}) \)
\( \implies \sin 2\alpha = \frac{20}{65} + \frac{36}{65} \)
\( \implies \sin 2\alpha = \frac{20 + 36}{65} \)
\( \implies \sin 2\alpha = \frac{56}{65} \).
In simple words: First, we use the given cosine and sine values to find the missing sine and cosine values for \( (\alpha + \beta) \) and \( (\alpha - \beta) \) using right triangles or the Pythagorean identity. Then, we think of \( 2\alpha \) as the sum of \( (\alpha + \beta) \) and \( (\alpha - \beta) \). We use the \( \sin(X+Y) \) formula, put in all the values we found, and add them up to get \( \frac{56}{65} \).

🎯 Exam Tip: When dealing with sums/differences of angles, convert all given information into sine and cosine ratios. To find a double angle like \( 2\alpha \), remember the trick of expressing it as \( (\alpha + \beta) + (\alpha – \beta) \) and then applying sum/difference formulas. Always remember to check quadrant rules for signs if angles are not strictly in the first quadrant.

Question 20. If \( \tan A = \frac{1}{2} \) and \( \tan B = \frac{1}{3} \) then \( \tan(2A + B) \) is equal to:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
We need to find \( \tan(2A + B) \). We can break this into two steps.
First, find \( \tan 2A \) using the double angle identity for tangent:
\( \tan 2A = \frac{2 \tan A}{1-\tan^2 A} \)
Given \( \tan A = \frac{1}{2} \):
\( \tan 2A = \frac{2 \times \frac{1}{2}}{1 - (\frac{1}{2})^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \).
Now, use the sum identity for tangent, \( \tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y} \), where \( X = 2A \) and \( Y = B \).
We have \( \tan 2A = \frac{4}{3} \) and we are given \( \tan B = \frac{1}{3} \).
\( \tan(2A + B) = \frac{\tan 2A + \tan B}{1 - \tan 2A \tan B} \)
\( \implies \tan(2A + B) = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \frac{4}{3} \times \frac{1}{3}} \)
\( \implies \tan(2A + B) = \frac{\frac{5}{3}}{1 - \frac{4}{9}} \)
\( \implies \tan(2A + B) = \frac{\frac{5}{3}}{\frac{9-4}{9}} \)
\( \implies \tan(2A + B) = \frac{\frac{5}{3}}{\frac{5}{9}} \)
\( \implies \tan(2A + B) = \frac{5}{3} \times \frac{9}{5} \)
\( \implies \tan(2A + B) = 3 \).
In simple words: First, we find \( \tan 2A \) using its double angle formula, which gives \( \frac{4}{3} \). Then, we use the formula for \( \tan(X+Y) \) where X is \( 2A \) and Y is B. We plug in the values for \( \tan 2A \) and \( \tan B \) and simplify the fraction to get the final answer, which is 3.

🎯 Exam Tip: Break down complex angle expressions (like \( 2A+B \)) into simpler parts that can be handled by known identities (like \( \tan 2A \) and \( \tan(X+Y) \)). Perform calculations step-by-step to avoid errors, especially when dealing with fractions.

Question 21. \( \tan(\frac{\pi}{4} – x) \) is:
(a) \( \left(\frac{1+\tan x}{1-\tan x}\right) \)
(b) \( \left(\frac{1-\tan x}{1+\tan x}\right) \)
(c) \( 1 - \tan x \)
(d) \( 1 + \tan x \)
Answer: (b) \( \left(\frac{1-\tan x}{1+\tan x}\right) \)
We need to simplify \( \tan(\frac{\pi}{4} – x) \).
We use the tangent difference identity, which states:
\( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
In this problem, \( A = \frac{\pi}{4} \) and \( B = x \).
We know that \( \tan(\frac{\pi}{4}) = \tan(45^{\circ}) = 1 \).
Substitute these values into the identity:
\( \tan(\frac{\pi}{4} – x) = \frac{\tan(\frac{\pi}{4}) - \tan x}{1 + \tan(\frac{\pi}{4}) \tan x} \)
\( \implies \tan(\frac{\pi}{4} – x) = \frac{1 - \tan x}{1 + 1 \times \tan x} \)
\( \implies \tan(\frac{\pi}{4} – x) = \frac{1 - \tan x}{1 + \tan x} \).
In simple words: To simplify \( \tan(\frac{\pi}{4} – x) \), we use the formula for \( \tan(A-B) \). We know that \( \tan(\frac{\pi}{4}) \) is 1. When we put this value into the formula, it simplifies to \( \frac{1 - \tan x}{1 + \tan x} \).

🎯 Exam Tip: Memorize the tangent sum and difference identities, especially involving \( \frac{\pi}{4} \) (or \( 45^{\circ} \)), as \( \tan(\frac{\pi}{4}) = 1 \) leads to very common and useful simplified forms like \( \frac{1 \pm \tan x}{1 \mp \tan x} \).

