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प्रश्न
Prove the following trigonometric identities.
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
Prove that:
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
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उत्तर
We have to prove `(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
We know that, sin2 θ + cos2 θ = 1
Multiplying both numerator and denominator by (1 − sin θ), we have
`(1 - sin θ)/(1 + sin θ) = ((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`
`= (1 - sin θ)^2/(1 - sin^2 θ)`
`= ((1 - sin θ)/cos θ)^2`
`= (1/cos θ - sin θ/cos θ)^2`
`= (sec θ - tan θ)^2`
संबंधित प्रश्न
If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`
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If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
If sec θ = `25/7`, then find the value of tan θ.
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
sec2θ – tan2θ = ?
Which is not correct formula?
