AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.3 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.3

### 10th Class Maths 11th Lesson Trigonometry Ex 11.3 Textbook Questions and Answers

Question 1.

Evaluate:

i) \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)

Answer:

Given that \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)

= \(\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}\) [∵ cot (90 – θ) = tan θ]

= \(\frac{\tan 36^{\circ}}{\tan 36^{\circ}}\)

= 1

ii) cos 12° – sin 78°

Answer:

Given that cos 12° – sin 78°

= cos 12° – sin(90° – 12°) [∵ sin (90 – θ) = cos θ]

= cos 12° – cos 12° = 0

iii) cosec 31° – sec 59°

Answer:

Given that cosec 31° – sec 59°

= cosec 31° – sec (90° – 31°) [∵ sec (90 – θ) = cosec θ]

= cosec 31° – cosec 31° = 0

iv) sin 15° sec 75°

Answer:

Given that sin 15° sec 75°

= sin 15° . sec (90° – 15°)

= sin 15° . cosec 15° [∵ sec (90 – θ) = cosec θ]

= sin 15° . \(\frac{1}{\sin 15^{\circ}}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]

= 1

v) tan 26° tan 64°

Answer:

Given that tan 26° tan 64°

= tan 26° . tan (90° – 26°)

= tan 26° . cot 26° [∵ tan (90 – θ) = cot θ]

= tan 26° . \(\frac{1}{\tan 26^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]

= 1

Question 2.

Show that

i) tan 48° tan 16° tan 42° tan 74° = 1

Answer:

L.H.S. = tan 48° tan 16° tan 42° tan 74°

= tan 48°. tan 16° . tan(90° – 48°) . tan(90° – 16°)

= tan 48° . tan 16° . cot 48° . cot 16° [∵ tan (90 – θ) = cot θ]

= tan 48° . tan 16° . \(\frac{1}{\tan 48^{\circ}}\) . \(\frac{1}{\tan 16^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]

= 1 = R.H.S.

∴ L.H.S. = R.H.S.

ii) cos 36° cos 54° – sin 36° sin 54° = 0

Answer:

L.H.S. = cos 36° cos 54° – sin 36° sin54°

= cos (90° – 54°) . cos (90° – 36°) – sin 36° . sin 54° [∵ cos (90 – θ) = sin θ]

= sin 54° . sin 36° – sin 36° . sin 54°

= 0 = R.H.S.

∴ L.H.S. = R.H.S.

Question 3.

If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A.

Answer:

Given that tan 2A = cot (A – 18°)

⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)]

⇒ 90° – 2A = A – 18°

⇒ 108° = 3A

⇒ A = \(\frac{108^{\circ}}{3}\) = 36°

Hence the value of A is 36°.

Question 4.

If tan A = cot B where A and B. are acute angles, prove that A + B = 90°.

Answer:

Given that tan A = cot B

⇒ cot (90° – A) = cot B [∵ tan θ = cot (90 – θ)]

⇒ 90° – A = B

⇒ A + B = 90°

Question 5.

If A, B and C are interior angles of a triangle ABC, then show that \(\tan \left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)=\cot \frac{\mathbf{C}}{2}\)

Answer:

Given A, B and C are interior angles of right angle triangle ABC then A + B + C = 180°.

On dividing the above equation by ‘2’ on both sides, we get 180°

On taking tan ratio on both sides

Hence proved.

Question 6.

Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0° and 45°.

Answer:

We have sin 75° + cos 65°

= sin (90° – 15°) + cos (90° – 25°)

= cos 15° + sin 25° [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]