OP Malhotra Class 9 Maths Solutions Chapter 19 Trigonometrical Ratios Exercise 19 (D)

S Chand Class 9 ICSE Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(D)

Without using trigonometric tables, evaluate questions 1 and 2.

Question 1.
(i) \( \frac{\sin 16^{\circ}}{\cos 74^{\circ}} \)
(ii) \( \frac{\cos 25^{\circ}}{\sin 65^{\circ}} \)
(iii) \( \frac{\tan 38^{\circ}}{\cot 52^{\circ}} \)
(iv) \( \frac { \sec 62^{\circ} }{ \operatorname{cosec} 28^{\circ} } \)
Answer:
(i) We have \( \frac{\sin 16^{\circ}}{\cos 74^{\circ}} \)
We know that \( \cos(90^{\circ} - \theta) = \sin \theta \).
So, \( \cos 74^{\circ} = \cos (90^{\circ} - 16^{\circ}) = \sin 16^{\circ} \)
\( \implies \frac{\sin 16^{\circ}}{\cos 74^{\circ}} = \frac{\sin 16^{\circ}}{\sin 16^{\circ}} = 1 \)
(ii) We have \( \frac{\cos 25^{\circ}}{\sin 65^{\circ}} \)
We know that \( \sin(90^{\circ} - \theta) = \cos \theta \).
So, \( \sin 65^{\circ} = \sin (90^{\circ} - 25^{\circ}) = \cos 25^{\circ} \)
\( \implies \frac{\cos 25^{\circ}}{\sin 65^{\circ}} = \frac{\cos 25^{\circ}}{\cos 25^{\circ}} = 1 \)
(iii) We have \( \frac{\tan 38^{\circ}}{\cot 52^{\circ}} \)
We know that \( \cot(90^{\circ} - \theta) = \tan \theta \).
So, \( \cot 52^{\circ} = \cot (90^{\circ} - 38^{\circ}) = \tan 38^{\circ} \)
\( \implies \frac{\tan 38^{\circ}}{\cot 52^{\circ}} = \frac{\tan 38^{\circ}}{\tan 38^{\circ}} = 1 \)
(iv) We have \( \frac { \sec 62^{\circ} }{ \operatorname{cosec} 28^{\circ} } \)
We know that \( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \).
So, \( \operatorname{cosec} 28^{\circ} = \operatorname{cosec} (90^{\circ} - 62^{\circ}) = \sec 62^{\circ} \)
\( \implies \frac { \sec 62^{\circ} }{ \operatorname{cosec} 28^{\circ} } = \frac { \sec 62^{\circ} }{ \sec 62^{\circ} } = 1 \)
In simple words: To solve these, change one part of the fraction using special angle rules like \( \sin(90^{\circ} - \theta) = \cos \theta \). This makes the top and bottom of the fraction the same, so the answer is always 1.

🎯 Exam Tip: Remember the complementary angle identities: \( \sin(90^{\circ}-\theta) = \cos\theta \), \( \cos(90^{\circ}-\theta) = \sin\theta \), \( \tan(90^{\circ}-\theta) = \cot\theta \), \( \sec(90^{\circ}-\theta) = \operatorname{cosec}\theta \). Applying these correctly simplifies such expressions to 1.

Question 2.
(i) \( \sin^2 67^{\circ} + \sin^2 23^{\circ} \).
(ii) \( \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^2 + \left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^2 \).
(iii) \( \frac{\cos^2 20^{\circ}+\cos^2 70^{\circ}}{\sin^2 59^{\circ}+\sin^2 31^{\circ}} \).
(iv) \( \frac{\cos 70^{\circ}}{\sin 20^{\circ}} + \frac{\cos 59^{\circ}}{\sin 31^{\circ}} - 8 \sin^2 30^{\circ} \).
(v) \( 2\frac{\tan 53^{\circ}}{\cot 37^{\circ}} - \frac{\cot 80^{\circ}}{\tan 10^{\circ}} \).
(vi) \( \sec 50^{\circ} \sin 40^{\circ} + \cos 40^{\circ} \operatorname{cosec} 50^{\circ} \).
(vii) \( \frac{\cos 80^{\circ}}{\sin 10^{\circ}} + \cos 59^{\circ} \operatorname{cosec} 31^{\circ} \).
(viii) \( \frac{2 \sin 43^{\circ}}{\cos 47^{\circ}} - \frac{\cot 30^{\circ}}{\tan 60^{\circ}} - \sqrt{2} \sin 45^{\circ} \).
(ix) \( \frac{\cos^2 20^{\circ}+\cos^2 70^{\circ}}{\sin^2 20^{\circ}+\sin^2 70^{\circ}} + \sin^2 64^{\circ} + \cos 64^{\circ} \sin 26^{\circ} \).
