NCERT Solutions For Class 10 Math Chapter – 8 Exercise – 8.3
Q1. Evaluate:
- Sin 180/ cos 720
- Tan 260/ cot 640
- Cos 480– sin 420
- Cosec 310– sec 590
Solution:-
- Sin 180/ cos 720
= sin ( 900 – 180) / cos 720
= 1
- Tan 260/ cot 640
= tan (900 – 360)/ cot 640
= cot 640 / cot 640
= 1
- Cos 480– sin 420
Cos(900 – 420) – sin 420
= sin 420 – sin 420 = 0
( iv ) Cosec 310 – sec 590
= cosec ( 900 – 590) – sec 420
= sec 590 – sec 590
= 0
Q2. Show that:
- Tan 480tan 230 tan 420 tan 670 = 1
- Cos 380cos 520 – sin 380 sin 520= 0
Solution:-
- Tan 480tan 230 tan 420 tan 670
= tan (900 – 420) tan (900 – 670) tan 420 tan 670
= cot 420 cot 670 tan 420 tan 670
= (cot 420 tan 420) (cot 670 tan 670)
= 1 x 1
= 1
- Cos 380cos 520 – sin 380 sin 520
= cos ( 900 – 520) cos ( 900 – 380) – sin 380 sin 520
= sin 520 sin 380 – sin 380 sin 520
= 0
Q3. If tan 2a = cot ( A – 180) where 2a is an an acute angle find the value of A.
Solution:-
Tan 2A = Cot ( A – 180)
Cot (900 – 2a) = cot ( a -180)
A= 360
Q4. If tan a – cot b prove that a + b = 900
Solution:-
Tan a = cot b
Tan a = tan (900 – B)
A = 900 – b
A + b = 900
Q5. If sec 4A = Cosec ( A – 200) where 4A is an acute angle find the value of A.
Solution:-
Sec 4A = Cosec ( A – 200)
Cosec ( 900 – 4A) = Cosec ( A – 200)
900 – 4A = A – 200
1100 = 5A
A = 220
Q6. If A , B and C are interior angle of a triangle ABC , then show that sin (B+C/2) = Cos A/2
Solution:-
A + B + C = 1800
B + C = 1800 – A
(B+C)/2 = (1800 – A)/2
Sin (B+C)/2 = Cos A/2
Q7. Express sin 670 + cos 750 in terms of trigonometric ratios of angle between 00 and 450
Solution:-
Sin 670 + cos 750
Sin (900 – 230) + cos ( 900– 150)
Cos 230+ sin 150