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प्रश्न
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
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उत्तर
LHS = sin( 90° - θ ) sin θ cot θ
= cos θ . sin θ . `cos θ/sin θ`
= cos2θ
= RHS
Hence proved.
संबंधित प्रश्न
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
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Prove the following identity :
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Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
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sec2θ = `square/576`
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If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
