Ncert Solutions For Class 10 Maths Ex 8.3

Ncert Solutions For Class 10 Maths Chapter 8 Ex 8.3

1) Calculate:

                (i) sin18cos72

                (ii) tan26cot64

                (iii) cos 48° – sin 42°

                (iv) cosec 31° – sec 59°

Ans:-

(i) sin18cos72

= sin(9018)cos72

= cos72cos72=1

 

(ii) tan26cot64

= tan(9036)cot64

cot64cot64=1

 

(iii) cos 48° – sin42°

= 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

 

2) Show that :

 (i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Ans:-

(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

 

3) We have 2P = cot ( P – 18 ° ), where 2P is an acute angle, calculate the value of P.

Ans:-     According to question,
tan 2P = cot (P- 18°)
=>cot (90° – 2P) = cot (P -18°)
Equating angles,
=>90° – 2P = P- 18°

=>108° = 3P
=> P = 36

 

4) If tan P = cot Q, prove that P + Q = 90°.

 AnswerAccording to question,

tanP = cot Q
=>tan P = tan (90° – Q)
=>P = 90° – Q
=>P + Q = 90°

 

5) If the value of sec 4P = cosec (P – 20°), in which 4P is an acute angle, find the value of P.

Ans:-According to question

sec 4P = cosec (P – 20°)

=> cosec (90° – 4P) = cosec (P – 20°)

Equating angles,
=> 90° – 4P= P- 20°
=> 110° = 5P
=> P = 22°

 

Q6) If X,Y and Z are interior angles of a triangle XYZ, then show that

    sin (Y+Z/2) = cos X2

Answer

In a triangle, sum of all the interior angles

X + Y + Z = 180

Y + Z = 180 – X

Y+Z2 = (180X)2

Y+Z2 = (90X2)

sin (Y+Z2) = sin (90X2)

sin (Y+Z2) = cosX2

 

Q7) Express sin 67 + cos 75 in terms of trigonometric ratios of angles between 0 and 45.

Answer

sin 67 + cos 75

= sin (9023) + cos (9015)
= cos 23 + sin 15

 

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