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प्रश्न
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
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उत्तर
We have to prove `(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
We know that, `sin^2 theta + cos^2 theta = 1`
`(1 + sec theta)/sec theta = (1 + 1/cos theta)/(1/cos theta)`
`= ((cos theta + 1)/cos theta)/(1/cos theta)`
`= (1 + cos theta)/1`
Multiplying the numerator and denominator by `(1 - cos theta)` we have
`(1 + sec theta)/sec theta = ((1 + cos theta)(1 - cos theta))/(1- cos theta)`
`= (1 - cos^2 theta)/(1- cos theta)`
`= sin^2 theta/(1 - cos theta)`
संबंधित प्रश्न
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Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
