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Mathematics

1cot θ + tan θ\dfrac{1}{\text{cot θ + tan θ}} is equal to :

  1. sin θ + cos θ

  2. sin θ. cos θ

  3. 1sin θ. cos θ\dfrac{1}{\text{sin θ. cos θ}}

  4. 1sin θ + cos θ\dfrac{1}{\text{sin θ + cos θ}}

Trigonometric Identities

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Answer

Solving,

⇒1cot θ + tan θ⇒1cos θsin θ+sin θcos θ⇒1cos2θ+sin2θcos θ. sin θ⇒cos θ. sin θcos2θ+sin2θ\Rightarrow \dfrac{1}{\text{cot θ + tan θ}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{cos θ}}{\text{sin θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{cos}^2 θ + \text{sin}^2 θ}{\text{cos θ. sin θ}}} \\[1em] \Rightarrow \dfrac{\text{cos θ. sin θ}}{\text{cos}^2 θ + \text{sin}^2 θ}

Substituting, sin2 θ + cos2 θ = 1, we get :

⇒ sin θ. cos θ

Hence, Option 2 is the correct option.

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