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

Question 1. Express the following as functions of angles less than 45° :
(i) sin 194°
(ii) sin 348°
(iii) cos 189°
(iv) sin (- 1785°)
(v) tan (3598°)
(vi) cot (- 1952°)
(vii) cosec (- 7498°).
Answer:
(i) We need to express \( \sin 194^\circ \) as a function of an angle less than \( 45^\circ \). We can write \( 194^\circ \) as \( 180^\circ + 14^\circ \).
So, \( \sin 194^\circ = \sin (180^\circ + 14^\circ) \)
\( \implies \sin 194^\circ = -\sin 14^\circ \) (Since \( \sin(180^\circ + \theta) = -\sin \theta \)).

(ii) For \( \sin 348^\circ \), we can write \( 348^\circ \) as \( 360^\circ - 12^\circ \).
So, \( \sin 348^\circ = \sin (360^\circ - 12^\circ) \)
\( \implies \sin 348^\circ = -\sin 12^\circ \) (Since \( \sin(360^\circ - \theta) = -\sin \theta \)).

(iii) To express \( \cos 189^\circ \), we write \( 189^\circ \) as \( 180^\circ + 9^\circ \).
So, \( \cos 189^\circ = \cos (180^\circ + 9^\circ) \)
\( \implies \cos 189^\circ = -\cos 9^\circ \) (Since \( \cos(180^\circ + \theta) = -\cos \theta \)).

(iv) For \( \sin (-1785^\circ) \), first we use \( \sin (-\theta) = -\sin \theta \).
\( \sin (-1785^\circ) = -\sin (1785^\circ) \).
Now, we find the equivalent angle for \( 1785^\circ \). \( 1785^\circ = 4 \times 360^\circ + 345^\circ = 1440^\circ + 345^\circ \).
So, \( \sin (1785^\circ) = \sin (360^\circ \times 4 + 345^\circ) = \sin 345^\circ \) (Since \( \sin(360^\circ \times n + \theta) = \sin \theta \)).
Then, we can write \( 345^\circ \) as \( 360^\circ - 15^\circ \).
\( \sin 345^\circ = \sin (360^\circ - 15^\circ) \)
\( \implies \sin 345^\circ = -\sin 15^\circ \) (Since \( \sin(360^\circ - \theta) = -\sin \theta \)).
Therefore, \( \sin (-1785^\circ) = -(-\sin 15^\circ) = \sin 15^\circ \).

(v) For \( \tan (3598^\circ) \), we can write \( 3598^\circ \) as \( 10 \times 360^\circ - 2^\circ = 3600^\circ - 2^\circ \).
So, \( \tan (3598^\circ) = \tan (10 \times 360^\circ - 2^\circ) \)
\( \implies \tan (3598^\circ) = \tan (-2^\circ) \) (Since \( \tan(360^\circ \times n - \theta) = \tan(-\theta) \)).
Then, using \( \tan (-\theta) = -\tan \theta \), we get
\( \tan (-2^\circ) = -\tan 2^\circ \).

(vi) For \( \cot (-1952^\circ) \), first use \( \cot (-\theta) = -\cot \theta \).
\( \cot (-1952^\circ) = -\cot (1952^\circ) \).
Now, express \( 1952^\circ \) as \( 5 \times 360^\circ + 152^\circ = 1800^\circ + 152^\circ \).
So, \( \cot (1952^\circ) = \cot (5 \times 360^\circ + 152^\circ) = \cot 152^\circ \) (Since \( \cot(360^\circ \times n + \theta) = \cot \theta \)).
Then, write \( 152^\circ \) as \( 180^\circ - 28^\circ \).
\( \cot 152^\circ = \cot (180^\circ - 28^\circ) \)
\( \implies \cot 152^\circ = -\cot 28^\circ \) (Since \( \cot(180^\circ - \theta) = -\cot \theta \)).
Therefore, \( \cot (-1952^\circ) = -(-\cot 28^\circ) = \cot 28^\circ \).

(vii) For \( \text{cosec } (-7498^\circ) \), first use \( \text{cosec } (-\theta) = -\text{cosec } \theta \).
\( \text{cosec } (-7498^\circ) = -\text{cosec } (7498^\circ) \).
Now, express \( 7498^\circ \) as \( 20 \times 360^\circ + 298^\circ = 7200^\circ + 298^\circ \).
So, \( \text{cosec } (7498^\circ) = \text{cosec } (20 \times 360^\circ + 298^\circ) = \text{cosec } 298^\circ \) (Since \( \text{cosec}(360^\circ \times n + \theta) = \text{cosec } \theta \)).
Then, write \( 298^\circ \) as \( 360^\circ - 62^\circ \).
\( \text{cosec } 298^\circ = \text{cosec } (360^\circ - 62^\circ) \)
\( \implies \text{cosec } 298^\circ = -\text{cosec } 62^\circ \) (Since \( \text{cosec}(360^\circ - \theta) = -\text{cosec } \theta \)).
Therefore, \( \text{cosec } (-7498^\circ) = -(-\text{cosec } 62^\circ) = \text{cosec } 62^\circ \). When working with trigonometric functions, identifying the quadrant of the angle helps determine the sign of the function. For example, 194° is in the third quadrant, where sine is negative.
In simple words: To change a big angle into a small one (less than 45 degrees), we use rules about adding or subtracting 180 or 360 degrees. We also remember if the sign changes based on which quarter of the circle the angle falls into.

🎯 Exam Tip: Always remember the quadrant rules for trigonometric functions (ASTC rule) and the periodicity of \( 360^\circ \) when simplifying angles. Pay attention to negative angles as well, as functions like sine, tangent, and cosecant are odd functions, while cosine and secant are even functions.

Question 2. Without using tables, give the value of each of the following :
(i) sin 120°
(ii) cot 330°
(iii) sec 210°
(iv) cos 315°
(v) cosec 675°
(vi) cos 855°
(vii) sin 4530°
(viii) sec \( \frac { 15\pi }{ 4 } \)
Answer:
(i) To find \( \sin 120^\circ \), we can write \( 120^\circ \) as \( 180^\circ - 60^\circ \).
\( \sin 120^\circ = \sin (180^\circ - 60^\circ) \)
\( \implies \sin 120^\circ = \sin 60^\circ \) (Since \( \sin(180^\circ - \theta) = \sin \theta \)).
The value of \( \sin 60^\circ \) is \( \frac{\sqrt{3}}{2} \).

(ii) To find \( \cot 330^\circ \), we write \( 330^\circ \) as \( 360^\circ - 30^\circ \).
\( \cot 330^\circ = \cot (360^\circ - 30^\circ) \)
\( \implies \cot 330^\circ = -\cot 30^\circ \) (Since \( \cot(360^\circ - \theta) = -\cot \theta \)).
The value of \( \cot 30^\circ \) is \( \sqrt{3} \).
Therefore, \( \cot 330^\circ = -\sqrt{3} \).

(iii) For \( \sec 210^\circ \), we write \( 210^\circ \) as \( 180^\circ + 30^\circ \).
\( \sec 210^\circ = \sec (180^\circ + 30^\circ) \)
\( \implies \sec 210^\circ = -\sec 30^\circ \) (Since \( \sec(180^\circ + \theta) = -\sec \theta \)).
The value of \( \sec 30^\circ \) is \( \frac{2}{\sqrt{3}} \).
Therefore, \( \sec 210^\circ = -\frac{2}{\sqrt{3}} \).

