Math, asked by si2mritik2ythamodimp, 1 year ago

Sin ( 180 + theta ) cos ( 90 + theta ) tan ( 270 + theta )cot ( 360 + theta) / sin ( 360 - theta ) cos ( 360 + theta ) cosec ( -theta ) sin ( 270 + theta )

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Answered by ARoy
77
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)
=1
Answered by sachinprasad78pck8qh
39

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