Question

Question 1 Evaluate the following
(i) sin60° cos30° + sin30° cos 60°
(ii) 2tan^2 45° + cos^2 30° − sin^2 60°
(iii) cos45 / (sec30 + cosec30)
(iv) (sin 30 + tan 45 - cosec 60) / (sec 30 + cos 60 + cot 45)
(v) (5cos^2 60 + 4sec^2 30 - tan^2 45) / (sin^2 30 + cos^2 30)

Class 10 - Math - Introduction to Trigonometry Page 187

Answer 1

(1) sin60.cos30 + sin30.cos60
use, formula,
sin( A + B) = sinA.cosB + cosA. sinB
hence,
sin60.cos30 + sin30.cos60 = sin(30+60)
= sin90 = 1

(2) 2tan^2 45° + cos^2 30° − sin^2 60°
= 2 (1)² + (√3/2)² - (√3/2)²
= 2

(3) cos45 / (sec30 + cosec30)
= 1/√2/(2/√3 + 2 )
= √3/2√2( 1 + √3)
= √6(√3 - 1)/8

(4) (sin 30 + tan 45 - cosec 60) / (sec 30 + cos 60 + cot 45)
= (1/2 + 1 - 2/√3)/( 2/√3 + 1/2 + 1)
= (3/2 - 2/√3)/( 2/√3 + 3/2)
=(3√3 - 4)/(4 + 3√3)

(v) (5cos^2 60 + 4sec^2 30 - tan^2 45) / (sin^2 30 + cos^2 30)
= ( 5 ×1/4 + 4 × 4/3 - 1)/( 1/4 + 3/4)
= ( 5/4 + 13/3 )
= 67/12

Answer 2

Here Your Answer :

I) sin60° cos30° + sin30° cos 60°

(:- Sin(A + B ) = SinA.CosB + CosA.sinB)

So
Sin60°Cos30° + Cos30°Sin60 = sin ( 60 +30)

Sin90° = 1

(ii) 2tan² 45° + cos² 30° − sin² 60°

= 2 (1)² +( 2/√3)² + ( 2/√3)²

= 2 + 2/3 + 2/3 = 2

iii ) cos45 / (sec30 + cosec30)

= 1/√2 / ( 2+√3/2 )

= √3/2 √2(1+ √3)

= √6( √3 - 1)/8

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

= ( 1/2 + 1 - 2√3 )/( 2√3 + 1/2 + 1)

= ( 3/2 - 2/√3)/( 2/√3 + 3/2 )

= (3√3 -4 )/( 4+ 3√3)

(v) (5cos²60 + 4sec² 30 - tan²45) / (sin²30 + cos²30)

= 5 ( 1/2)² + 4(2/√3)² - ( 1)² / ( 1/2)² + (2/√3)²

= 5/4 + 4 × 4/3 - 1/ 1/4 + 4/3

= (5/4 + 13/3)

= ( 65/12)