(iv) For \( \cos 315^\circ \), we write \( 315^\circ \) as \( 360^\circ - 45^\circ \).
\( \cos 315^\circ = \cos (360^\circ - 45^\circ) \)
\( \implies \cos 315^\circ = \cos 45^\circ \) (Since \( \cos(360^\circ - \theta) = \cos \theta \)).
The value of \( \cos 45^\circ \) is \( \frac{1}{\sqrt{2}} \).

(v) For \( \text{cosec } 675^\circ \), we find the equivalent angle by subtracting multiples of \( 360^\circ \).
\( 675^\circ = 360^\circ + 315^\circ \).
So, \( \text{cosec } 675^\circ = \text{cosec } (360^\circ + 315^\circ) = \text{cosec } 315^\circ \).
Then, write \( 315^\circ \) as \( 360^\circ - 45^\circ \).
\( \text{cosec } 315^\circ = \text{cosec } (360^\circ - 45^\circ) \)
\( \implies \text{cosec } 315^\circ = -\text{cosec } 45^\circ \) (Since \( \text{cosec}(360^\circ - \theta) = -\text{cosec } \theta \)).
The value of \( \text{cosec } 45^\circ \) is \( \sqrt{2} \).
Therefore, \( \text{cosec } 675^\circ = -\sqrt{2} \).

(vi) For \( \cos 855^\circ \), we find the equivalent angle.
\( 855^\circ = 2 \times 360^\circ + 135^\circ = 720^\circ + 135^\circ \).
So, \( \cos 855^\circ = \cos (720^\circ + 135^\circ) = \cos 135^\circ \).
Then, write \( 135^\circ \) as \( 180^\circ - 45^\circ \).
\( \cos 135^\circ = \cos (180^\circ - 45^\circ) \)
\( \implies \cos 135^\circ = -\cos 45^\circ \) (Since \( \cos(180^\circ - \theta) = -\cos \theta \)).
The value of \( \cos 45^\circ \) is \( \frac{1}{\sqrt{2}} \).
Therefore, \( \cos 855^\circ = -\frac{1}{\sqrt{2}} \).

(vii) For \( \sin 4530^\circ \), we find the equivalent angle.
\( 4530^\circ = 12 \times 360^\circ + 210^\circ = 4320^\circ + 210^\circ \).
So, \( \sin 4530^\circ = \sin (12 \times 360^\circ + 210^\circ) = \sin 210^\circ \) (Since \( \sin(360^\circ \times n + \theta) = \sin \theta \)).
Then, write \( 210^\circ \) as \( 180^\circ + 30^\circ \).
\( \sin 210^\circ = \sin (180^\circ + 30^\circ) \)
\( \implies \sin 210^\circ = -\sin 30^\circ \) (Since \( \sin(180^\circ + \theta) = -\sin \theta \)).
The value of \( \sin 30^\circ \) is \( \frac{1}{2} \).
Therefore, \( \sin 4530^\circ = -\frac{1}{2} \).

(viii) For \( \sec \frac{15\pi}{4} \), we convert the angle to a simpler form.
\( \frac{15\pi}{4} = \frac{16\pi - \pi}{4} = 4\pi - \frac{\pi}{4} \).
So, \( \sec \frac{15\pi}{4} = \sec \left(4\pi - \frac{\pi}{4}\right) \).
Since \( \sec(2n\pi - \theta) = \sec(-\theta) = \sec \theta \), we have
\( \sec \left(4\pi - \frac{\pi}{4}\right) = \sec \left(-\frac{\pi}{4}\right) \)
\( \implies \sec \left(-\frac{\pi}{4}\right) = \sec \frac{\pi}{4} \).
The value of \( \sec \frac{\pi}{4} \) (which is \( \sec 45^\circ \)) is \( \sqrt{2} \).
In simple words: To find these values, we break down large angles into smaller, known angles using \( 180^\circ \) or \( 360^\circ \) rules. We always remember the basic trigonometric values for angles like 30, 45, and 60 degrees.

🎯 Exam Tip: Practice converting angles into reference angles in the first quadrant and applying the correct sign based on the original angle's quadrant. Memorizing common trigonometric values for \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \) is essential for these types of questions.

Question 3. With the help of tables, find the values, correct to four places of decimals, of each of the following:
(i) cos 116°
(ii) sin 267°
(iii) sin (- 263°)
(iv) cos 280° 10′
(v) tan (2015° 24').
Answer:
(i) To find \( \cos 116^\circ \), we write \( 116^\circ \) as \( 180^\circ - 64^\circ \).
\( \cos 116^\circ = \cos (180^\circ - 64^\circ) \)
\( \implies \cos 116^\circ = -\cos 64^\circ \).
Using trigonometric tables, \( \cos 64^\circ \approx 0.4384 \).
Therefore, \( \cos 116^\circ \approx -0.4384 \).

(ii) To find \( \sin 267^\circ \), we write \( 267^\circ \) as \( 270^\circ - 3^\circ \).
\( \sin 267^\circ = \sin (270^\circ - 3^\circ) \)
\( \implies \sin 267^\circ = -\cos 3^\circ \) (Since \( \sin(270^\circ - \theta) = -\cos \theta \)).
Using trigonometric tables, \( \cos 3^\circ \approx 0.9986 \).
Therefore, \( \sin 267^\circ \approx -0.9986 \).

(iii) To find \( \sin (-263^\circ) \), first use \( \sin (-\theta) = -\sin \theta \).
\( \sin (-263^\circ) = -\sin (263^\circ) \).
Then, write \( 263^\circ \) as \( 270^\circ - 7^\circ \).
\( \sin 263^\circ = \sin (270^\circ - 7^\circ) \)
\( \implies \sin 263^\circ = -\cos 7^\circ \) (Since \( \sin(270^\circ - \theta) = -\cos \theta \)).
So, \( \sin (-263^\circ) = -(-\cos 7^\circ) = \cos 7^\circ \).
Using trigonometric tables, \( \cos 7^\circ \approx 0.9925 \).
Therefore, \( \sin (-263^\circ) \approx 0.9925 \).

(iv) To find \( \cos 280^\circ 10' \), we write \( 280^\circ 10' \) as \( 270^\circ + 10^\circ 10' \).
\( \cos 280^\circ 10' = \cos (270^\circ + 10^\circ 10') \)
\( \implies \cos 280^\circ 10' = \sin (10^\circ 10') \) (Since \( \cos(270^\circ + \theta) = \sin \theta \)).
Using trigonometric tables, \( \sin (10^\circ 10') \approx 0.1766 \).
Therefore, \( \cos 280^\circ 10' \approx 0.1766 \).

(v) To find \( \tan (2015^\circ 24') \), we first reduce the angle.
\( 2015^\circ 24' = 5 \times 360^\circ + 215^\circ 24' = 1800^\circ + 215^\circ 24' \).
So, \( \tan (2015^\circ 24') = \tan (215^\circ 24') \) (Since \( \tan(360^\circ \times n + \theta) = \tan \theta \)).
Then, we write \( 215^\circ 24' \) as \( 180^\circ + 35^\circ 24' \).
\( \tan (215^\circ 24') = \tan (180^\circ + 35^\circ 24') \)
\( \implies \tan (215^\circ 24') = \tan (35^\circ 24') \) (Since \( \tan(180^\circ + \theta) = \tan \theta \)).
Using trigonometric tables, \( \tan (35^\circ 24') \approx 0.7107 \).
Therefore, \( \tan (2015^\circ 24') \approx 0.7107 \). It's important to use the correct reduction formulas and reference angles, especially when working with angles in different quadrants, to ensure the sign is correct for the final decimal value.
In simple words: We first change the given angle to a simpler angle using 180 or 360 degree rules. Then, we look up the value of that simpler angle in a trigonometry table. We are careful to make sure the answer has the right positive or negative sign.

