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Mathematics

Prove that :

tan2 θ + cos2 θ - 1 = tan2 θ. sin2 θ

Trigonometric Identities

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Answer

Solving L.H.S. of the above equation, we get :

⇒ tan2 θ + cos2 θ - 1

⇒ sin2θcos2θ\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - (1 - cos2 θ)

⇒ sin2θcos2θ\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - sin2 θ

⇒ sin2 θ (1cos2θ−1)\Big(\dfrac{1}{\text{cos}^2 θ} - 1\Big)

⇒ sin2θ(1−cos2θ)cos2θ\dfrac{\text{sin}^2 θ(1 - \text{cos}^2 θ)}{\text{cos}^2 θ}

⇒ sin2θcos2θ\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ}. sin2 θ

⇒ tan2 θ. sin2 θ.

Since. L.H.S. = R.H.S.

Hence, proved that tan2 θ + cos2 θ - 1 = tan2 θ. sin2 θ

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