NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3
NCERT Solutions for Maths Chapter 8, Exercise 8.3 involve complete answers for each question in the exercise 8.3. The solutions provide students a strategic methods to prepare for their exam. Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Chapter 8 Introduction to Trigonometry Exercise 8.3 prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way.
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 2:(I) tan 48° tan 23° tan 42° tan 67° = 1
Taking LHS,
tan 48° tan 23° tan 42° tan 67°------------Equation (1)
We know that tan (90° – A) = tan A
By manipulating the Equation (1) using the property above,
= tan (90° − 42°) tan (90° − 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°) (By rearranging)
= (1) (1) [As cot A. tan A = 1]
= 1
(II)cos 38° cos 52° − sin 38° sin 52°
Consider LHS :
cos 38° cos 52° − sin 38° sin 52° --------------Equation (1)
= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52° [As, Cos (90 – θ) = Sin θ]
= sin 52° sin 38° − sin 38° sin 52°
= 0
Question 3:
If tan 2A = cot (A − 18°), where 2A is an acute angle, find the value of A.
Solution 3:
Given that,
tan 2A = cot (A − 18°) -------------Equation (1)
We know that tan 2A = cot (90 – 2A) by substituting this in Equation (1)
cot (90° − 2A) = cot (A −18°)
Hence by Equating,
90° − 2A = A− 18°
A + 2A =90° + 18°
3A= 108°
A = 36°
Question 4:
If tan A = cot B, prove that A + B = 90°
Solution 4:
Given,
tan A = cot B---------Equation (1),
We know that tan A = cot (90 – A) by substituting this in Equation (1)
tan A = tan (90° − B)
By Equating,
A = 90° − B
A + B = 90° (By Transposing)
Question 5:
If sec 4A = cosec (A − 20°), where 4A is an acute angle, find the value of A.
Solution 5:
Given,
sec 4A = cosec (A − 20°) ---------Equation (1),
We know that Sec A = Cosec (90-A) by substituting this in Equation (1)
cosec (90° − 4A) = cosec (A − 20°)
By Equating,
90° − 4A= A− 20°
110° = 5A (By Transposing)
A = 22°
Question 7:
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution 7:
Sin 67° + cos 75°
Since, Cos (90 – θ ) = Sin θ and Sin (90 – θ) = Cos θ
= sin (90° − 23°) + cos (90° − 15°)
= cos 23° + sin 15°
