NCERT solutions for class 10 Maths
Chapter 8
Coordinate Geometry
Exercise 8.3
Ex 8.3 Question 1.
Evaluate :
(i) sin 18°/cos 72°
(ii) tan 26°/cot 64°
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Solution:
(i) sin 18°/cos 72°
sin 18° = cos(90° – 18°) = cos 72° [cos 90° – θ = sin θ ]
= cos 72° /cos 72° = 1
(ii) tan 26°/cot 64°
tan 36° = cot(90° – 36°) = cot 64° [cot 90° – θ = tan θ ]
= cot 64°/cot 64° = 1
(iii) cos 48° – sin 42°
cos (90° – 42°) = sin 42° [cos 90° – θ = sin θ ]
= sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
cosec (90° – 59°) – sec 59° [cosec 90° – θ = sec θ ]
= sec 59° – sec 59° = 0
Ex 8.3 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) tan 48° tan 23° tan 42° tan 67°
tan 48° = tan (90° – 42°) = cot 42°
tan 23° = tan (90° – 67°) = cot 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
Putting the values
= 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 38° = cos (90° – 52°) = sin 52°
cos 52°= cos (90°-38°) = sin 38°
= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
Putting the values
= sin 52° sin 38° – sin 38° sin 52° = 0
Ex 8.3 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°)
tan 2A = cot (90° – 2A)
⇒ cot (90° – 2A) = cot (A -18°) [cot 90° – θ = tan θ]
⇒ 90° – 2A = A- 18° ⇒ 108° = 3A
A = 108° / 3
Hence, A = 36°
Ex 8.3 Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
tan A = cot B
cot B = tan (90° – B)
tan A = tan (90° – B)
A = 90° – B
A + B = 90°
Ex 8.3 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°)
sec 4A = cosec (90° – 4A)
cosec (90° – 4A) = cosec (A – 20°)
90° – 4A= A- 20°
110° = 5A
A = 110°/ 5 = 22°
Hence, A = 22°
Ex 8.3 Question 6.
If A, B and C are interior angles of a triangle ABC, then show that
sin (B+C/2) = cos A/2
Solution:
L.H.S. = sin (B+C/2)
= sin (180 – A /2 ) [A + B + C = 180°]
sin (90° – A/2)
cos A/2 = R.H.S. [sin 90° – θ = cos θ ]
Ex 8.3 Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75°
sin θ = cos (90° – θ)
& cos θ = sin (90° – θ )
sin 67° + cos 75° = cos (90° – 67°) + sin (90° – 15°)
= cos 23° + sin 15°