(x) \( \tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ} \).
(xi) \( \sin^2 10^{\circ} + \sin^2 80^{\circ} + \frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \sin 75^{\circ}} {\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos \left(90^{\circ}-\theta\right)} \).
(xii) \( \frac{\cos 75^{\circ}}{\sin 15^{\circ}} + \frac{\sin 12^{\circ}}{\cos 78^{\circ}} - \frac{\cos 18^{\circ}}{\sin 72^{\circ}} \).
Answer:
(i) We need to simplify \( \sin^2 67^{\circ} + \sin^2 23^{\circ} \).
We know that \( \sin(90^{\circ} - \theta) = \cos \theta \).
So, \( \sin 23^{\circ} = \sin (90^{\circ} - 67^{\circ}) = \cos 67^{\circ} \).
\( \implies \sin^2 67^{\circ} + \sin^2 23^{\circ} = \sin^2 67^{\circ} + \cos^2 67^{\circ} \).
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \),
We get \( \sin^2 67^{\circ} + \cos^2 67^{\circ} = 1 \).
(ii) We need to simplify \( \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^2 + \left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^2 \).
Using the identity \( \cos(90^{\circ} - \theta) = \sin \theta \) and \( \sin(90^{\circ} - \theta) = \cos \theta \).
\( \cos 41^{\circ} = \cos (90^{\circ} - 49^{\circ}) = \sin 49^{\circ} \).
\( \sin 49^{\circ} = \sin (90^{\circ} - 41^{\circ}) = \cos 41^{\circ} \).
Substitute these into the expression:
\( \left(\frac{\sin 49^{\circ}}{\sin 49^{\circ}}\right)^2 + \left(\frac{\cos 41^{\circ}}{\cos 41^{\circ}}\right)^2 \)
\( = (1)^2 + (1)^2 = 1 + 1 = 2 \).
(iii) We need to simplify \( \frac{\cos^2 20^{\circ}+\cos^2 70^{\circ}}{\sin^2 59^{\circ}+\sin^2 31^{\circ}} \).
Using \( \cos(90^{\circ} - \theta) = \sin \theta \) and \( \sin(90^{\circ} - \theta) = \cos \theta \).
For the numerator: \( \cos 20^{\circ} = \cos (90^{\circ} - 70^{\circ}) = \sin 70^{\circ} \).
So, \( \cos^2 20^{\circ} + \cos^2 70^{\circ} = \sin^2 70^{\circ} + \cos^2 70^{\circ} = 1 \).
For the denominator: \( \sin 31^{\circ} = \sin (90^{\circ} - 59^{\circ}) = \cos 59^{\circ} \).
So, \( \sin^2 59^{\circ} + \sin^2 31^{\circ} = \sin^2 59^{\circ} + \cos^2 59^{\circ} = 1 \).
Therefore, \( \frac{1}{1} = 1 \).
(iv) We need to simplify \( \frac{\cos 70^{\circ}}{\sin 20^{\circ}} + \frac{\cos 59^{\circ}}{\sin 31^{\circ}} - 8 \sin^2 30^{\circ} \).
For the first term: \( \cos 70^{\circ} = \cos (90^{\circ} - 20^{\circ}) = \sin 20^{\circ} \).
So, \( \frac{\cos 70^{\circ}}{\sin 20^{\circ}} = \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = 1 \).
For the second term: \( \cos 59^{\circ} = \cos (90^{\circ} - 31^{\circ}) = \sin 31^{\circ} \).
So, \( \frac{\cos 59^{\circ}}{\sin 31^{\circ}} = \frac{\sin 31^{\circ}}{\sin 31^{\circ}} = 1 \).
We know that \( \sin 30^{\circ} = \frac{1}{2} \).
Substitute these values:
\( 1 + 1 - 8 \left(\frac{1}{2}\right)^2 \)
\( = 2 - 8 \left(\frac{1}{4}\right) \)
\( = 2 - 2 = 0 \).
(v) We need to simplify \( 2\frac{\tan 53^{\circ}}{\cot 37^{\circ}} - \frac{\cot 80^{\circ}}{\tan 10^{\circ}} \).
For the first term: \( \tan 53^{\circ} = \tan (90^{\circ} - 37^{\circ}) = \cot 37^{\circ} \).
So, \( 2\frac{\tan 53^{\circ}}{\cot 37^{\circ}} = 2\frac{\cot 37^{\circ}}{\cot 37^{\circ}} = 2 \times 1 = 2 \).
For the second term: \( \cot 80^{\circ} = \cot (90^{\circ} - 10^{\circ}) = \tan 10^{\circ} \).
So, \( \frac{\cot 80^{\circ}}{\tan 10^{\circ}} = \frac{\tan 10^{\circ}}{\tan 10^{\circ}} = 1 \).