🎯 Exam Tip: Always state the transformation of the angle clearly (e.g., \( 180^\circ - \theta \), \( 270^\circ + \theta \)) and explicitly mention the rule for the sign change. This helps in avoiding errors and shows clear working for full marks.

Question 4. Find the value of sin 750° cos 300° + cos 1470° sin (- 1020°).
Answer: We need to evaluate each trigonometric term separately.
First, for \( \sin 750^\circ \):
\( 750^\circ = 2 \times 360^\circ + 30^\circ = 720^\circ + 30^\circ \).
So, \( \sin 750^\circ = \sin (720^\circ + 30^\circ) = \sin 30^\circ = \frac{1}{2} \).

Next, for \( \cos 300^\circ \):
\( 300^\circ = 360^\circ - 60^\circ \).
So, \( \cos 300^\circ = \cos (360^\circ - 60^\circ) = \cos 60^\circ = \frac{1}{2} \).

Now, for \( \cos 1470^\circ \):
\( 1470^\circ = 4 \times 360^\circ + 30^\circ = 1440^\circ + 30^\circ \).
So, \( \cos 1470^\circ = \cos (1440^\circ + 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2} \).

Finally, for \( \sin (-1020^\circ) \):
First, use \( \sin (-\theta) = -\sin \theta \).
\( \sin (-1020^\circ) = -\sin (1020^\circ) \).
Now, for \( \sin 1020^\circ \):
\( 1020^\circ = 2 \times 360^\circ + 300^\circ = 720^\circ + 300^\circ \).
So, \( \sin 1020^\circ = \sin (720^\circ + 300^\circ) = \sin 300^\circ \).
Then, for \( \sin 300^\circ \):
\( 300^\circ = 360^\circ - 60^\circ \).
So, \( \sin 300^\circ = \sin (360^\circ - 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2} \).
Therefore, \( \sin (-1020^\circ) = -(-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} \).

Now substitute these values back into the original expression:
\( \sin 750^\circ \cos 300^\circ + \cos 1470^\circ \sin (-1020^\circ) \)
\( = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{1}{4} + \frac{3}{4} \)
\( = \frac{4}{4} \)
\( = 1 \). Evaluating expressions by breaking them into smaller, manageable parts is a good strategy for complex trigonometric problems.
In simple words: We find the value of each part of the math problem by changing the big angles into smaller, familiar angles. Then, we put all those found values back into the original problem and do the simple addition and multiplication.

🎯 Exam Tip: When dealing with multiple trigonometric functions in one expression, simplify each term individually first, reducing large angles to their principal values and determining the correct sign. Then, substitute these simplified values back into the main expression.

Question 5. Evaluate \( \frac{\cos 3 \theta-2 \cos 4 \theta}{\sin 3 \theta+2 \sin 4 \theta} \), when \( \theta = 150^\circ \).
Answer: We need to substitute \( \theta = 150^\circ \) into the expression and evaluate each trigonometric term.
Given \( \theta = 150^\circ \).

First, calculate \( \cos 3\theta \):
\( \cos 3\theta = \cos (3 \times 150^\circ) = \cos 450^\circ \).
We can write \( 450^\circ = 360^\circ + 90^\circ \).
So, \( \cos 450^\circ = \cos (360^\circ + 90^\circ) = \cos 90^\circ = 0 \).

Next, calculate \( \cos 4\theta \):
\( \cos 4\theta = \cos (4 \times 150^\circ) = \cos 600^\circ \).
We can write \( 600^\circ = 360^\circ + 240^\circ \).
So, \( \cos 600^\circ = \cos (360^\circ + 240^\circ) = \cos 240^\circ \).
Then, \( 240^\circ = 180^\circ + 60^\circ \).
\( \cos 240^\circ = \cos (180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \).

Now, calculate \( \sin 3\theta \):
\( \sin 3\theta = \sin (3 \times 150^\circ) = \sin 450^\circ \).
We can write \( 450^\circ = 360^\circ + 90^\circ \).
So, \( \sin 450^\circ = \sin (360^\circ + 90^\circ) = \sin 90^\circ = 1 \).

Finally, calculate \( \sin 4\theta \):
\( \sin 4\theta = \sin (4 \times 150^\circ) = \sin 600^\circ \).
We can write \( 600^\circ = 360^\circ + 240^\circ \).
So, \( \sin 600^\circ = \sin (360^\circ + 240^\circ) = \sin 240^\circ \).
Then, \( 240^\circ = 180^\circ + 60^\circ \).
\( \sin 240^\circ = \sin (180^\circ + 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2} \).

Now, substitute these values into the given expression:
\[ \frac{\cos 3 \theta-2 \cos 4 \theta}{\sin 3 \theta+2 \sin 4 \theta} = \frac{0 - 2\left(-\frac{1}{2}\right)}{1 + 2\left(-\frac{\sqrt{3}}{2}\right)} \]
\[ \implies = \frac{0 + 1}{1 - \sqrt{3}} \]
\[ \implies = \frac{1}{1 - \sqrt{3}} \] To rationalize the denominator, multiply the numerator and denominator by the conjugate \( (1 + \sqrt{3}) \):
\[ = \frac{1}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \]
\[ \implies = \frac{1 + \sqrt{3}}{1^2 - (\sqrt{3})^2} \]
\[ \implies = \frac{1 + \sqrt{3}}{1 - 3} \]
\[ \implies = \frac{1 + \sqrt{3}}{-2} \]
\[ \implies = -\frac{1}{2}(1 + \sqrt{3}) \text{ or } -\frac{1 + \sqrt{3}}{2} \] This result is derived by carefully evaluating each trigonometric term at the given angle, making sure to simplify the angle first.In simple words: First, we replace the letter 'theta' with 150 degrees in all the terms. Then, we find the value of each `cos` and `sin` part. Finally, we put all these values back into the main fraction and solve it.

🎯 Exam Tip: Remember to simplify angles by using the periodicity of trigonometric functions and quadrant rules before calculating their values. When rationalizing the denominator, always multiply by the conjugate to get rid of the square root from the bottom of the fraction.

Question 6. Simplify: \( \frac{\cos (-\theta)}{\sin \left(90^{\circ}+\theta\right)} \).
Answer: We need to simplify the given trigonometric expression.
First, let's simplify the numerator and the denominator using trigonometric identities:
For the numerator, \( \cos (-\theta) \):
Since cosine is an even function, \( \cos (-\theta) = \cos \theta \).
For the denominator, \( \sin (90^\circ + \theta) \):
Using the co-function identity for \( 90^\circ + \theta \), \( \sin (90^\circ + \theta) = \cos \theta \).

Now, substitute these simplified terms back into the expression:
\[ \frac{\cos (-\theta)}{\sin \left(90^{\circ}+\theta\right)} = \frac{\cos \theta}{\cos \theta} \]
\[ \implies = 1 \] This simplification demonstrates how fundamental trigonometric identities can reduce complex expressions to a simple value.In simple words: We know that `cos` of a negative angle is the same as `cos` of the positive angle. We also know that `sin(90 + theta)` is the same as `cos(theta)`. So, the top and bottom of the fraction become the same, which means the whole fraction simplifies to 1.

🎯 Exam Tip: Always recall the basic trigonometric identities: even/odd functions like \( \cos(-\theta) = \cos\theta \) and \( \sin(-\theta) = -\sin\theta \), and co-function identities like \( \sin(90^\circ+\theta) = \cos\theta \) and \( \cos(90^\circ+\theta) = -\sin\theta \). These are key to quick simplification.

