Question

1÷secA+tanA= 1-sinA÷CosA​

Answer 1

Given,

$\bold{\frac{1}{secA+tanA}} = \bold{\frac{1-sinA}{cosA}}$

⇒ $\bold{\frac{1}{\frac{1}{cosA}}} + \bold{\frac{sinA}{cosA}} = \bold{\frac{1-sinA}{cosA}}$

⇒ $\bold{\frac{1}{\frac{1+sinA}{cosA}}} = \bold{\frac{1-sinA}{cosA}}$

⇒ $\bold{\frac{cosA}{1+sinA}} = \bold{\frac{1-sinA}{cosA}}$

Now,

Cross multiply  the terms,

⇒ $\bold{cos^2A} = \bold{1-sin^2A}$

⇒ $\bold{cos^2A} = \bold{cos^2A}$

LHS = RHS

Hence Proved !!!!