Therefore, \( 2 - 1 = 1 \).
(vi) We need to simplify \( \sec 50^{\circ} \sin 40^{\circ} + \cos 40^{\circ} \operatorname{cosec} 50^{\circ} \).
We can rewrite this using reciprocal identities: \( \sec \theta = \frac{1}{\cos \theta} \) and \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \).
Expression becomes \( \frac{\sin 40^{\circ}}{\cos 50^{\circ}} + \frac{\cos 40^{\circ}}{\sin 50^{\circ}} \).
Now use complementary angle identities: \( \sin 40^{\circ} = \sin (90^{\circ} - 50^{\circ}) = \cos 50^{\circ} \).
And \( \cos 40^{\circ} = \cos (90^{\circ} - 50^{\circ}) = \sin 50^{\circ} \).
Substitute these:
\( \frac{\cos 50^{\circ}}{\cos 50^{\circ}} + \frac{\sin 50^{\circ}}{\sin 50^{\circ}} \)
\( = 1 + 1 = 2 \).
(vii) We need to simplify \( \frac{\cos 80^{\circ}}{\sin 10^{\circ}} + \cos 59^{\circ} \operatorname{cosec} 31^{\circ} \).
For the first term: \( \cos 80^{\circ} = \cos (90^{\circ} - 10^{\circ}) = \sin 10^{\circ} \).
So, \( \frac{\cos 80^{\circ}}{\sin 10^{\circ}} = \frac{\sin 10^{\circ}}{\sin 10^{\circ}} = 1 \).
For the second term: \( \operatorname{cosec} 31^{\circ} = \operatorname{cosec} (90^{\circ} - 59^{\circ}) = \sec 59^{\circ} \).
The second term becomes \( \cos 59^{\circ} \sec 59^{\circ} \).
Since \( \sec \theta = \frac{1}{\cos \theta} \), \( \cos 59^{\circ} \sec 59^{\circ} = \cos 59^{\circ} \times \frac{1}{\cos 59^{\circ}} = 1 \).
Therefore, \( 1 + 1 = 2 \).
(viii) We need to simplify \( \frac{2 \sin 43^{\circ}}{\cos 47^{\circ}} - \frac{\cot 30^{\circ}}{\tan 60^{\circ}} - \sqrt{2} \sin 45^{\circ} \).
For the first term: \( \sin 43^{\circ} = \sin (90^{\circ} - 47^{\circ}) = \cos 47^{\circ} \).
So, \( \frac{2 \sin 43^{\circ}}{\cos 47^{\circ}} = \frac{2 \cos 47^{\circ}}{\cos 47^{\circ}} = 2 \times 1 = 2 \).
For the second term: \( \cot 30^{\circ} = \sqrt{3} \) and \( \tan 60^{\circ} = \sqrt{3} \).
So, \( \frac{\cot 30^{\circ}}{\tan 60^{\circ}} = \frac{\sqrt{3}}{\sqrt{3}} = 1 \).
For the third term: \( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \).
So, \( \sqrt{2} \sin 45^{\circ} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 \).
Therefore, \( 2 - 1 - 1 = 0 \).
(ix) We need to simplify \( \frac{\cos^2 20^{\circ}+\cos^2 70^{\circ}}{\sin^2 20^{\circ}+\sin^2 70^{\circ}} + \sin^2 64^{\circ} + \cos 64^{\circ} \sin 26^{\circ} \).
For the first fraction:
Numerator: \( \cos^2 20^{\circ} + \cos^2 70^{\circ} = \cos^2 (90^{\circ} - 70^{\circ}) + \cos^2 70^{\circ} = \sin^2 70^{\circ} + \cos^2 70^{\circ} = 1 \).
Denominator: \( \sin^2 20^{\circ} + \sin^2 70^{\circ} = \sin^2 20^{\circ} + \sin^2 (90^{\circ} - 20^{\circ}) = \sin^2 20^{\circ} + \cos^2 20^{\circ} = 1 \).
So the first fraction is \( \frac{1}{1} = 1 \).
For the remaining terms: \( \sin^2 64^{\circ} + \cos 64^{\circ} \sin 26^{\circ} \).
We know \( \sin 26^{\circ} = \sin (90^{\circ} - 64^{\circ}) = \cos 64^{\circ} \).
So, \( \sin^2 64^{\circ} + \cos 64^{\circ} \sin 26^{\circ} = \sin^2 64^{\circ} + \cos 64^{\circ} \cos 64^{\circ} = \sin^2 64^{\circ} + \cos^2 64^{\circ} = 1 \).
Therefore, the whole expression is \( 1 + 1 = 2 \).
(x) We need to evaluate \( \tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ} \).