Question 7. Simplify: \( \frac{\tan (-\theta)}{\sin \left(540^{\circ}+\theta\right)} \).
Answer: We need to simplify the given trigonometric expression.
First, let's simplify the numerator and the denominator separately:
For the numerator, \( \tan (-\theta) \):
Since tangent is an odd function, \( \tan (-\theta) = -\tan \theta \).
For the denominator, \( \sin (540^\circ + \theta) \):
We can write \( 540^\circ \) as \( 360^\circ + 180^\circ \).
So, \( \sin (540^\circ + \theta) = \sin (360^\circ + 180^\circ + \theta) \).
Using the periodicity of sine, \( \sin (360^\circ + \phi) = \sin \phi \), so \( \sin (360^\circ + (180^\circ + \theta)) = \sin (180^\circ + \theta) \).
Now, using the identity \( \sin (180^\circ + \theta) = -\sin \theta \).
So, \( \sin (540^\circ + \theta) = -\sin \theta \).

Now, substitute these simplified terms back into the expression:
\[ \frac{\tan (-\theta)}{\sin \left(540^{\circ}+\theta\right)} = \frac{-\tan \theta}{-\sin \theta} \]
\[ \implies = \frac{\tan \theta}{\sin \theta} \] We know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substitute this:
\[ = \frac{\frac{\sin \theta}{\cos \theta}}{\sin \theta} \]
\[ \implies = \frac{\sin \theta}{\cos \theta} \times \frac{1}{\sin \theta} \]
\[ \implies = \frac{1}{\cos \theta} \] We also know that \( \frac{1}{\cos \theta} = \sec \theta \).
Therefore, \( \frac{\tan (-\theta)}{\sin \left(540^{\circ}+\theta\right)} = \sec \theta \). This step-by-step simplification highlights the power of trigonometric reduction formulas in transforming complex expressions.
In simple words: First, we change `tan` of a negative angle to negative `tan` of the positive angle. Then, we simplify `sin(540 + theta)` by breaking 540 degrees into `360 + 180` degrees. This becomes negative `sin(theta)`. After that, we put these new parts into the fraction and use the rule `tan(theta) = sin(theta)/cos(theta)` to simplify it all the way to `sec(theta)`.

🎯 Exam Tip: When an angle is greater than \( 360^\circ \), subtract multiples of \( 360^\circ \) to bring it to a simpler angle within \( 0^\circ \) to \( 360^\circ \). Then apply reduction formulas based on the quadrant. Always convert `tan` to `sin/cos` if it helps simplify further.

Question 8. Simplify: \( \frac{\sin \left(90^{\circ}-\theta\right) \sec \left(180^{\circ}-\theta\right) \sin (-\theta)}{\sin \left(180^{\circ}+\theta\right) \cot \left(360^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}+\theta\right)} \).
Answer: We will simplify each trigonometric term in the expression.
For the numerator:
1. \( \sin (90^\circ - \theta) = \cos \theta \)
2. \( \sec (180^\circ - \theta) = -\sec \theta \)
3. \( \sin (-\theta) = -\sin \theta \)
So, the numerator becomes \( (\cos \theta) \times (-\sec \theta) \times (-\sin \theta) = \cos \theta \times \sec \theta \times \sin \theta \).
Since \( \sec \theta = \frac{1}{\cos \theta} \), then \( \cos \theta \times \sec \theta = 1 \).
Thus, the numerator simplifies to \( 1 \times \sin \theta = \sin \theta \).

For the denominator:
1. \( \sin (180^\circ + \theta) = -\sin \theta \)
2. \( \cot (360^\circ - \theta) = -\cot \theta \)
3. \( \operatorname{cosec} (90^\circ + \theta) = \sec \theta \)
So, the denominator becomes \( (-\sin \theta) \times (-\cot \theta) \times (\sec \theta) = \sin \theta \times \cot \theta \times \sec \theta \).
We know that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \).
Substitute these into the denominator expression:
\( \sin \theta \times \frac{\cos \theta}{\sin \theta} \times \frac{1}{\cos \theta} \).
The \( \sin \theta \) terms cancel out, and the \( \cos \theta \) terms cancel out.
Thus, the denominator simplifies to \( 1 \).

Now, substitute the simplified numerator and denominator back into the main expression:
\[ \frac{\sin \left(90^{\circ}-\theta\right) \sec \left(180^{\circ}-\theta\right) \sin (-\theta)}{\sin \left(180^{\circ}+\theta\right) \cot \left(360^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}+\theta\right)} = \frac{\sin \theta}{1} \]
\[ \implies = \sin \theta \] This complete breakdown of each term is crucial for simplifying complex trigonometric expressions efficiently.
In simple words: We simplify each part of the fraction separately. We use rules like `sin(90-theta) = cos(theta)`, `sec(180-theta) = -sec(theta)`, and `sin(-theta) = -sin(theta)`. Then, we do the same for the bottom part. After simplifying all the terms, we put them back into the fraction. We use the fact that `sec(theta)` is `1/cos(theta)` and `cot(theta)` is `cos(theta)/sin(theta)` to cancel out terms and get the final simple answer.

🎯 Exam Tip: When simplifying complex fractions involving trigonometric functions, break down the problem by simplifying each individual term in the numerator and denominator using reduction formulas and reciprocal identities. Combine terms carefully, cancelling common factors to reach the simplest form.

Question 9. Evaluate \( \frac{\sin 150^{\circ}-5 \cos 300^{\circ}+7 \tan 225^{\circ}}{\tan 135^{\circ}+3 \sin 210^{\circ}} \).
Answer: We need to evaluate each trigonometric term:
1. \( \sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2} \).
2. \( \cos 300^\circ = \cos (360^\circ - 60^\circ) = \cos 60^\circ = \frac{1}{2} \).
3. \( \tan 225^\circ = \tan (180^\circ + 45^\circ) = \tan 45^\circ = 1 \).
4. \( \tan 135^\circ = \tan (180^\circ - 45^\circ) = -\tan 45^\circ = -1 \).
5. \( \sin 210^\circ = \sin (180^\circ + 30^\circ) = -\sin 30^\circ = -\frac{1}{2} \).

Now, substitute these values into the expression:
\[ \frac{\sin 150^{\circ}-5 \cos 300^{\circ}+7 \tan 225^{\circ}}{\tan 135^{\circ}+3 \sin 210^{\circ}} \]
\[ = \frac{\frac{1}{2} - 5\left(\frac{1}{2}\right) + 7(1)}{-1 + 3\left(-\frac{1}{2}\right)} \]
\[ = \frac{\frac{1}{2} - \frac{5}{2} + 7}{-1 - \frac{3}{2}} \]
\[ = \frac{\frac{1-5}{2} + 7}{\frac{-2-3}{2}} \]
\[ = \frac{-\frac{4}{2} + 7}{-\frac{5}{2}} \]
\[ = \frac{-2 + 7}{-\frac{5}{2}} \]
\[ = \frac{5}{-\frac{5}{2}} \]
\[ = 5 \times \left(-\frac{2}{5}\right) \]
\[ = -2 \] This problem shows how important it is to correctly evaluate each term before combining them to avoid errors.
In simple words: We first find the value of each `sin`, `cos`, and `tan` part in the problem. Then, we put all these values back into the big fraction. After that, we do the addition, subtraction, and multiplication in the top and bottom parts separately. Finally, we divide the top result by the bottom result.

🎯 Exam Tip: Carefully evaluate each trigonometric term, ensuring the correct sign based on the quadrant. Be meticulous with fraction arithmetic, especially when combining terms in the numerator and denominator before performing the final division.

