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1). 02). 13). cosec θ4). sec θ

Answer»

[(sec2θ + 1)√(SEC$2$θ – 1)] × 1/2 × (cotθ – tanθ)

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⇒$ [(sec2θ + 1)√(TAN$2$θ)] × 1/2 × (cosθ/sinθ – sinθ/cosθ)

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⇒$ [(sec2θ + 1).tanθ] × 1/2 × [(cos2θ – sin2θ) / (sinθ.cosθ)]

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⇒$ 1/2 × [(sec2θ + 1).tanθ] × [cos2θ/(sinθ.cosθ)]

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⇒$ 1/2 × [(1 + cos2θ).tanθ] / (sinθ.cosθ)

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⇒$ 1/2 × [(1 + 2cos2θ – 1) × sinθ/cosθ] / (sinθ.conθ)

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⇒$ 1/2 × 2cos2θ × sinθ/cosθ × 1/(sinθ.cosθ)

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⇒$ 1$



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