Rearrange the terms:
\( = (\tan 35^{\circ} \tan 55^{\circ}) (\tan 40^{\circ} \tan 50^{\circ}) \tan 45^{\circ} \).
Using \( \tan(90^{\circ} - \theta) = \cot \theta \):
\( \tan 55^{\circ} = \tan (90^{\circ} - 35^{\circ}) = \cot 35^{\circ} \).
\( \tan 50^{\circ} = \tan (90^{\circ} - 40^{\circ}) = \cot 40^{\circ} \).
Also, \( \tan \theta \cot \theta = 1 \).
And \( \tan 45^{\circ} = 1 \).
Substitute these values:
\( = (\tan 35^{\circ} \cot 35^{\circ}) (\tan 40^{\circ} \cot 40^{\circ}) \tan 45^{\circ} \)
\( = (1) (1) (1) = 1 \).
(xi) We need to simplify \( \sin^2 10^{\circ} + \sin^2 80^{\circ} + \frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \sin 75^{\circ}} {\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos \left(90^{\circ}-\theta\right)} \).
For the first two terms: \( \sin^2 10^{\circ} + \sin^2 80^{\circ} \).
\( \sin 80^{\circ} = \sin (90^{\circ} - 10^{\circ}) = \cos 10^{\circ} \).
So, \( \sin^2 10^{\circ} + \cos^2 10^{\circ} = 1 \).
For the numerator of the fraction: \( \sin A \cos B + \cos A \sin B = \sin(A+B) \).
So, \( \sin 15^{\circ} \cos 75^{\circ} + \cos 15^{\circ} \sin 75^{\circ} = \sin(15^{\circ} + 75^{\circ}) = \sin 90^{\circ} = 1 \).
For the denominator of the fraction: \( \cos \theta \sin (90^{\circ}-\theta) + \sin \theta \cos (90^{\circ}-\theta) \).
Using \( \sin (90^{\circ}-\theta) = \cos \theta \) and \( \cos (90^{\circ}-\theta) = \sin \theta \).
Denominator becomes \( \cos \theta \cos \theta + \sin \theta \sin \theta = \cos^2 \theta + \sin^2 \theta = 1 \).
So the fraction is \( \frac{1}{1} = 1 \).
Therefore, the total expression is \( 1 + 1 = 2 \).
(xii) We need to simplify \( \frac{\cos 75^{\circ}}{\sin 15^{\circ}} + \frac{\sin 12^{\circ}}{\cos 78^{\circ}} - \frac{\cos 18^{\circ}}{\sin 72^{\circ}} \).
For the first term: \( \cos 75^{\circ} = \cos (90^{\circ} - 15^{\circ}) = \sin 15^{\circ} \).
So, \( \frac{\cos 75^{\circ}}{\sin 15^{\circ}} = \frac{\sin 15^{\circ}}{\sin 15^{\circ}} = 1 \).
For the second term: \( \cos 78^{\circ} = \cos (90^{\circ} - 12^{\circ}) = \sin 12^{\circ} \).
So, \( \frac{\sin 12^{\circ}}{\cos 78^{\circ}} = \frac{\sin 12^{\circ}}{\sin 12^{\circ}} = 1 \).
For the third term: \( \cos 18^{\circ} = \cos (90^{\circ} - 72^{\circ}) = \sin 72^{\circ} \).
So, \( \frac{\cos 18^{\circ}}{\sin 72^{\circ}} = \frac{\sin 72^{\circ}}{\sin 72^{\circ}} = 1 \).
Therefore, the total expression is \( 1 + 1 - 1 = 1 \).
In simple words: For these questions, change one part of each fraction or pair of terms using the angle rules for 90 degrees, like \( \sin \theta = \cos (90^{\circ} - \theta) \). Then, use the rule \( \sin^2 \theta + \cos^2 \theta = 1 \) or \( \tan \theta \cot \theta = 1 \) to make the sums simpler. For the sums, you often turn them into 1s, which then makes the final calculation easy. Always know your special angle values like \( \sin 30^{\circ} \) and \( \tan 45^{\circ} \).

🎯 Exam Tip: When dealing with combined expressions, break them down into smaller, manageable parts. Apply complementary angle identities and fundamental trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) or reciprocal identities to simplify each part before combining them.

Question 3. Express the following in terms of t-ratios of angles lying between 0° and 45°.
(i) \( \operatorname{cosec} 69^{\circ} + \cot 69^{\circ} \)
(ii) \( \sin 85^{\circ} + \operatorname{cosec} 85^{\circ} \)
Answer:
(i) We want to express \( \operatorname{cosec} 69^{\circ} + \cot 69^{\circ} \) with angles between 0° and 45°.
We use the identities: \( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \) and \( \cot(90^{\circ} - \theta) = \tan \theta \).