Question 10. If \( \sin (7\Phi + 9^\circ) = \cos 2\Phi \), find a value of \( \Phi \).
Answer: We are given the equation \( \sin (7\Phi + 9^\circ) = \cos 2\Phi \).
We know that \( \cos x = \sin (90^\circ - x) \). So, we can rewrite \( \cos 2\Phi \) as \( \sin (90^\circ - 2\Phi) \).
Substitute this into the equation:
\( \sin (7\Phi + 9^\circ) = \sin (90^\circ - 2\Phi) \).
This implies that the angles must be equal, or related by the property of sine functions where \( \sin A = \sin B \implies A = n \times 180^\circ + (-1)^n B \). For the simplest solution, we can consider the angles equal if they are acute or within the same range.
\( 7\Phi + 9^\circ = 90^\circ - 2\Phi \).
Now, we solve for \( \Phi \):
Add \( 2\Phi \) to both sides:
\( 7\Phi + 2\Phi + 9^\circ = 90^\circ \)
\( 9\Phi + 9^\circ = 90^\circ \)
Subtract \( 9^\circ \) from both sides:
\( 9\Phi = 90^\circ - 9^\circ \)
\( 9\Phi = 81^\circ \)
Divide by 9:
\( \Phi = \frac{81^\circ}{9} \)
\( \implies \Phi = 9^\circ \). This method relies on converting one trigonometric function into another, which is a common strategy in solving such equations.In simple words: We change `cos(2Phi)` into `sin(90 - 2Phi)` because `cos(x)` is the same as `sin(90 - x)`. Then, we set the angles inside the `sin` functions equal to each other. We solve this simple equation to find the value of `Phi`.

🎯 Exam Tip: When solving trigonometric equations involving different functions (e.g., sine and cosine), convert one function into the other using co-function identities like \( \sin x = \cos (90^\circ - x) \) or \( \cos x = \sin (90^\circ - x) \) to make the equation solvable.

Question 11. Find the values of \( \theta \) lying between \( 0^\circ \) and \( 360^\circ \) when
(i) \( \sin \theta = \frac { 1 }{ 2 } \)
(ii) \( \tan \theta = -1 \)
(iii) \( \sec \theta = -2 \)
(iv) \( \sin \theta = \sin 21^\circ \)
(v) \( \tan \theta = -2.0145 \)
(vi) \( \sin \theta = \cos 317^\circ \)
(vii) \( \cos \theta = \sin 285^\circ \)
Answer: We need to find the values of \( \theta \) within the given range for each condition.
(i) Given \( \sin \theta = \frac{1}{2} \).
We know that \( \sin 30^\circ = \frac{1}{2} \). Sine is positive in the first and second quadrants.
First quadrant solution: \( \theta = 30^\circ \).
Second quadrant solution: \( \theta = 180^\circ - 30^\circ = 150^\circ \).
So, \( \theta = 30^\circ \) or \( 150^\circ \).

(ii) Given \( \tan \theta = -1 \).
We know that \( \tan 45^\circ = 1 \). Tangent is negative in the second and fourth quadrants.
Second quadrant solution: \( \theta = 180^\circ - 45^\circ = 135^\circ \).
Fourth quadrant solution: \( \theta = 360^\circ - 45^\circ = 315^\circ \).
So, \( \theta = 135^\circ \) or \( 315^\circ \).

(iii) Given \( \sec \theta = -2 \).
This means \( \cos \theta = \frac{1}{\sec \theta} = \frac{1}{-2} = -\frac{1}{2} \).
We know that \( \cos 60^\circ = \frac{1}{2} \). Cosine is negative in the second and third quadrants.
Second quadrant solution: \( \theta = 180^\circ - 60^\circ = 120^\circ \).
Third quadrant solution: \( \theta = 180^\circ + 60^\circ = 240^\circ \).
So, \( \theta = 120^\circ \) or \( 240^\circ \).

(iv) Given \( \sin \theta = \sin 21^\circ \).
Since \( \sin \theta = \sin \alpha \), then \( \theta = \alpha \) or \( \theta = 180^\circ - \alpha \).
So, \( \theta = 21^\circ \) or \( \theta = 180^\circ - 21^\circ = 159^\circ \).
Thus, \( \theta = 21^\circ \) or \( 159^\circ \).

(v) Given \( \tan \theta = -2.0145 \).
First, find the reference angle for \( \tan \alpha = 2.0145 \). Using a calculator or tables, \( \alpha \approx 63^\circ 36' \).
Since \( \tan \theta \) is negative, \( \theta \) lies in the second or fourth quadrants.
Second quadrant solution: \( \theta = 180^\circ - 63^\circ 36' = 116^\circ 24' \).
Fourth quadrant solution: \( \theta = 360^\circ - 63^\circ 36' = 296^\circ 24' \).
So, \( \theta = 116^\circ 24' \) or \( 296^\circ 24' \).

(vi) Given \( \sin \theta = \cos 317^\circ \).
First, simplify \( \cos 317^\circ \). We can write \( 317^\circ = 270^\circ + 47^\circ \).
\( \cos 317^\circ = \cos (270^\circ + 47^\circ) = \sin 47^\circ \).
So, the equation becomes \( \sin \theta = \sin 47^\circ \).
This implies \( \theta = 47^\circ \) or \( \theta = 180^\circ - 47^\circ = 133^\circ \).
So, \( \theta = 47^\circ \) or \( 133^\circ \).

(vii) Given \( \cos \theta = \sin 285^\circ \).
First, simplify \( \sin 285^\circ \). We can write \( 285^\circ = 270^\circ + 15^\circ \).
\( \sin 285^\circ = \sin (270^\circ + 15^\circ) = -\cos 15^\circ \).
So, the equation becomes \( \cos \theta = -\cos 15^\circ \).
For \( \cos \theta = -\cos \alpha \), \( \theta = 180^\circ - \alpha \) or \( \theta = 180^\circ + \alpha \).
So, \( \theta = 180^\circ - 15^\circ = 165^\circ \) or \( \theta = 180^\circ + 15^\circ = 195^\circ \).
So, \( \theta = 165^\circ \) or \( 195^\circ \). In solving these trigonometric equations, remembering the signs of functions in each quadrant and the relationships between angles (like \( 180^\circ - \theta \) or \( 360^\circ - \theta \)) is critical.
In simple words: For each problem, we find the basic angle first. Then, we use the rules about which "quarter" of the circle (quadrant) the angle is in to find all possible angles between 0 and 360 degrees. For example, if `sin` is positive, the angle can be in the first or second quarter.

🎯 Exam Tip: Always identify the reference angle and then determine the quadrants where the trigonometric function has the specified sign. This will give you all possible solutions within the given range. For problems involving mixed functions like \( \sin \theta = \cos \alpha \), convert them to the same function first.

Question 12. If \( 0^\circ < \theta < 90^\circ \) and \( \cos \theta = \frac { 4 }{ 5 } \), find the values of
(i) cos (90° + θ)
(ii) cosec (180° + θ)
(iii) tan (360° - θ)
(iv) sin (270° - θ).
Answer: Given \( \cos \theta = \frac{4}{5} \) and \( 0^\circ < \theta < 90^\circ \) (which means \( \theta \) is in the first quadrant).
First, let's find the values of \( \sin \theta \) and \( \tan \theta \).
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{25-16}{25} = \frac{9}{25} \).
Since \( \theta \) is in the first quadrant, \( \sin \theta \) is positive.
So, \( \sin \theta = \sqrt{\frac{9}{25}} = \frac{3}{5} \).

