Class 10 - Mathematics
Introduction To Trigonometry - Exercise 8.3
Top Block 1
Exercise 8.3
Question : 1:Evaluate :
(i) sin 18o/cos 72o (ii) tan 26/cot 64o (iii) cos 48o – sin 42o (iv) cosec 31o – sec 59o
Answer :
(i) sin 18o/cos 72o = cos(90o – 18o)/cos 720 [sin(90o – θ) = cos θ]
= cos 72o/cos 72o
= 1
(ii) tan 26o/cot 64o = cot(90o – 64o)/cot 640 [tan(90o – θ) = cot θ]
= cot 64o/cot 64o
= 1
(iii) cos 48o – sin 42o = sin(90o – 42o) – sin 420 [sin(90o – θ) = cos θ]
= sin 42o – sin 42o
= 0
(iv) cosec 31o – sec 59o = sec(90o – 59o) – sec 590 [sec(90o – θ) = cosec θ]
= sec 59o – sec 59o
= 0
Question : 2:Show that :
(i) tan 48o tan 23o tan 42o tan 67o = 1
(ii) cos 38o cos 52o – sin 38o sin 52o = 0
Answer :
(i) tan 48o tan 23o tan 42o tan 67o = tan 48o tan 23o cot(90o – 42o) cot(90o – 23o)
= tan 48o tan 23o cot 48o cot 23o [tan(90o – θ) = cot θ]
= tan 48o tan 23o (1/tan 48o )(1/tan 230 )
= 1
(ii) cos 38o cos 52o – sin 38o sin 52o = cos 38o cos 52o – cos(90o – 38o) cos(90o – 520 )
= cos 38o cos 52o – cos 38o cos 52o
= 0
Question : 3:If tan 2A = cot (A – 18o), where 2A is an acute angle, find the value of A.
Answer :
Given, tan 2A = cot (A – 18o)
⇒ cot (90o – 2A) = cot (A – 18o) [tan(90o – θ) = cot θ]
⇒ 90o – 2A = A – 18o
⇒ 2A + A = 90o + 18o
⇒ 3A = 108o
⇒ A = 108o/3
⇒ A = 36o
Question : 4:If tan A = cot B, prove that A + B = 90o.
Answer :
Given, tan A = cot B
⇒ cot(90o – A) = cot B [tan(90o – θ) = cot θ]
⇒ 90o – A = B
⇒ A + B = 90o
Mddle block 1
Question : 5:If sec 4A = cosec (A – 20o), where 4A is an acute angle, find the value of A.
Answer :
Given, sec 4A = cosec (A – 20o)
⇒ cosec(90o – 4A) = cosec (A – 20o) [sec(90o – θ) = cosec θ]
⇒ 90o – 4A = A – 20o
⇒ 4A + A = 90o + 20o
⇒ 5A = 110o
⇒ A = 110o/5
⇒ A = 22o
Question : 6:If A, B and C are interior angles of a triangle ABC, then show that
sin (B + C)/2 = cos A/2
Answer :
Given, A, B and C are interior angles of a triangle ABC.
⇒ A + B + C = 180o
⇒ (A + B + C)/2 = 180o/2
⇒ (A + B + C)/2 = 90o
⇒ A/2 + (B + C)/2 = 90o
⇒ (B + C)/2 = 90o – A/2
Now, sin{(B + C)/2} = sin (90o – A/2)
⇒ sin{(B + C)/2} = cos A/2 [sin(90o – A) = cos A]
Question : 7:Express sin 67o + cos 75o in terms of trigonometric ratios of angles between 0o and 45o.
Answer :
We know that sin θ = cos(90o – θ) and cos θ = sin(90o – θ)
Therefore, sin 67 + cos 75 = cos(90o – θ) + sin(90o – θ)
= cos 23o + sin 15o