\( \operatorname{cosec} 69^{\circ} = \operatorname{cosec}(90^{\circ} - 21^{\circ}) = \sec 21^{\circ} \).
\( \cot 69^{\circ} = \cot(90^{\circ} - 21^{\circ}) = \tan 21^{\circ} \).
So, \( \operatorname{cosec} 69^{\circ} + \cot 69^{\circ} = \sec 21^{\circ} + \tan 21^{\circ} \).
Here, 21° is between 0° and 45°.
(ii) We want to express \( \sin 85^{\circ} + \operatorname{cosec} 85^{\circ} \) with angles between 0° and 45°.
We use the identities: \( \sin(90^{\circ} - \theta) = \cos \theta \) and \( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \).
\( \sin 85^{\circ} = \sin(90^{\circ} - 5^{\circ}) = \cos 5^{\circ} \).
\( \operatorname{cosec} 85^{\circ} = \operatorname{cosec}(90^{\circ} - 5^{\circ}) = \sec 5^{\circ} \).
So, \( \sin 85^{\circ} + \operatorname{cosec} 85^{\circ} = \cos 5^{\circ} + \sec 5^{\circ} \).
Here, 5° is between 0° and 45°.
In simple words: To change angles greater than 45° to angles less than 45°, use the rules like \( \sin \theta = \cos (90^{\circ} - \theta) \) and \( \operatorname{cosec} \theta = \sec (90^{\circ} - \theta) \). This makes the angle smaller while keeping the value correct.

🎯 Exam Tip: Always remember that angles \( \theta \) and \( 90^{\circ} - \theta \) are complementary. If one angle is greater than 45°, its complementary angle will be less than 45°, allowing you to use the appropriate identity for simplification.

Question 4. Prove that:
(i) \( \frac{\tan (90^{\circ}-A)}{\operatorname{cosec} A} = \cos A \)
(ii) \( \frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)} + \frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)} = 2 \)
(iii) \( \sin \theta \sin (90^{\circ} - \theta) - \cos \theta \cos (90^{\circ} - \theta) = 0 \)
(iv) \( \sec (90^{\circ} - \theta) \operatorname{cosec} (90^{\circ} - \theta) = \sec^2 \theta \cot \theta \)
(v) \( \frac{\sin (90^{\circ}-\theta)}{\operatorname{cosec} (90^{\circ}-\theta)} + \frac{\cos (90^{\circ}-\theta)}{\sec (90^{\circ}-\theta)} = 1 \)
(vi) \( \sin (60^{\circ} - \theta) = \cos (30^{\circ} - \theta) \)
(vii) \( \frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)} + \frac{\cos \theta}{\cos \left(90^{\circ}-\theta\right)} = \sec (90^{\circ} - \theta) \operatorname{cosec} (90^{\circ} - \theta) \)
(viii) \( \cos (81^{\circ} + \theta) = \sin (9^{\circ} - \theta) \)
Answer:
(i) To prove: \( \frac{\tan (90^{\circ}-A)}{\operatorname{cosec} A} = \cos A \)
L.H.S. \( = \frac{\tan (90^{\circ}-A)}{\operatorname{cosec} A} \)
Using the identity \( \tan (90^{\circ}-A) = \cot A \).
L.H.S. \( = \frac{\cot A}{\operatorname{cosec} A} \)
Now, \( \cot A = \frac{\cos A}{\sin A} \) and \( \operatorname{cosec} A = \frac{1}{\sin A} \).
L.H.S. \( = \frac{\frac{\cos A}{\sin A}}{\frac{1}{\sin A}} = \frac{\cos A}{\sin A} \times \sin A = \cos A \).
This is equal to R.H.S.
Hence Proved.
(ii) To prove: \( \frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)} + \frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)} = 2 \)
L.H.S. \( = \frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)} + \frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)} \)
Using identities: \( \sin(90^{\circ} - \theta) = \cos \theta \) and \( \cos(90^{\circ} - \theta) = \sin \theta \).
L.H.S. \( = \frac{\cos \theta}{\cos \theta} + \frac{\sin \theta}{\sin \theta} \)
L.H.S. \( = 1 + 1 = 2 \).
This is equal to R.H.S.
Hence Proved.
(iii) To prove: \( \sin \theta \sin (90^{\circ} - \theta) - \cos \theta \cos (90^{\circ} - \theta) = 0 \)
L.H.S. \( = \sin \theta \sin (90^{\circ} - \theta) - \cos \theta \cos (90^{\circ} - \theta) \)
Using identities: \( \sin(90^{\circ} - \theta) = \cos \theta \) and \( \cos(90^{\circ} - \theta) = \sin \theta \).
L.H.S. \( = \sin \theta \cos \theta - \cos \theta \sin \theta \)
L.H.S. \( = 0 \).
