Q14 of 268 Page 4

Prove that :
sin (90^o- θ) tan θ=sin θ

Taking LHS

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

⇒ cos θ tan θ

[∵ tan θ ]

= sin θ = RHS

Hence Proved

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

Prove that :

sin 70^o. sec 20^o=1

Prove that :

tan 63^o. tan 27^o=1

Prove that :

Prove that :sin (90o- θ) tan θ=sin θ