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Questions
Prove the following trigonometric identities.
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
Prove the following:
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
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Solution
We need to prove `tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
Now using cot θ = `1/tan θ` in the LHS, we get
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = tan θ/(1 - 1/tan θ) + (1/tan θ)/(1 - tan θ)`
`= tan θ/(((tan θ - 1)/tan θ)) + 1/(tan θ(1 - tan θ))`
`= (tan θ)/(tan θ - 1)(tan θ) + 1/(tan θ(1 - tan θ)`
`= tan^2 θ/(tan θ - 1) - 1/(tan θ(tan θ - 1))`
`= (tan^3 θ - 1)/(tan θ(tan θ - 1))`
Further using the identity `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`, we get
`(tan^3 θ - 1)/(tan(tan θ - 1)) = ((tan θ - 1)(tan^2 θ + tan θ + 1))/(tan θ (tan θ - 1))`
`= (tan^2 θ + tan θ + 1)/(tan θ)`
`= tan^2 θ/tan θ+ tan θ/tan θ + 1/tan θ`
= tan θ + 1 + cot θ
Hence `tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
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L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
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