Question

Trigonometry
Solve Question 3 a) pls

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\underline{\textsf{To Prove : }} \\ \\ \sf \implies \dfrac{Sec \: A }{Sec \: A \: + \: 1 } \: + \: \dfrac{Sec \: A}{Sec \: A \: - \: 1 } \: = \: 2 \: Cosec^2 \: A$



$\sf LHS \: = \: \dfrac{Sec \: A }{Sec \: A \: + \: 1 } \: + \: \dfrac{Sec \: A}{Sec \: A \: - \: 1 } \\ \\ \\ \\ \sf \qquad \: = \: \dfrac{Sec \: A ( Sec \: A \: - \: 1 ) \: + \: Sec \: A ( Sec \: A + \: 1)}{(Sec \: A \: + \: 1 )(Sec \: A \: - \: 1)} \\ \\ \\ \\ \sf \qquad \: = \: \dfrac{Sec {}^{2} A - \cancel{ Sec \: A }\: + \: Sec^{2} A \: + \: \cancel{Sec \: A }}{Sec^{2} A \: - \: 1 }$




$\sf \qquad \: = \dfrac{2 \: Sec^2 A}{Tan^2 A} \qquad \: \: \: \: \qquad\{ Sec^2 \theta \: - \: 1 \: = \: Tan^2 \theta \} \\ \\ \\ \sf \qquad \: = \dfrac{2}{Cos^2A} \: \div \: \dfrac{Sin^2A}{Cos^2A} \quad \left \{ Sec^2A \: = \: \dfrac{1}{Cos^2A} , Tan^2A \: = \: \dfrac{Sin^2A}{Cos^2A} \right\}$



$\sf \qquad \: = \dfrac{2}{\cancel{Cos^2A} }\: \times \: \dfrac{\cancel{Cos^2A}}{Sin^2A} \\ \\ \\ \sf \qquad \: = \dfrac{2}{Sin^2A} \\ \\ \\ \sf \qquad \: = 2 \: Cosec^2A \qquad \qquad \quad \left\{ Sin^2A \: = \: \dfrac{1}{Cosec^2A} \right \} \\ \\ \\ \sf \qquad \: = LHS \\ \\ \sf \underline{\underline{\mathsf{\Large{Proved !! }}}}$

Answer 2

Hii Poorvi

Here's the answer