Q23 of 268 Page 4

Prove that :
sin θ .cos (90° - θ) + cos θ sin (90° - θ).

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

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

⇒ cos² θ + sin² θ [∵ cos² θ + sin² θ = 1]

= 1 = RHS

Hence Proved

Prove that :

sin²40^o+sin²50^o=1

Prove that :

sin²29^o + sin²61^o=1

Prove that :

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

Prove that :

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

Prove that :sin θ .cos (90° - θ) + cos θ sin (90° - θ).