Short Answer Questions 1 — Question 25

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

⇒ tan² θ + cos² θ - 1

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

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

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

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

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