Q10 of 268 Page 4

Prove that

Taking LHS,

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

⇒ 1+ 1

= 2 = RHS

Hence Proved

If A + B = 90^o, sin A = a, sin B = b, then prove that

(a) a² + b² = 1

(b)

Show that sin(50° + θ) – cos (40° – θ) = 0.

In a ∆ABC prove that