Q13 of 268 Page 4

Prove that :
sin 70^o. sec 20^o=1

Taking LHS

sin 70° sec 20°

[∵ cos θ = sin (90° - θ)]

= 1 = RHS

Hence Proved

If tan 3 θ =cot (θ +18^o), where 3 θ and θ +18^o are acute angles, find the value of θ.

If sec 5 θ =cosec (θ -36^o), where 5 θ is an acute angle, find the value of θ.

Prove that :

sin (90^o- θ) tan θ=sin θ

Prove that :

tan 63^o. tan 27^o=1

Prove that :sin 70o. sec 20o=1