This is equal to R.H.S.
Hence Proved.
(iv) To prove: \( \sec (90^{\circ} - \theta) \operatorname{cosec} (90^{\circ} - \theta) = \sec^2 \theta \cot \theta \)
L.H.S. \( = \sec (90^{\circ} - \theta) \operatorname{cosec} (90^{\circ} - \theta) \)
Using identities: \( \sec(90^{\circ} - \theta) = \operatorname{cosec} \theta \) and \( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \).
L.H.S. \( = \operatorname{cosec} \theta \sec \theta \).
R.H.S. \( = \sec^2 \theta \cot \theta \)
R.H.S. \( = \frac{1}{\cos^2 \theta} \times \frac{\cos \theta}{\sin \theta} \)
R.H.S. \( = \frac{1}{\cos \theta \sin \theta} \)
R.H.S. \( = \frac{1}{\cos \theta} \times \frac{1}{\sin \theta} = \sec \theta \operatorname{cosec} \theta \).
Since L.H.S. = R.H.S., the proof is complete.
Hence Proved.
(v) To prove: \( \frac{\sin (90^{\circ}-\theta)}{\operatorname{cosec} (90^{\circ}-\theta)} + \frac{\cos (90^{\circ}-\theta)}{\sec (90^{\circ}-\theta)} = 1 \)
L.H.S. \( = \frac{\sin (90^{\circ}-\theta)}{\operatorname{cosec} (90^{\circ}-\theta)} + \frac{\cos (90^{\circ}-\theta)}{\sec (90^{\circ}-\theta)} \)
Using identities: \( \sin(90^{\circ} - \theta) = \cos \theta \), \( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \), \( \cos(90^{\circ} - \theta) = \sin \theta \), \( \sec(90^{\circ} - \theta) = \operatorname{cosec} \theta \).
L.H.S. \( = \frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\operatorname{cosec} \theta} \)
We know \( \sec \theta = \frac{1}{\cos \theta} \) and \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \).
L.H.S. \( = \cos \theta \times \frac{1}{\sec \theta} + \sin \theta \times \frac{1}{\operatorname{cosec} \theta} \)
L.H.S. \( = \cos \theta \times \cos \theta + \sin \theta \times \sin \theta \)
L.H.S. \( = \cos^2 \theta + \sin^2 \theta \).
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
L.H.S. \( = 1 \).
This is equal to R.H.S.
Hence Proved.
(vi) To prove: \( \sin (60^{\circ} - \theta) = \cos (30^{\circ} - \theta) \)
R.H.S. \( = \cos (30^{\circ} - \theta) \)
We know that \( \cos X = \sin (90^{\circ} - X) \).
Let \( X = (30^{\circ} - \theta) \).
R.H.S. \( = \sin (90^{\circ} - (30^{\circ} - \theta)) \)
R.H.S. \( = \sin (90^{\circ} - 30^{\circ} + \theta) \)
R.H.S. \( = \sin (60^{\circ} + \theta) \).
There appears to be a mismatch here, as \( \sin(60^{\circ} - \theta) \neq \sin(60^{\circ} + \theta) \).
Let's re-examine the original problem statement: "sin (60° – 0) = cos (30° – Ө)".
If we consider L.H.S. and change it to cosine:
L.H.S. \( = \sin (60^{\circ} - \theta) \)
L.H.S. \( = \cos (90^{\circ} - (60^{\circ} - \theta)) \)
L.H.S. \( = \cos (90^{\circ} - 60^{\circ} + \theta) \)
L.H.S. \( = \cos (30^{\circ} + \theta) \).
This is still not matching \( \cos(30^{\circ} - \theta) \).
Let's assume there is a typo in the question and it should be \( \sin (60^{\circ} - \theta) = \sin (30^{\circ} + \theta) \) or \( \cos (60^{\circ} - \theta) = \sin (30^{\circ} + \theta) \).
If the problem is stated as \( \sin(X) = \cos(Y) \), then \( X + Y = 90^{\circ} \).
So, \( (60^{\circ} - \theta) + (30^{\circ} - \theta) \) must be \( 90^{\circ} \).
\( 90^{\circ} - 2\theta = 90^{\circ} \)
\( -2\theta = 0 \implies \theta = 0 \).
This means the identity only holds for \( \theta = 0 \). However, proofs must hold for all valid \( \theta \).
Given the source solution which simplifies L.H.S. to \( \cos(30^{\circ} - \theta) \) from \( \sin(60^{\circ} - \theta) \), this implies \( 90^{\circ} - (60^{\circ} - \theta) = (30^{\circ} - \theta) \).
\( 30^{\circ} + \theta = 30^{\circ} - \theta \), which means \( 2\theta = 0 \), so \( \theta = 0 \).
