Q24 of 268 Page 4

Prove that :
cos θ . cos(90° – θ) + sin θ sin (90° – θ) = 0

Taking LHS = cos θ cos ( 90° - θ) + sin θ sin (90° - θ)

⇒ cos θ × sin θ – sinθ × cos θ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]

= 0 = RHS

Hence Proved

Prove that :

sin²29^o + sin²61^o=1

Prove that :

sin θ .cos (90° - θ) + cos θ sin (90° - θ).

Prove that :

sin 42°. cos 48° + cos 42° . sin 48° = 1

Prove that :

Prove that :cos θ . cos(90° – θ) + sin θ sin (90° – θ) = 0