Question

1/(secA + tan A) = sec(A) - tanA​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Given LHS =$\frac{1}{secA-tanA}$

On rationalizing we get,

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

$\frac{secA+tanA}{(secA-tanA)(secA+tanA)} </p><p>(secA−tanA)(secA+tanA)$

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

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

secA+tanA.

LHS = RHS.

Hope this helps!