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प्रश्न
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
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उत्तर
LHS = `(tanθ + sinθ)/(tanθ - sinθ)`
= `(sinθ/cosθ + sinθ)/(sinθ/cosθ - sinθ) = (sinθ + sinθcosθ)/(sinθ + sinθcosθ)`
= `(sinθ(1 + cosθ))/sin(1 + cosθ) = (1 + cosθ)/(1 - cosθ)`
= `(1 + 1/secθ)/(1 - 1/secθ) = ((secθ + 1)/secθ)/((secθ - 1)/secθ)`
= `(secθ + 1)/(secθ - 1)`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
Prove that `sqrt((1 + cos theta)/(1 - cos theta)) + sqrt((1 - cos theta)/(1 + cos theta)) = 2 cosec theta`
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
`sec theta (1- sin theta )( sec theta + tan theta )=1`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - A))/(cos A)`.
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
