GSEB Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.3

Question 1.
(i) \( \frac{\sin 18^{\circ}}{\cos 72^{\circ}} \)
(ii) \( \frac{\tan 26^{\circ}}{\cot 64^{\circ}} \)
(iii) \( \cos 48^{\circ} - \sin 42^{\circ} \)
(iv) \( \operatorname{cosec} 31^{\circ} - \sec 59^{\circ} \)
Answer:
(i) \( \frac{\sin 18^{\circ}}{\cos 72^{\circ}} = \frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)} \)
\( = \frac{\sin 18^{\circ}}{\sin 18^{\circ}} = 1 \) [Since \( \cos (90^{\circ} - \theta) = \sin \theta \)]
(ii) \( \frac{\tan 26^{\circ}}{\cot 64^{\circ}} = \frac{\tan 26^{\circ}}{\cot \left(90^{\circ}-26^{\circ}\right)} \)
\( = \frac{\tan 26^{\circ}}{\tan 26^{\circ}} = 1 \) [Since \( \cot (90^{\circ} - \theta) = \tan \theta \)]
(iii) \( \cos 48^{\circ} - \sin 42^{\circ} \)
\( = \cos (90^{\circ} - 42^{\circ}) - \sin 42^{\circ} \)
\( = \sin 42^{\circ} - \sin 42^{\circ} = 0 \) [Since \( \cos (90^{\circ} - \theta) = \sin \theta \)]
(iv) \( \operatorname{cosec} 31^{\circ} - \sec 59^{\circ} \)
\( = \operatorname{cosec} (90^{\circ} - 59^{\circ}) - \sec 59^{\circ} \)
\( = \sec 59^{\circ} - \sec 59^{\circ} = 0 \) [Since \( \operatorname{cosec} (90^{\circ} - \theta) = \sec \theta \)]
In simple words: For each problem, we use trigonometric identities that relate angles which add up to 90 degrees. This helps us change one part of the expression to match the other, making them cancel out or simplify to 1.

Exam Tip: Remember the complementary angle identities: \( \sin(90^\circ - \theta) = \cos \theta \), \( \cos(90^\circ - \theta) = \sin \theta \), \( \tan(90^\circ - \theta) = \cot \theta \), \( \cot(90^\circ - \theta) = \tan \theta \), \( \sec(90^\circ - \theta) = \operatorname{cosec} \theta \), and \( \operatorname{cosec}(90^\circ - \theta) = \sec \theta \).

Question 2. Show that
(i) \( \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1 \)
(ii) \( \cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0 \)
Answer:
(i) LHS \( = \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} \)
\( = \tan 48^{\circ} \tan 42^{\circ} \tan 23^{\circ} \tan 67^{\circ} \)
\( = \tan (90^{\circ} - 42^{\circ}) \tan 42^{\circ} \tan (90^{\circ} - 67^{\circ}) \tan 67^{\circ} \)
\( = \cot 42^{\circ} \tan 42^{\circ} \cot 67^{\circ} \tan 67^{\circ} \)
\( = \frac{1}{\tan 42^{\circ}} \times \tan 42^{\circ} \times \frac{1}{\tan 67^{\circ}} \times \tan 67^{\circ} \)
\( = 1 = \text{RHS} \)
(ii) LHS \( = \cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} \)
\( = \cos 38^{\circ} \cos (90^{\circ} - 38^{\circ}) - \sin 38^{\circ} \sin (90^{\circ} - 38^{\circ}) \)
\( = \cos 38^{\circ} \sin 38^{\circ} - \sin 38^{\circ} \cos 38^{\circ} = 0 = \text{RHS} \)
In simple words: For the first part, we rearrange the terms and use the identity that \( \tan \theta \cot \theta = 1 \) after changing angles using the \( 90^{\circ} - \theta \) rule. For the second part, we convert one of the angles in each pair using the \( 90^{\circ} - \theta \) rule, making both terms identical so they subtract to zero.

Exam Tip: When proving identities involving products, look for pairs of angles that sum to \( 90^{\circ} \) so you can use complementary identities. For subtraction problems, try to make the terms identical.

Question 3. If \( \tan 2A = \cot (A - 18^{\circ}) \), where 2A is an acute angle, find the value of A. (CBSE 2012)
Answer:
We have
\( \tan 2A = \cot (A - 18^{\circ}) \)
We know that \( \cot \theta = \tan (90^{\circ} - \theta) \). So, \( \cot (A - 18^{\circ}) = \tan (90^{\circ} - (A - 18^{\circ})) \)
\( \tan 2A = \tan (90^{\circ} - A + 18^{\circ}) \)
\( 2A = 90^{\circ} - A + 18^{\circ} \)
\( 2A = 108^{\circ} - A \)
\( 2A + A = 108^{\circ} \)
\( 3A = 108^{\circ} \)
\( A = \frac{108^{\circ}}{3} \)
\( A = 36^{\circ} \)
In simple words: We changed the cotangent into a tangent using the rule \( \cot \theta = \tan (90^{\circ} - \theta) \). Once both sides were tangent functions, we could set the angles equal to each other to solve for A.

