EXERCISE 8.3
Q 1 (i & ii), Ex 8.3 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Q 1 (iii & iv), Ex 8.3 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Page No 189:
Question 1:
Evaluate(I)
(II)
(III) cos 48° − sin 42°
(IV)cosec 31° − sec 59°
Answer:
(I)
(II)
(III)cos 48° − sin 42° = cos (90°− 42°) − sin 42°
= sin 42° − sin 42°
= 0
(IV) cosec 31° − sec 59° = cosec (90° − 59°) − sec 59°
= sec 59° − sec 59°
= 0
Q 2 (i & ii), Ex 8.3 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Question 2:
Show that(I) tan 48° tan 23° tan 42° tan 67° = 1
(II)cos 38° cos 52° − sin 38° sin 52° = 0
Answer:
(I) tan 48° tan 23° tan 42° tan 67°= 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°)
= (1) (1)
= 1
(II) cos 38° cos 52° − sin 38° sin 52°
= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°
= sin 52° sin 38° − sin 38° sin 52°
= 0
Q 3, Ex 8.3 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
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°)
90° − 2A = A− 18°
108° = 3A
A = 36°
Question 4:
If tan A = cot B, prove that A + B = 90°Answer:
Given that,tan A = cot B
tan A = tan (90° − B)
A = 90° − B
A + B = 90°
Q 5, Ex 8.3 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Question 5:
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.Answer:
Given that,sec 4A = cosec (A − 20°)
cosec (90° − 4A) = cosec (A − 20°)
90° − 4A= A− 20°
110° = 5A
A = 22°
Page No 190:
Question 6:
If A, Band C are interior angles of a triangle ABC then show that
Answer:
We know that for a triangle ABC,∠ A + ∠B + ∠C = 180°
∠B + ∠C= 180° − ∠A
Q 7, Ex 8.3 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Question 7:
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.Answer:
sin 67° + cos 75°= sin (90° − 23°) + cos (90° − 15°)
= cos 23° + sin 15°
0 comments:
Post a Comment