Q9 of 268 Page 4

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

LHS = sin (50° + θ) – cos (40° - θ)

We know that,

Sin A = cos (90° - A)

Here, A = 50° + θ

⇒ cos {90° -( 50° + θ)} – cos (40° - θ)

⇒ cos (40° - θ) – cos (40° - θ)

= 0 = RHS

Hence Proved

If X + Y = 90^o, then express cos X in terms of simplest trigonometric ratio of Y.

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

(a) a² + b² = 1

(b)

Prove that

In a ∆ABC prove that