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प्रश्न
Prove the following trigonometric identities.
`tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + tan theta + cot theta`
उत्तर
We need to prove `tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + tan theta + cot theta`
Now using cot theta = `1/tan theta` in the LHS we get
`tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = tan theta/(1 - 1/tan theta) + (1/tan theta)/(1 - tan theta)`
`= tan theta/(((tan theta - 1)/tan theta)) + 1/(tan theta(1 - tan theta))`
`= (tan theta)/(tan theta - 1)(tan theta) + 1/(tan theta(1 - tan theta)`
`= tan^2 theta/(tan theta - 1) - 1/(tan theta(tan theta - 1))`
`= (tan^3 theta - 1)/(tan theta(tan theta - 1))`
Further using the identity `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`, we get
`(tan^3 theta - 1)/(tan(tan theta - 1)) = ((tan theta - 1)(tan^2 theta + tan theta + 1))/(tan theta (tan theta - 1))`
`= (tan^2 theta + tan theta + 1)/(tan theta)`
`= tan^2 theta/tan theta + tan theta/tan theta + 1/tan theta`
`= tan theta + 1 + cot theta`
Hence `tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + tan theta + cot theta`
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= `1/(sinθ xx cosθ)` ............... `square`
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= cosecθ × secθ
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∴ cotθ + tanθ = cosecθ × secθ
