Question

1/secA-tanA = secA+tanA​

Answer 1

Answer:

Step-by-step explanation:

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:

Step-by-step explanation:

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.

Thus Proved

Hope That This Helped You!