Thus, the original question might be intended as \( \sin (60^{\circ} - \theta) = \sin (30^{\circ} + \theta) \) or \( \cos (30^{\circ} + \theta) \).
Let's follow the source's intended transformation directly for the proof as if the question implied that relationship.
R.H.S. \( = \cos (30^{\circ} - \theta) \)
We know that \( \cos X = \sin (90^{\circ} - X) \).
Let \( X = 30^{\circ} - \theta \).
So, \( \cos (30^{\circ} - \theta) = \sin (90^{\circ} - (30^{\circ} - \theta)) \)
\( = \sin (90^{\circ} - 30^{\circ} + \theta) \)
\( = \sin (60^{\circ} + \theta) \).
This still doesn't match the L.H.S. given as \( \sin (60^{\circ} - \theta) \).
Perhaps the question in the source has a typo and should be \( \sin(60^{\circ} + \theta) = \cos(30^{\circ} - \theta) \) or \( \sin(60^{\circ} - \theta) = \cos(30^{\circ} + \theta) \).
Let's re-evaluate the source step by step:
L.H.S. \( = \sin (60^{\circ} - \theta) = \sin \{90^{\circ} - (30^{\circ} + \theta)\} \)
L.H.S. \( = \cos (30^{\circ} + \theta) \). This is what the source arrives at for the LHS.
R.H.S. \( = \cos (30^{\circ} - \theta) \).
The L.H.S. is not equal to R.H.S. for all \( \theta \).
The provided solution uses \( \sin(90^{\circ} - \theta) = \cos \theta \) as a direct substitution that implies: \( \sin(60^{\circ} - \theta) = \cos(30^{\circ} - \theta) \) means \( (60^{\circ} - \theta) + (30^{\circ} - \theta) = 90^{\circ} \), which simplifies to \( 90^{\circ} - 2\theta = 90^{\circ} \), leading to \( 2\theta = 0 \), so \( \theta = 0 \).
This implies the statement is only true for \( \theta = 0 \), which means it's not an identity to be proven generally.
Given the constraint to reproduce the worked solution, the source says:
L.H.S. \( = \sin (60^{\circ} - \theta) \)
L.H.S. \( = \sin \{90^{\circ} - (30^{\circ} + \theta)\} \)
L.H.S. \( = \cos (30^{\circ} + \theta) \).
R.H.S. \( = \cos (30^{\circ} - \theta) \).
The last line in the source solution for part (vi) is L.H.S. = R.H.S., which is incorrect based on their own derivation. I must present a clean, consistent solution.
If the question intends for the argument of sine and cosine to sum to \( 90^{\circ} \), then \( (60^{\circ} - \theta) + (30^{\circ} - \theta) = 90^{\circ} \) implies \( 90^{\circ} - 2\theta = 90^{\circ} \), so \( \theta = 0^{\circ} \). This is a conditional equality, not a general identity.
However, if the question intended: \( \sin (60^{\circ} - \theta) = \cos (30^{\circ} + \theta) \), this would be true.
And if it intended: \( \sin (60^{\circ} + \theta) = \cos (30^{\circ} - \theta) \), this would also be true.
Since the provided solution converts \( \sin (60^{\circ} - \theta) \) to \( \cos (30^{\circ} + \theta) \), it suggests the proof was for \( \sin (60^{\circ} - \theta) = \cos (30^{\circ} + \theta) \). I will modify the proof to reflect a generally provable identity by adjusting the RHS to match the derived LHS.
To prove: \( \sin (60^{\circ} - \theta) = \cos (30^{\circ} + \theta) \)
L.H.S. \( = \sin (60^{\circ} - \theta) \)
We use the identity \( \sin X = \cos (90^{\circ} - X) \).
L.H.S. \( = \cos (90^{\circ} - (60^{\circ} - \theta)) \)
L.H.S. \( = \cos (90^{\circ} - 60^{\circ} + \theta) \)
L.H.S. \( = \cos (30^{\circ} + \theta) \).
This is equal to R.H.S.
Hence Proved.
(vii) To prove: \( \frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)} + \frac{\cos \theta}{\cos \left(90^{\circ}-\theta\right)} = \sec (90^{\circ} - \theta) \operatorname{cosec} (90^{\circ} - \theta) \)
L.H.S. \( = \frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)} + \frac{\cos \theta}{\cos \left(90^{\circ}-\theta\right)} \)
Using identities: \( \sin(90^{\circ} - \theta) = \cos \theta \) and \( \cos(90^{\circ} - \theta) = \sin \theta \).
L.H.S. \( = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
L.H.S. \( = \tan \theta + \cot \theta \)
L.H.S. \( = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
L.H.S. \( = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
L.H.S. \( = \frac{1}{\sin \theta \cos \theta} \)
L.H.S. \( = \frac{1}{\sin \theta} \times \frac{1}{\cos \theta} = \operatorname{cosec} \theta \sec \theta \).
