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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Exercise – 8.3)

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°


 

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