Prove that: 1 / ( secA - tanA) = secA + tanA

Answer:


When you see the questions having either
secA - tanA or secA + tanA on LHS or either of them in the RHS then remember you need to use the Trigonometric identity:

Sec^2 A - tan^2 A = 1

We have there, a^2 - b^2
Which we can open as the factor:
(a+ b) (a-b)
So,
(Sec A+ tanA) (secA- tanA)

Then we cancel the equal terms from the numerator and denominator.
And we get the LHS.

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