Math, asked by rajesh9366, 1 year ago


 \cot( \beta  \beta  \\  \\ e \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \\  \sin( \sin( \tan( \\  \\  \\  \\  \\  \\  \\  \\  \\  \\ ) ) ) )

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Answered by Anonymous
0

Given, Sin 72° = p

⇒ Sin (90°-18°) = p

⇒ Cos 18° = p ∵ [ Cosθ = Sin(90°-θ) ]

Now, Sin 18° = √1 - Cos² 18° ∵ [Sinθ = √1 - Cos²θ ]

⇒ Sin 18° = √1 - p²

Tan 18° = Sin 18° / Cos 18° ∵ [ Tanθ = Sinθ / Cosθ ]

= (√1 - p²) / p

∴ Tan 18° = (√1 - p²) / p

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