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NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Here, Below you all know about NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 Question Answer. I know many of you confuse about finding Chapter 8 Introduction to Trigonometry Ex 8.3 Of Class 10 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 10
SubjectMaths
ChapterChapter 8
ExerciseClass 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3
Number of Questions Solved7
NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

NCERT TEXTBOOK EXERCISES

Question 1.

Solution:

Question 2. Show that:

(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:

(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° tan (90° – 48°) tan (90° – 23°)
= tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° .$\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}$
= 1 = RHS

(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin 38° sin (90° – 38°)
= cos 38° sin 38°- sin 38° cos 38° = 0 = RHS

Question 3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵cot (90° – θ) = tan θ]
⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = 108°/3
∴ ∠ A = 36°

Question 4. If tan A = cot B, prove that A + B = 90°.

Solution:

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

Question 5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:

sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°) [cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20° ⇒ 5A = 110°
A = 110°/5
A = 22°
∴ ∠ A = 22°

Question 6. If A, Band Care interior angles of a triangle ABC, then show that: sin $\frac { B+C }{ 2 }$ = cos A/2

Solution:

Question 7. Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°) = cos 23° + sin 15°

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 PDF

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