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प्रश्न
`Prove the following trigonometric identities.
`(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
उत्तर
We need to prove `(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
Here, we will first solve the L.H.S.
Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta` we get
`(sec A - tan A)^2 = (1/cos A - sin A/cos A)^2`
`= ((1 -sin A)/cos A)^2`
`= (1 - sin A)^2/(cos A)^2`
Further using the property `sin^2 theta + cos^2 theta = 1` we get
`((1 - sin A)^2/(cos A)) = (1 - sin A)^2/(1 - sin^2 A)`
`= (1 - sin A)^2/((1 - sin A)(1 + sin A))` (using `a^2 - b^2 = (a + b)(a - b))`
`= (1 - sin A)/(1 + sin A)`
henc e proved
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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