Exam Tip: When solving trigonometric equations, try to express both sides of the equation using the same trigonometric ratio (e.g., both as tan, both as sin) to easily equate the angles.

Question 4. If \( \tan A = \cot B \), prove that \( A+ B = 90^{\circ} \).
Answer:
We have
\( \tan A = \cot B \)
We know that \( \cot B = \tan (90^{\circ} - B) \) [Since \( \cot \theta = \tan (90^{\circ} - \theta) \)]
So, \( \tan A = \tan (90^{\circ} - B) \)
Therefore, \( A = 90^{\circ} - B \)
This gives us \( A + B = 90^{\circ} \)
In simple words: Given that the tangent of angle A is equal to the cotangent of angle B, we can change cot B into tan (90 degrees minus B). Since the tangent values are equal, the angles themselves must be equal, showing that A plus B equals 90 degrees.

Exam Tip: For proofs involving complementary angles, always start by converting one of the trigonometric ratios using the \( 90^{\circ} - \theta \) identity to match the other ratio.

Question 5. If \( \sec 4A = \operatorname{cosec} (A - 20^{\circ}) \) where 4A is an acute angle, find the value of A.
Answer:
We have
\( \sec 4A = \operatorname{cosec} (A - 20^{\circ}) \)
We know that \( \sec \theta = \operatorname{cosec} (90^{\circ} - \theta) \). So, \( \sec 4A = \operatorname{cosec} (90^{\circ} - 4A) \)
\( \operatorname{cosec} (90^{\circ} - 4A) = \operatorname{cosec} (A - 20^{\circ}) \)
Therefore, \( 90^{\circ} - 4A = A - 20^{\circ} \)
\( 90^{\circ} + 20^{\circ} = A + 4A \)
\( 110^{\circ} = 5A \)
\( A = \frac{110^{\circ}}{5} \)
\( A = 22^{\circ} \)
In simple words: We use the identity that secant of an angle is the cosecant of (90 degrees minus that angle). This lets us set the angles equal and solve for A by combining like terms.

Exam Tip: When an equation has secant and cosecant, transform one into the other using their complementary angle relationship. This simplifies the equation to a linear algebraic form, making it easy to solve for the unknown angle.

Question 6. If A, B and C are interior angles of a triangle ABC, then show that \( \sin \frac{B + C}{2} = \cos \frac{A}{2} \).
Answer:
In a triangle, the sum of angles is \( 180^{\circ} \).
So, \( A + B + C = 180^{\circ} \)
Divide by 2 on both sides:
\( \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{180^{\circ}}{2} \)
\( \frac{A}{2} + \frac{B + C}{2} = 90^{\circ} \)
\( \frac{B + C}{2} = 90^{\circ} - \frac{A}{2} \)
Taking sine on both sides:
\( \sin \left(\frac{B + C}{2}\right) = \sin \left(90^{\circ} - \frac{A}{2}\right) \)
We know that \( \sin (90^{\circ} - \theta) = \cos \theta \).
So, \( \sin \left(\frac{B + C}{2}\right) = \cos \frac{A}{2} \)
In simple words: Because A, B, and C are angles in a triangle, they add up to 180 degrees. If we divide everything by two, we can rearrange the equation to show that (B+C)/2 is equal to 90 degrees minus A/2. Then, taking the sine of both sides, we use a trigonometric identity to convert sine (90 minus an angle) into cosine of that angle, proving the statement.

Exam Tip: For problems involving triangle angles and trigonometric ratios, always start with the angle sum property of a triangle \( (A+B+C = 180^\circ) \) and then perform algebraic manipulations to get the desired form before applying trigonometric identities.

Question 7. Express \( \sin 67^{\circ} + \cos 75^{\circ} \) in terms of trigonometric ratios of angles between \( 0^{\circ} \) and \( 45^{\circ} \).
Answer:
We have
\( \sin 67^{\circ} + \cos 75^{\circ} \)
We know that \( \sin \theta = \cos (90^{\circ} - \theta) \) and \( \cos \theta = \sin (90^{\circ} - \theta) \).
So, \( \sin 67^{\circ} = \sin (90^{\circ} - 23^{\circ}) = \cos 23^{\circ} \)
And \( \cos 75^{\circ} = \cos (90^{\circ} - 15^{\circ}) = \sin 15^{\circ} \)
Therefore, \( \sin 67^{\circ} + \cos 75^{\circ} = \cos 23^{\circ} + \sin 15^{\circ} \)
Here, \( 23^{\circ} \) and \( 15^{\circ} \) are both between \( 0^{\circ} \) and \( 45^{\circ} \).
In simple words: We convert each trigonometric ratio into its complementary form. Sine 67 degrees becomes cosine of (90 minus 67), which is cosine 23 degrees. Cosine 75 degrees becomes sine of (90 minus 75), which is sine 15 degrees. Both 23 degrees and 15 degrees fall within the 0 to 45-degree range.

Exam Tip: To express trigonometric ratios of angles greater than \( 45^{\circ} \) in terms of angles between \( 0^{\circ} \) and \( 45^{\circ} \), use the complementary angle identities. For example, \( \sin \theta = \cos (90^{\circ} - \theta) \) and \( \cos \theta = \sin (90^{\circ} - \theta) \).