Now, we can find \( \tan \theta \):
\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4} \).

Now, let's evaluate each part:
(i) \( \cos (90^\circ + \theta) \):
Using the identity \( \cos (90^\circ + \theta) = -\sin \theta \).
Substitute \( \sin \theta = \frac{3}{5} \).
So, \( \cos (90^\circ + \theta) = -\frac{3}{5} \).

(ii) \( \operatorname{cosec} (180^\circ + \theta) \):
Using the identity \( \operatorname{cosec} (180^\circ + \theta) = -\operatorname{cosec} \theta \).
We know \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{1}{3/5} = \frac{5}{3} \).
So, \( \operatorname{cosec} (180^\circ + \theta) = -\frac{5}{3} \).

(iii) \( \tan (360^\circ - \theta) \):
Using the identity \( \tan (360^\circ - \theta) = -\tan \theta \).
Substitute \( \tan \theta = \frac{3}{4} \).
So, \( \tan (360^\circ - \theta) = -\frac{3}{4} \).

(iv) \( \sin (270^\circ - \theta) \):
Using the identity \( \sin (270^\circ - \theta) = -\cos \theta \).
Substitute \( \cos \theta = \frac{4}{5} \).
So, \( \sin (270^\circ - \theta) = -\frac{4}{5} \). This problem emphasizes understanding fundamental trigonometric identities and their application across different quadrants.
In simple words: First, we use `cos(theta) = 4/5` to find `sin(theta)` and `tan(theta)`, remembering that `theta` is a small angle in the first quarter. Then, for each part, we use special rules to change the expression (like `cos(90+theta)`) into `sin(theta)` or `cos(theta)`, making sure to get the right positive or negative sign. Finally, we put in the values we found for `sin(theta)` and `cos(theta)`.

🎯 Exam Tip: Always start by finding all six trigonometric ratios for the given base angle if not provided. Then, for each required expression, apply the correct angle reduction formula and the appropriate sign based on the quadrant of the transformed angle. This ensures accuracy.

Question 13. Find six angles for which \( \sin \theta = - \frac{\sqrt{3}}{2} \).
Answer: We need to find six angles \( \theta \) such that \( \sin \theta = -\frac{\sqrt{3}}{2} \).
First, we know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Since \( \sin \theta \) is negative, \( \theta \) must lie in the third or fourth quadrants.

**Angles in the range \( 0^\circ \) to \( 360^\circ \):**
1. Third quadrant: \( \theta = 180^\circ + 60^\circ = 240^\circ \).
2. Fourth quadrant: \( \theta = 360^\circ - 60^\circ = 300^\circ \).

**Angles by adding \( 360^\circ \) (one full rotation):**
3. Add \( 360^\circ \) to \( 240^\circ \): \( \theta = 240^\circ + 360^\circ = 600^\circ \).
4. Add \( 360^\circ \) to \( 300^\circ \): \( \theta = 300^\circ + 360^\circ = 660^\circ \).

**Angles by adding \( 720^\circ \) (two full rotations):**
5. Add \( 720^\circ \) to \( 240^\circ \): \( \theta = 240^\circ + 720^\circ = 960^\circ \).
6. Add \( 720^\circ \) to \( 300^\circ \): \( \theta = 300^\circ + 720^\circ = 1020^\circ \).

Therefore, six angles for which \( \sin \theta = -\frac{\sqrt{3}}{2} \) are \( 240^\circ, 300^\circ, 600^\circ, 660^\circ, 960^\circ, 1020^\circ \). Finding multiple solutions for trigonometric equations often involves adding or subtracting full rotations to the principal values.
In simple words: First, we find the basic angle where `sin` gives `sqrt(3)/2`. Since we need negative `sin`, we look in the third and fourth quarters of the circle. This gives us two main angles. To find more angles, we simply add 360 degrees (a full circle) to these two angles again and again.

🎯 Exam Tip: To find multiple solutions for trigonometric equations, first find the principal values in the \( 0^\circ \)- \( 360^\circ \) range based on the function's sign and reference angle. Then, generate additional solutions by adding or subtracting multiples of \( 360^\circ \) (or \( 2\pi \) radians) to these principal values.

Question 14. Find all the angles between \( 0^\circ \) and \( 720^\circ \) whose tangent is \( - \frac{1}{\sqrt{3}} \).
Answer: We need to find all angles \( \theta \) between \( 0^\circ \) and \( 720^\circ \) such that \( \tan \theta = -\frac{1}{\sqrt{3}} \).
First, we know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Since \( \tan \theta \) is negative, \( \theta \) must lie in the second or fourth quadrants.

**Angles in the range \( 0^\circ \) to \( 360^\circ \):**
1. Second quadrant: \( \theta = 180^\circ - 30^\circ = 150^\circ \).
2. Fourth quadrant: \( \theta = 360^\circ - 30^\circ = 330^\circ \).

**Angles in the range \( 360^\circ \) to \( 720^\circ \) (by adding \( 360^\circ \) to the previous solutions):**
3. Add \( 360^\circ \) to \( 150^\circ \): \( \theta = 150^\circ + 360^\circ = 510^\circ \).
4. Add \( 360^\circ \) to \( 330^\circ \): \( \theta = 330^\circ + 360^\circ = 690^\circ \).

All these angles are within the specified range of \( 0^\circ \) to \( 720^\circ \).
Therefore, the angles are \( 150^\circ, 330^\circ, 510^\circ, 690^\circ \). It is important to systematically add full rotations to cover the entire given range for solutions.
In simple words: We first find the basic angle where `tan` gives `1/sqrt(3)`. Because we need negative `tan`, we look in the second and fourth quarters of the circle. This gives us two angles between 0 and 360 degrees. To find angles up to 720 degrees, we simply add 360 degrees to each of these two angles.

🎯 Exam Tip: When finding angles within a specific range, always start by identifying the reference angle and the quadrants where the function has the required sign. Then, systematically add multiples of \( 360^\circ \) to find all solutions within the given interval.

Question 15. Find the values of \( \theta \) between \( 0^\circ \) and \( 360^\circ \) which satisfy the equations
(i) \( \sin^2 \theta = \frac { 3 }{ 4 } \)
(ii) \( \cos 3\theta = \frac { 1 }{ 2 } \).
Answer: We need to solve each equation for \( \theta \) between \( 0^\circ \) and \( 360^\circ \).
(i) Given \( \sin^2 \theta = \frac{3}{4} \).
Taking the square root of both sides, we get \( \sin \theta = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \).
**Case 1: \( \sin \theta = \frac{\sqrt{3}}{2} \)**
We know \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). Sine is positive in the first and second quadrants.
First quadrant: \( \theta = 60^\circ \).
Second quadrant: \( \theta = 180^\circ - 60^\circ = 120^\circ \).
**Case 2: \( \sin \theta = -\frac{\sqrt{3}}{2} \)**
Sine is negative in the third and fourth quadrants.
Third quadrant: \( \theta = 180^\circ + 60^\circ = 240^\circ \).
Fourth quadrant: \( \theta = 360^\circ - 60^\circ = 300^\circ \).
Therefore, the values of \( \theta \) are \( 60^\circ, 120^\circ, 240^\circ, 300^\circ \).

(ii) Given \( \cos 3\theta = \frac{1}{2} \).
We know \( \cos 60^\circ = \frac{1}{2} \). Cosine is positive in the first and fourth quadrants.
For \( 3\theta \), the general solution for \( \cos x = \cos \alpha \) is \( x = 2n\pi \pm \alpha \). So, \( 3\theta = 2n \times 180^\circ \pm 60^\circ \).
Thus, \( \theta = \frac{2n \times 180^\circ \pm 60^\circ}{3} = 2n \times 60^\circ \pm 20^\circ \).