Question 22. \( \sin \left(\cos ^{-1} \frac{3}{5}\right) \) is:
(a) \( \frac{3}{5} \)
(b) \( \frac{5}{3} \)
(c) \( \frac{4}{5} \)
(d) \( \frac{5}{4} \)
Answer: (c) \( \frac{4}{5} \)
Let \( A = \cos^{-1} (\frac{3}{5}) \).
This means \( \cos A = \frac{3}{5} \).
Since the range of \( \cos^{-1} x \) is \( [0, \pi] \), and \( \frac{3}{5} \) is positive, \( A \) must be in the first quadrant where sine is also positive.
We need to find \( \sin A \). We can use the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \).
\( \sin^2 A = 1 - \cos^2 A \)
\( \implies \sin^2 A = 1 - (\frac{3}{5})^2 \)
\( \implies \sin^2 A = 1 - \frac{9}{25} \)
\( \implies \sin^2 A = \frac{25 - 9}{25} \)
\( \implies \sin^2 A = \frac{16}{25} \)
\( \implies \sin A = \sqrt{\frac{16}{25}} \)
\( \implies \sin A = \frac{4}{5} \) (We take the positive root because A is in the first quadrant).

So, \( \sin \left(\cos^{-1} \frac{3}{5}\right) = \sin A = \frac{4}{5} \).
In simple words: First, we let A be the angle whose cosine is \( \frac{3}{5} \). This means we have a right triangle where the adjacent side is 3 and the hypotenuse is 5. We can find the opposite side using the Pythagorean theorem, which will be 4. Then, sine of angle A is the opposite side (4) divided by the hypotenuse (5), giving \( \frac{4}{5} \).

🎯 Exam Tip: When dealing with expressions like \( \sin(\cos^{-1} x) \) or \( \tan(\sin^{-1} x) \), let the inverse function be an angle (e.g., \( \theta = \cos^{-1} x \)). This allows you to convert the problem into finding a trigonometric ratio of \( \theta \), which can be solved using a right-angled triangle or Pythagorean identities.

Question 23. The value of \( \frac{1}{\operatorname{cosec}(-45^{\circ})} \) is:
(a) \( \frac{-1}{\sqrt{2}} \)
(b) \( \frac{1}{\sqrt{2}} \)
(c) \( \sqrt{2} \)
(d) \( -\sqrt{2} \)
Answer: (a) \( \frac{-1}{\sqrt{2}} \)
The given expression is \( \frac{1}{\operatorname{cosec}(-45^{\circ})} \).
We know the reciprocal identity: \( \frac{1}{\operatorname{cosec} \theta} = \sin \theta \).
So, the expression can be rewritten as:
\( \sin(-45^{\circ}) \)
Now, we use the property that sine is an odd function, meaning \( \sin(-\theta) = -\sin \theta \).
\( \sin(-45^{\circ}) = -\sin(45^{\circ}) \)
We know that the value of \( \sin 45^{\circ} \) is \( \frac{1}{\sqrt{2}} \).
Therefore, \( -\sin(45^{\circ}) = -\frac{1}{\sqrt{2}} \).
In simple words: We first change \( \frac{1}{\operatorname{cosec}(-45^{\circ})} \) into \( \sin(-45^{\circ}) \) because cosecant is the inverse of sine. Then, we use the rule that the sine of a negative angle is the negative of the sine of the positive angle. So \( \sin(-45^{\circ}) \) becomes \( -\sin(45^{\circ}) \), which is \( -\frac{1}{\sqrt{2}} \).

🎯 Exam Tip: Always remember the reciprocal identities (e.g., \( \frac{1}{\operatorname{cosec} \theta} = \sin \theta \), \( \frac{1}{\sec \theta} = \cos \theta \)) and the even/odd function properties (\(\sin(-\theta) = -\sin \theta\), \(\cos(-\theta) = \cos \theta\), \(\tan(-\theta) = -\tan \theta\)). These are fundamental for simplifying trigonometric expressions.

Question 24. If \( p \sec 50^{\circ} = \tan 50^{\circ} \) then \( p \) is:
(a) \( \cos 50^{\circ} \)
(b) \( \sin 50^{\circ} \)
(c) \( \tan 50^{\circ} \)
(d) \( \sec 50^{\circ} \)
Answer: (b) \( \sin 50^{\circ} \)
We are given the equation \( p \sec 50^{\circ} = \tan 50^{\circ} \).
We need to solve for \( p \). To do this, we can express \( \sec 50^{\circ} \) and \( \tan 50^{\circ} \) in terms of sine and cosine.
We know that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
Substitute these into the equation:
\( p \times \frac{1}{\cos 50^{\circ}} = \frac{\sin 50^{\circ}}{\cos 50^{\circ}} \)
To isolate \( p \), we can multiply both sides of the equation by \( \cos 50^{\circ} \):
\( p = \frac{\sin 50^{\circ}}{\cos 50^{\circ}} \times \cos 50^{\circ} \)
The \( \cos 50^{\circ} \) terms cancel out.
\( p = \sin 50^{\circ} \).
In simple words: We start with the equation \( p \sec 50^{\circ} = \tan 50^{\circ} \). We rewrite \( \sec 50^{\circ} \) as \( \frac{1}{\cos 50^{\circ}} \) and \( \tan 50^{\circ} \) as \( \frac{\sin 50^{\circ}}{\cos 50^{\circ}} \). After putting these into the equation, we can cancel \( \cos 50^{\circ} \) from both sides, which leaves us with \( p = \sin 50^{\circ} \).

