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Prove the Following Trigonometric Identities. tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ

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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`

Theorem
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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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Chapter 18: Trigonometric identities - Exercise 18A [Page 424]

APPEARS IN

Nootan Mathematics [English] Class 10 ICSE
Chapter 18 Trigonometric identities
Exercise 18A | Q 22. (i) | Page 424
R.D. Sharma Mathematics [English] Class 10
Chapter 11 Trigonometric Identities
Exercise 11.1 | Q 30 | Page 44
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