Question

,
3. Prove that
sin (180° + 0) cos (90° + 0) tan (270° - O) cot (360° - 0)
sin (360° - 0) cos ( 360° + ) cosec (-6) sin (270° + 0)​

Answer 1

hello mates ⚘

sin(180°+θ)cos(90°°+θ)tan(270°+θ)cot(360°+θ)/sin(360°-θ)cos(360°+θ)cosec(-θ)sin(270°+θ)

=-sinθ.sinθ.(-cotθ).cotθ/(-sinθ).cosθ.(-cosecθ)(-cosθ)

=sin²θ(-cot²θ)/sinθ.cosecθ.(-cos²θ)

={(sin²θ)(cos²θ/sin²θ)}/(sinθ)(1/sinθ)(cos²θ)

=(cos²θ)/(cos²θ)