🎯 Exam Tip: When solving equations involving different trigonometric functions, convert all functions into their sine and cosine forms. This often simplifies the equation and makes it easier to solve for the unknown variable.

Question 25. \( (\frac{\cos x}{\operatorname{cosec} x})-\sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x} \) is:
(a) \( \cos^2 x \sin^2 x \)
(b) \( \sin^2 x - \cos^2 x \)
(c) 1
(d) 0
Answer: (d) 0
The given expression is \( \frac{\cos x}{\operatorname{cosec} x} - \sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x} \).
Let's simplify each part of the expression.
**Part 1:** \( \frac{\cos x}{\operatorname{cosec} x} \)
We know that \( \operatorname{cosec} x = \frac{1}{\sin x} \).
So, \( \frac{\cos x}{\operatorname{cosec} x} = \frac{\cos x}{\frac{1}{\sin x}} = \cos x \sin x \).
**Part 2:** \( \sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x} \)
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can deduce:
\( 1 - \sin^2 x = \cos^2 x \)
\( 1 - \cos^2 x = \sin^2 x \)
So, \( \sqrt{1-\sin ^{2} x} = \sqrt{\cos^2 x} = |\cos x| \)
And \( \sqrt{1-\cos ^{2} x} = \sqrt{\sin^2 x} = |\sin x| \)
Assuming \( x \) is in a quadrant where \( \sin x \) and \( \cos x \) are positive (or that the product \( \sin x \cos x \) is positive, such as in the first or third quadrant, which is a common simplification in such problems unless specified otherwise), we can write \( \sqrt{\cos^2 x} = \cos x \) and \( \sqrt{\sin^2 x} = \sin x \).
Therefore, \( \sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x} = \cos x \sin x \).
Now, combine both parts back into the original expression:
\( \cos x \sin x - \cos x \sin x \)
\( \implies 0 \).
In simple words: We first simplify the left part of the expression by changing \( \operatorname{cosec} x \) to \( \frac{1}{\sin x} \), which makes it \( \cos x \sin x \). For the right part, we use the rule \( \sin^2 x + \cos^2 x = 1 \) to change \( \sqrt{1-\sin^2 x} \) to \( \cos x \) and \( \sqrt{1-\cos^2 x} \) to \( \sin x \). So, the right part also becomes \( \cos x \sin x \). Subtracting \( \cos x \sin x \) from \( \cos x \sin x \) gives us 0.

🎯 Exam Tip: Always convert all trigonometric functions into their simplest forms (sine and cosine). Remember the Pythagorean identities \( \sin^2 x + \cos^2 x = 1 \) and its rearrangements. Be careful with square roots, as \( \sqrt{a^2} = |a| \), but for simplicity in MCQs, it often implies the positive value when not explicitly specified otherwise for quadrant considerations.

Question 25. The value of \( \left(\frac{\cos x}{\operatorname{cosec} x}\right)-\sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x} \) is:
(a) cos² x sin² x
(b) sin² x - cos² x
(c) 1
(d) 0
Answer: (d) 0
\( \left(\frac{\cos x}{\operatorname{cosec} x}\right)-\sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x} \)
We know that \( \operatorname{cosec} x = \frac{1}{\sin x} \), so \( \frac{\cos x}{\operatorname{cosec} x} = \cos x \times \sin x \).
Also, using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\( \sqrt{1-\sin ^{2} x} = \sqrt{\cos ^{2} x} = \cos x \)
\( \sqrt{1-\cos ^{2} x} = \sqrt{\sin ^{2} x} = \sin x \)
Substitute these back into the expression:
\( = \cos x \times \sin x - (\cos x \times \sin x) \)
\( = 0 \)
This shows how basic trigonometric identities can simplify complex expressions into a simple value.
In simple words: We changed the given math problem using simple rules for sine and cosine. After changing it, the problem became two identical parts that were subtracted from each other, making the final answer zero.

🎯 Exam Tip: Always remember fundamental trigonometric identities like \( \operatorname{cosec} x = \frac{1}{\sin x} \) and \( \sin^2 x + \cos^2 x = 1 \) to simplify complex expressions efficiently.