AP Board Class 10 Maths Chapter 11 Trigonometry Ex 11.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 11 Trigonometry Ex 11.3 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

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AP Board Class 10 Maths Chapter 11 Trigonometry Ex 11.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 11 Trigonometry Ex 11.3 Book Answers

AP Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 10th Maths Chapter 11 Trigonometry Ex 11.3 book solutions pdf online from this page.

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Textbooks Solutions PDF

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Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 11 Trigonometry Ex 11.3
Provider	Hsslive

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AP Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Textbooks Solutions with Answer PDF Download

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

Question 1.
Evaluate:
i) tan36∘cot54∘
Answer:
Given that tan36∘cot54∘
= tan36∘cot(90∘−36∘) [∵ cot (90 – θ) = tan θ]
= tan36∘tan36∘
= 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° . 1sin15∘ [∵ cosec θ = 1sin𝜃]
= 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° . 1tan26∘ [∵ cot θ = 1tan𝜃]
= 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° . 1tan48∘ . 1tan16∘ [∵ cot θ = 1tan𝜃]
= 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 = 108∘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(𝐀+𝐁2)=cot𝐂2
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]

AP Board Textbook Solutions PDF for Class 10th Maths

AP Board Class 10 Maths Chapter 1 Real Numbers Textbook Solutions PDF

AP Board Class 10 Maths Chapter 2 Sets Textbook Solutions PDF

AP Board Class 10 Maths Chapter 3 Polynomials Textbook Solutions PDF

AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Textbook Solutions PDF

AP Board Class 10 Maths Chapter 5 Quadratic Equations Textbook Solutions PDF

AP Board Class 10 Maths Chapter 6 Progressions Textbook Solutions PDF

AP Board Class 10 Maths Chapter 7 Coordinate Geometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 8 Similar Triangles Textbook Solutions PDF

AP Board Class 10 Maths Chapter 9 Tangents and Secants to a Circle Textbook Solutions PDF

AP Board Class 10 Maths Chapter 10 Mensuration Textbook Solutions PDF

AP Board Class 10 Maths Chapter 11 Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 12 Applications of Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 13 Probability Textbook Solutions PDF

AP Board Class 10 Maths Chapter 14 Statistics Textbook Solutions PDF

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 exam preparation. The AP Board STD 10th Maths Chapter 11 Trigonometry Ex 11.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Books State Board syllabus with maximum efficiency.

FAQs Regarding Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.3 Textbook Solutions

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Important Terms

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