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प्रश्न
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
उत्तर
LHS = `sqrt((1 + sin θ)/(1 - sin θ) xx (1 + sin θ)/(1 + sin θ))`
= `sqrt((1 + sin θ)^2/(1 - sin^2θ))`
= `sqrt((1 + sin θ)^2/(cos^2θ)`
= `(1 + sin θ)/cos θ = 1/cos θ + sin θ/cos θ`
= sec θ + tan θ
= RHS
Hence proved.
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संबंधित प्रश्न
If x = a cos θ and y = b cot θ, show that:
`a^2/x^2 - b^2/y^2 = 1`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
If `sec theta = x ,"write the value of tan" theta`.
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
If sin θ = `1/2`, then find the value of θ.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