Let's find values for different integer values of \( n \):
**For \( n = 0 \):**
\( \theta = 2(0) \times 60^\circ \pm 20^\circ = \pm 20^\circ \).
Since \( \theta \) must be between \( 0^\circ \) and \( 360^\circ \), \( \theta = 20^\circ \).

**For \( n = 1 \):**
\( \theta = 2(1) \times 60^\circ \pm 20^\circ = 120^\circ \pm 20^\circ \).
So, \( \theta = 120^\circ + 20^\circ = 140^\circ \) or \( \theta = 120^\circ - 20^\circ = 100^\circ \).

**For \( n = 2 \):**
\( \theta = 2(2) \times 60^\circ \pm 20^\circ = 240^\circ \pm 20^\circ \).
So, \( \theta = 240^\circ + 20^\circ = 260^\circ \) or \( \theta = 240^\circ - 20^\circ = 220^\circ \).

**For \( n = 3 \):**
\( \theta = 2(3) \times 60^\circ \pm 20^\circ = 360^\circ \pm 20^\circ \).
So, \( \theta = 360^\circ - 20^\circ = 340^\circ \). (The value \( 360^\circ + 20^\circ = 380^\circ \) is outside the range).
Therefore, the values of \( \theta \) are \( 20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ, 340^\circ \). When solving trigonometric equations involving multiples of \( \theta \) (like \( 3\theta \)), remember to find all solutions within the range for the transformed angle (e.g., \( 3\theta \) up to \( 3 \times 360^\circ \)) and then divide by the multiple.
In simple words: For the first part, we find the angles where `sin(theta)` is either `sqrt(3)/2` or `-sqrt(3)/2`. This gives us four angles. For the second part, we first find the basic angles for `cos(3theta) = 1/2`. Then, we divide these angles by 3, making sure to find all possible angles within 0 to 360 degrees by considering multiple rotations.

🎯 Exam Tip: When solving equations like \( \sin^2 \theta = k \), remember to consider both positive and negative square roots. For equations like \( \cos n\theta = k \), find all general solutions for \( n\theta \) by considering multiple rotations up to \( n \times 360^\circ \), then divide by \( n \) to find \( \theta \).

Question 16. If \( \tan \theta = 0.4 \), when \( \theta \) lies between \( 0^\circ \) and \( 360^\circ \), write down the possible values of \( \theta \) and \( \sin \theta \).
Answer: Given \( \tan \theta = 0.4 \).
Since \( \tan \theta \) is positive, \( \theta \) lies in the first or third quadrants.
First, find the reference angle where \( \tan \alpha = 0.4 \). Using a calculator or tables, \( \alpha \approx 21^\circ 48' \).
**Possible values of \( \theta \):**
1. First quadrant: \( \theta = 21^\circ 48' \).
2. Third quadrant: \( \theta = 180^\circ + 21^\circ 48' = 201^\circ 48' \).
So, the possible values for \( \theta \) are \( 21^\circ 48' \) and \( 201^\circ 48' \).

Now, we need to find \( \sin \theta \) for these values.
We know that \( \sin \theta = \frac{\tan \theta}{\pm \sqrt{1 + \tan^2 \theta}} \).
Alternatively, we can use \( \tan \theta = 0.4 = \frac{4}{10} = \frac{2}{5} \).
Consider a right-angled triangle with opposite side 2 and adjacent side 5. The hypotenuse would be \( \sqrt{2^2 + 5^2} = \sqrt{4+25} = \sqrt{29} \).
So, \( \sin \alpha = \frac{2}{\sqrt{29}} \).

**For \( \theta = 21^\circ 48' \) (First quadrant):**
\( \sin \theta \) is positive.
\( \sin (21^\circ 48') = \frac{2}{\sqrt{29}} \approx \frac{2}{5.385} \approx 0.3714 \).

**For \( \theta = 201^\circ 48' \) (Third quadrant):**
\( \sin \theta \) is negative.
\( \sin (201^\circ 48') = \sin (180^\circ + 21^\circ 48') = -\sin (21^\circ 48') \).
So, \( \sin (201^\circ 48') \approx -0.3714 \). This problem combines finding angles from a given ratio and then finding another ratio for those angles, requiring attention to signs in different quadrants.
In simple words: First, we use `tan(theta) = 0.4` to find the basic angle. Since `tan` is positive, we find one angle in the first quarter and another in the third quarter of the circle. Then, for each of these angles, we find its `sin` value. Remember that `sin` is positive in the first quarter and negative in the third quarter.

🎯 Exam Tip: When given a trigonometric ratio and asked for other ratios or angles, always determine the quadrant(s) of the angle first. This is crucial for applying the correct sign to the calculated values and for finding all possible angles within the specified range.

Question 17. If \( \cos x^\circ = \sin 200^\circ \), find the possible values of \( x \) between \( -180^\circ \) and \( 360^\circ \).
Answer: We are given \( \cos x^\circ = \sin 200^\circ \).
First, simplify \( \sin 200^\circ \). We can write \( 200^\circ = 270^\circ - 70^\circ \).
So, \( \sin 200^\circ = \sin (270^\circ - 70^\circ) = -\cos 70^\circ \).
Now, the equation becomes \( \cos x^\circ = -\cos 70^\circ \).
For \( \cos A = -\cos B \), the general solutions for \( A \) are \( 180^\circ - B \) and \( 180^\circ + B \).

So, possible values for \( x \) are:
1. \( x = 180^\circ - 70^\circ = 110^\circ \).
2. \( x = 180^\circ + 70^\circ = 250^\circ \).

We also need to consider the negative range for \( x \), i.e., between \( -180^\circ \) and \( 0^\circ \).
Since \( \cos \theta = \cos (-\theta) \), we can also have \( x = -110^\circ \).
Let's check if \( x = -110^\circ \) works: \( \cos (-110^\circ) = \cos 110^\circ = -\cos 70^\circ \), which is true.
The value \( -250^\circ \) (from \( -(180^\circ + 70^\circ) \)) is outside the range.

Therefore, the possible values of \( x \) between \( -180^\circ \) and \( 360^\circ \) are \( -110^\circ, 110^\circ, 250^\circ \). This type of problem highlights the importance of using reduction formulas and the periodicity of cosine to find all solutions within a given range, including negative angles.
In simple words: First, we change `sin(200)` into `cos(-70)` using angle rules. So the problem becomes `cos(x) = -cos(70)`. Then we find the angles for `x` where `cos(x)` is negative `cos(70)`. These angles are `180-70`, `180+70`, and also `-110` because `cos(-angle)` is the same as `cos(angle)`. We pick only the angles within -180 and 360 degrees.

🎯 Exam Tip: When the range includes negative angles, always remember that \( \cos (-\theta) = \cos \theta \). This often provides additional solutions by taking the negative of the positive angles found. Also, ensure all transformations of functions and angles are done correctly, including signs based on quadrants.

Question 18. If A, B, C are the angles of a triangle, prove that \( \cos C = - \cos (A + B) \).
Answer: We are given that A, B, and C are the angles of a triangle.
In a triangle, the sum of the angles is \( 180^\circ \).
So, \( A + B + C = 180^\circ \).

We want to prove \( \cos C = -\cos (A + B) \).
From the sum of angles, we can write \( C = 180^\circ - (A + B) \).

Now, take the cosine of both sides:
\( \cos C = \cos (180^\circ - (A + B)) \).

