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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
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x2 – y2) = (a2 – b2).
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
cos θ . sec θ = ?
(sec θ + tan θ) . (sec θ – tan θ) = ?
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
