SSC CPO SI and ASI Model Paper 2

Question : 141 of 200

Marks: +1, -0

Find out the value of sin 2a – cos2 2a if, sin a × sec (30° + a) = 1 where (0° < a < 60°).

Solution:

(1) sin a × sec (30° + a) = 1

⇒sina/{cos( 30°+a)}=1

⇒sina/{sin (90°- 30° –a)}=1

⇒sina/{sin( 60°-a)}=1

⇒sinaa= sin a = sin (60° – a)

⇒ 2a = 60°

∴ sin 2a – cos^2 2a

= sin 60° – (cos 60°)^2

=√3 /2-1/4= {{2 √3}- 1}/4

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