Introduction to Trigonometry

Mathematics Class Tenth

10th-Math-home

NCERT Exercise 8.2 Solution part1

(i) sin 60o cos 30o + sin 30o cos 60o

Solution:

Given,

sin 60o cos 30o + sin 30o cos 60o

By substituting trigonometric ratios of sine of 60o and 30o, cosine of 30o and 60o, we get

[sin 30o = 1/2, sin 60o = 3/2, cos 30o = 3/2, and cos 60 = 1/2]

= (3/2 × 3/2) + (1/2 × 1/2)

= 3/4 + 1/4

= 3 + 1 /4

= 4/4 = 1 Answer

(ii) 2 tan 245o + cos230o– sin260o

Solution:

Given,

2 tan 245o + cos230o– sin260o

Now, by substituting values for trigonometric ratios of tan 45o, cos 30o and sin 60o, we get

[∵ tan 45o = 1, cos 30o 3/2, and sin 60o 3/2 ]

2 × (1)2 + (3/2)2 – (3/2)2

= 2 × 1 + 3/4 – 3/4

= 2 + 0 = 2

Thus, 2tan2 45o + cos2 30o – sin260o = 2 Answer

(iii) cos 45o/sec 30o + cosec 30o

Solution:

Given, cos 45o/sec 30o + cosec 30o

After substituting values of trigonometric ratios of cos 45o, sec 30o, and cosec 30o, we get

[∵ cos 45o = 1/2, sec 30o = 2/3, and cosec 30o = 2]

1/2/2/3 + 2

= 1/2/2 + 23/3

= 1/2 × 3/2 + 23

= 3/2 × (2 + 23)

= 3/22 + 26

After multiplying with 22 – 26/22 – 26 we get

= 3/22 + 26 × 22 – 26/22 – 26

= 3 (22 – 26)/(22 + 26) (22 – 26)

= 3 (22 – 26)/(22)2 – (26)2

= 26 – 218/8 – 24

= 26 – 218/– 16

By taking –2 as common

= – 2( –6 + √18)/– 16

= –6 + 18/8

= 18 – 6/8

= 9 × 2 – 6/8

= 32 – 6/8 Answer

(iv) sin 30o + tan 45o – cosec 60o/sec 30o + cos 60o + cot 45o

Solution:

Given,

sin 30o + tan 45o – cosec 60o/sec 30o + cos 60o + cot 45o

After substituting values of different angles

[sin30o = 1/2, tan45o = 1, cosec60o = 2/3, sec30o = 2/3, cos 60o = 1/2, and cot45o = 1]

= 1/2 + 1 – 2/3/2/3 + 1/2 + 1

= 3 + 23 – 4/23/4 + 3 + 23/23

= 3 + 23 – 4/23 × 23/4 + 3 + 23

= 3 + 23 – 4/4 + 3 + 23

= 33 – 4/33 + 4

After multiplying with 33 – 4/33 – 4 we get

= 33 – 4/33 + 4 × 33 – 4/33 – 4

= (33 – 4)2/(33)2 – (4)2

∵ (a + b) (a – b) = a2 – b2

= (33)2 + 42 – 2 × 4 × 33/27 – 16

= 27 + 16 – 24 3/11

= 43 – 243/11 Answer

(v) 5 cos2 60 0 + 4 sec2300 – tan2450/sin2300 + cos2300

Solution

Given

5 cos2 60 0 + 4 sec2300 – tan2450/sin2300 + cos2300

 

After substituting values of different angles

5 × (1/2)2 + 4 (2/3)2 – 1 2/ (1/2)2 + (3/2)2

= 5 × 1/4 + 4 × 4/3 – 1/1/4 + 3/4

= 5/4 + 16/3 – 1 / 1 + 3/4

= 15 + 64 – 12/12 / 4/4

= 67/12 Answer

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