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प्रश्न
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
उत्तर
We have to prove `tan theta - cot theta = (2 sin^2 theta - `1)/(sin theta cos theta)`
We know that. `sin^2 theta + cos^2 theta - 1`
So,
`tan theta - cot theta = sin theta/cos theta - cos theta/sin theta`
`= (sin^2 theta - cos^2 theta)/(sin theta cos theta)`
`= (sin^2 theta - (1 - sin^2 theta))/(sin theta cos theta)`
`= (sin^2 theta - (1 - sin^2 theta))/(sin theta cos theta)`
`= (2 sin^2 theta - 1)/(sin theta cos theta)`
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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
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
