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Mathematics

sin2θcos2θ−1cos2θ\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{ \text{cos}^2 θ} = x

Statement (1): x = 1

Statement (2): x = sin2θ−1cos2θ=−cos2θcos2θ\dfrac{\text{sin}^2 θ - 1}{{\text{cos}^2 θ}} = \dfrac{-\text{cos}^2 θ}{{\text{cos}^2 θ}} = -1

  1. Both the statement are true.

  2. Both the statement are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Trigonometric Identities

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Answer

Given, sin2θcos2θ−1cos2θ\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{ \text{cos}^2 θ} = x

⇒x=sin2θcos2θ−1cos2θ⇒x=sin2θ−1cos2θ⇒x=sin2θ−(sin2θ+cos2θ)cos2θ⇒x=sin2θ−sin2θ−cos2θcos2θ⇒x=−cos2θcos2θ⇒x=−1\Rightarrow x = \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{ \text{cos}^2 θ}\\[1em] \Rightarrow x = \dfrac{\text{sin}^2 θ - 1}{\text{cos}^2 θ}\\[1em] \Rightarrow x = \dfrac{\text{sin}^2 θ - (\text{sin}^2 θ + \text{cos}^2 θ)}{\text{cos}^2 θ} \\[1em] \Rightarrow x = \dfrac{\text{sin}^2 θ - \text{sin}^2 θ - \text{cos}^2 θ}{\text{cos}^2 θ}\\[1em] \Rightarrow x = \dfrac{- \text{cos}^2 θ}{\text{cos}^2 θ}\\[1em] \Rightarrow x = -1

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

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