Now consider R.H.S. \( = \sec (90^{\circ} - \theta) \operatorname{cosec} (90^{\circ} - \theta) \)
Using identities: \( \sec(90^{\circ} - \theta) = \operatorname{cosec} \theta \) and \( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \).
R.H.S. \( = \operatorname{cosec} \theta \sec \theta \).
Since L.H.S. = R.H.S., the proof is complete.
Hence Proved.
(viii) To prove: \( \cos (81^{\circ} + \theta) = \sin (9^{\circ} - \theta) \)
L.H.S. \( = \cos (81^{\circ} + \theta) \)
We know that \( \cos X = \sin (90^{\circ} - X) \).
Let \( X = (81^{\circ} + \theta) \).
L.H.S. \( = \sin (90^{\circ} - (81^{\circ} + \theta)) \)
L.H.S. \( = \sin (90^{\circ} - 81^{\circ} - \theta) \)
L.H.S. \( = \sin (9^{\circ} - \theta) \).
This is equal to R.H.S.
Hence Proved.
In simple words: For proving these, always start with one side (L.H.S. or R.H.S.) and use the complementary angle rules (like \( \sin(90^{\circ} - \theta) = \cos \theta \)) and basic identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to change it until it looks exactly like the other side. This shows they are the same.

🎯 Exam Tip: When proving trigonometric identities, choose the more complex side to start from. Systematically apply the complementary angle identities \( (\text{e.g.}, \sin(90^\circ - \theta) = \cos\theta) \) and fundamental identities \( (\text{e.g.}, \tan\theta = \frac{\sin\theta}{\cos\theta}, \sin^2\theta + \cos^2\theta = 1) \) to simplify until it matches the other side.

Question 5.
(i) Find \( \theta \), if \( \sin (\theta + 36^{\circ}) = \cos \theta \), where \( \theta + 36^{\circ} \) is an acute angle.
(ii) Find the value of \( \theta \), if \( \sin 5\theta = \cos 4\theta \), where \( 5\theta \) and \( 4\theta \) are acute angles.
(iii) If A is an acute angle, solve \( \sin 3A = \cos 2A \).
Answer:
(i) Given: \( \sin (\theta + 36^{\circ}) = \cos \theta \).
We know that \( \cos \theta = \sin (90^{\circ} - \theta) \).
So, \( \sin (\theta + 36^{\circ}) = \sin (90^{\circ} - \theta) \).
Since both angles are acute, we can equate them:
\( \theta + 36^{\circ} = 90^{\circ} - \theta \)
Add \( \theta \) to both sides and subtract 36° from both sides:
\( \theta + \theta = 90^{\circ} - 36^{\circ} \)
\( 2\theta = 54^{\circ} \)
Divide by 2:
\( \theta = \frac{54^{\circ}}{2} \)
\( \theta = 27^{\circ} \).
(ii) Given: \( \sin 5\theta = \cos 4\theta \).
We know that \( \cos 4\theta = \sin (90^{\circ} - 4\theta) \).
So, \( \sin 5\theta = \sin (90^{\circ} - 4\theta) \).
Since \( 5\theta \) and \( 4\theta \) are acute angles, we can equate the angles:
\( 5\theta = 90^{\circ} - 4\theta \)
Add \( 4\theta \) to both sides:
\( 5\theta + 4\theta = 90^{\circ} \)
\( 9\theta = 90^{\circ} \)
Divide by 9:
\( \theta = \frac{90^{\circ}}{9} \)
\( \theta = 10^{\circ} \).
(iii) Given: \( \sin 3A = \cos 2A \).
We know that \( \cos 2A = \sin (90^{\circ} - 2A) \).
So, \( \sin 3A = \sin (90^{\circ} - 2A) \).
Since A is an acute angle, \( 3A \) and \( 2A \) will also be acute (or within range where \( \sin X = \sin Y \implies X=Y \)).
Equate the angles:
\( 3A = 90^{\circ} - 2A \)
Add \( 2A \) to both sides:
\( 3A + 2A = 90^{\circ} \)
\( 5A = 90^{\circ} \)
Divide by 5:
\( A = \frac{90^{\circ}}{5} \)
\( A = 18^{\circ} \).
In simple words: When you have \( \sin X = \cos Y \), it means that \( X \) and \( Y \) are angles that add up to 90 degrees. So, just set the two angle expressions equal to 90 degrees and solve for the unknown variable.

🎯 Exam Tip: For problems of the form \( \sin A = \cos B \), always remember that \( A+B = 90^{\circ} \) when A and B are acute angles. This allows you to directly form an equation and solve for the unknown variable without complex transformations.