Question

Prove that 1/secA-tanA=secA+tanA

Answer 1

Given LHS = $\frac{1}{secA-tanA}$

On rationalizing we get,

$\frac{1}{secA-tanA} * \frac{secA+tanA}{secA+tanA}$

$\frac{secA+tanA}{(secA-tanA)(secA+tanA)}$

$\frac{secA+tanA}{sec^2A-tan^2A}$

$\frac{secA+tanA}{1}$

secA+tanA.


LHS = RHS.


Hope this helps!

Answer 2

Answer:

LHS=1/secA+tanA

1/secA+tanA*secA-tanA/secA-tanA

secA-tanA/sec^2A-tan^2A

secA-tanA/1

SecA-tanA

LHS=RHS

Step-by-step explanation: