Question

pls prove the following this and show all steps​

Answer 1

To prove :

$\tt \dfrac{(1 + {sin \theta)}^{2} + (1 - {sin \theta)}^{2} }{2 {cos}^{2} \theta } = \dfrac{1 + {sin}^{2} \theta}{1 - {sin}^{2} \theta}$

Proof :

Identities to be used :

• (a + b)² = a² + b² + 2ab
• sin²∅ + cos² ∅ = 1

Taking the LHS

Solving the brackets :

$\tt \dfrac{1 + {sin}^{2} \theta + 2sin \theta + (1 + {sin}^{2} \theta - 2sin \theta ) }{2 {cos}^{2} \theta }$

$\tt \implies \dfrac{1 + {sin}^{2} \theta + 2sin \theta + 1 + {sin}^{2} \theta - 2sin \theta }{2 {cos}^{2} \theta } \\ \\ \tt \implies \frac{1 + {sin}^{2} \theta + 1 + {sin}^{2} \theta }{2 {cos}^{2}\theta} \\ \\ \tt \implies \dfrac{2 + 2 {sin}^{2} \theta }{2 {cos}^{2} \theta } \\ \\ \tt \implies \frac{2(1 + {sin}^{2} \theta )}{2 {cos}^{2} \theta } \\ \\ \tt \implies \frac{1 + {sin}^{2} \theta}{ {cos}^{2} \theta} \\ \\ \bf \because {sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \bf {cos}^{2} \theta = 1 - {sin}^{2} \theta \\ \\ \boxed{\tt \implies \frac{1 + {sin}^{2} \theta }{1 - {sin}^{2} \theta }}$

LHS = RHS

Hence Proved .