Using the trigonometric identity \( \cos (180^\circ - \theta) = -\cos \theta \), where \( \theta = (A + B) \).
\( \cos (180^\circ - (A + B)) = -\cos (A + B) \).

Therefore, \( \cos C = -\cos (A + B) \). This identity is a fundamental property derived directly from the angle sum property of triangles.
In simple words: In any triangle, the three angles (A, B, C) add up to 180 degrees. So, angle C is 180 degrees minus the sum of angles A and B. When we take `cos` of C, it becomes `cos(180 - (A+B))`. We know a rule that `cos(180 - something)` is equal to `-cos(something)`. Using this rule, `cos(C)` becomes `-cos(A+B)`, which is what we needed to show.

🎯 Exam Tip: For problems involving angles of a triangle, always start by stating the fundamental property: \( A + B + C = 180^\circ \). This is the key to transforming one angle in terms of the others and then applying trigonometric identities.

Question 19. Prove that \( \tan \frac{B+C}{2} = \cot \frac { A }{ 2 } \).
Answer: We are given that A, B, and C are the angles of a triangle.
In a triangle, the sum of the angles is \( 180^\circ \).
So, \( A + B + C = 180^\circ \).

We want to prove \( \tan \frac{B+C}{2} = \cot \frac{A}{2} \).
From the sum of angles, we can write \( B + C = 180^\circ - A \).

Now, divide both sides by 2:
\( \frac{B+C}{2} = \frac{180^\circ - A}{2} = \frac{180^\circ}{2} - \frac{A}{2} = 90^\circ - \frac{A}{2} \).

Now, take the tangent of both sides:
\( \tan \left(\frac{B+C}{2}\right) = \tan \left(90^\circ - \frac{A}{2}\right) \).

Using the co-function identity \( \tan (90^\circ - \theta) = \cot \theta \), where \( \theta = \frac{A}{2} \).
\( \tan \left(90^\circ - \frac{A}{2}\right) = \cot \frac{A}{2} \).

Therefore, \( \tan \frac{B+C}{2} = \cot \frac{A}{2} \). This proof clearly shows the relationship between angles in a triangle and their trigonometric functions.
In simple words: Since A, B, and C are angles of a triangle, they add up to 180 degrees. So, (B+C) is equal to (180 - A). If we divide this by 2, (B+C)/2 becomes (90 - A/2). We know a special rule that `tan(90 - something)` is the same as `cot(something)`. So, `tan((B+C)/2)` becomes `cot(A/2)`, which is what we needed to show.

🎯 Exam Tip: When solving problems involving \( \frac{A+B}{2} \), \( \frac{B+C}{2} \), etc., always use the identity \( A+B+C = 180^\circ \) to express one sum in terms of the other angles, then divide by 2. This step leads directly to the application of co-function identities.

Question 20. Evaluate \( \frac{\tan (B+C)+\tan (C+A)+\tan (A+B)}{\tan (\pi-A)+\tan (2 \pi-B)+\tan (3 \pi-C)} \).
Answer: We are given that A, B, C are angles of a triangle. Thus, \( A+B+C = \pi \) (or \( 180^\circ \)).

Let's simplify the terms in the numerator:
1. \( B+C = \pi - A \)
\( \tan (B+C) = \tan (\pi - A) = -\tan A \).
2. \( C+A = \pi - B \)
\( \tan (C+A) = \tan (\pi - B) = -\tan B \).
3. \( A+B = \pi - C \)
\( \tan (A+B) = \tan (\pi - C) = -\tan C \).
So, the numerator is \( -\tan A - \tan B - \tan C \).

Now, let's simplify the terms in the denominator:
1. \( \tan (\pi - A) = -\tan A \).
2. \( \tan (2\pi - B) \): Since tangent has a period of \( \pi \), \( \tan (2\pi - B) = \tan (-B) = -\tan B \).
3. \( \tan (3\pi - C) \): Since tangent has a period of \( \pi \), \( \tan (3\pi - C) = \tan (\pi - C) = -\tan C \).
So, the denominator is \( -\tan A - \tan B - \tan C \).

Now, substitute the simplified numerator and denominator back into the expression:
\[ \frac{-\tan A - \tan B - \tan C}{-\tan A - \tan B - \tan C} \] Since the numerator and denominator are identical, assuming \( -\tan A - \tan B - \tan C \neq 0 \), the expression simplifies to:
\[ = 1 \] This problem showcases how the property of angles in a triangle, combined with periodicity and reduction formulas, can lead to simple results.In simple words: Since A, B, C are angles of a triangle, they add up to 180 degrees (or pi radians). We use this to change `B+C` to `180-A`, and so on. We also use rules like `tan(180-A) = -tan(A)` and `tan(2*180-B) = -tan(B)`. After changing every part in the top and bottom of the fraction, we find that the top part is exactly the same as the bottom part. So, the whole fraction simplifies to 1.

🎯 Exam Tip: For problems involving \( \tan(\pi - A) \) or \( \tan(2\pi - B) \), always remember the periodicity of tangent (\( \tan(n\pi \pm \theta) \)) and its behavior in different quadrants. This allows for quick simplification of terms.

Question 21. If A, B, C, D are the angles of a quadrilateral, prove that \( \cos\frac { 1 }{ 2 } (A + B) + \cos\frac { 1 }{ 2 } (C + D) = 0 \).
Answer: We are given that A, B, C, D are the angles of a quadrilateral.
In a quadrilateral, the sum of the angles is \( 360^\circ \).
So, \( A + B + C + D = 360^\circ \).

We want to prove \( \cos\frac{1}{2}(A + B) + \cos\frac{1}{2}(C + D) = 0 \).
From the sum of angles, we can write \( A + B = 360^\circ - (C + D) \).

Now, divide both sides by 2:
\( \frac{A+B}{2} = \frac{360^\circ - (C + D)}{2} = \frac{360^\circ}{2} - \frac{C + D}{2} = 180^\circ - \frac{C + D}{2} \).

Let \( X = \frac{C + D}{2} \). Then \( \frac{A+B}{2} = 180^\circ - X \).

Now, substitute this into the expression we want to prove:
\( \cos\frac{1}{2}(A + B) + \cos\frac{1}{2}(C + D) = \cos(180^\circ - X) + \cos X \).

Using the trigonometric identity \( \cos (180^\circ - \theta) = -\cos \theta \), where \( \theta = X \).
\( \cos(180^\circ - X) = -\cos X \).

So, the expression becomes:
\( -\cos X + \cos X \)
\( = 0 \).

Therefore, \( \cos\frac{1}{2}(A + B) + \cos\frac{1}{2}(C + D) = 0 \). This identity directly stems from the angle sum property of quadrilaterals and the cosine reduction formula.
In simple words: In any four-sided shape (quadrilateral), all the angles (A, B, C, D) add up to 360 degrees. So, the sum of `A+B` is `360 - (C+D)`. If we divide this by two, `(A+B)/2` becomes `180 - (C+D)/2`. When we put this into the problem, `cos((A+B)/2)` becomes `cos(180 - (C+D)/2)`. We know that `cos(180 - something)` is `-cos(something)`. So, the problem changes to `-cos((C+D)/2) + cos((C+D)/2)`, which clearly adds up to 0.

🎯 Exam Tip: Similar to triangles, the sum of angles in a quadrilateral is \( 360^\circ \). Utilize this property to relate sums of angles (e.g., \( A+B \) to \( C+D \)). Then, apply the angle reduction formulas for \( 180^\circ - \theta \) or \( 360^\circ - \theta \) to simplify the trigonometric expressions.