Prove that :sin229o + sin261o=1

Taking LHS= sin229o + sin261o⇒ cos2 (90° - 29°) + sin2 61° [∵ Sin θ = cos (90° - θ)]⇒ cos2 61° + sin2 61°= 1 =RHS [∵ cos2 θ + sin2 θ = 1]Hence Proved

Q22 of 268 Page 4

Prove that :
sin²29^o + sin²61^o=1

Taking LHS= sin²29^o + sin²61^o

⇒ cos² (90° - 29°) + sin² 61° [∵ Sin θ = cos (90° - θ)]

⇒ cos² 61° + sin² 61°

= 1 =RHS [∵ cos² θ + sin² θ = 1]

Hence Proved

Prove that :

cos 27+ sin51^o = sin63^o+cos 39^o

Prove that :

sin²40^o+sin²50^o=1

Prove that :

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

Prove that :

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