Question

If secA-tanA=x then prove that 1/x=secA+tanA

Answer 1

$\textbf{- Answer -}$

We know that,


sec²A - tan²A = 1

or, (secA - tanA) (secA + tanA) = 1

or, x (secA + tanA) = 1

or, 1/x = secA + tanA

Therefore, 1/x = secA + tanA (Proved)

#$\textbf{MarkAsBrainliest}$

Answer 2

Heya, As you have seen a method [above] ,

I have a new,

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x = SecA - tanA

Then,

$\frac{1}{x} = \frac{1}{ \sec(A) - \tan(A) }$

By rationalization,

$\frac{1}{x} = \frac{1}{ \sec(A) - \tan(A) } \times \frac{ \sec(A) + \tan(A) }{ \sec(A) + \tan(A) } \\ \\ \frac{1}{x} = \frac{ \sec(A) + \tan(A) }{ { \sec(A) }^{2} - { \tan(A) }^{2} }$

We know, sec²∅-tan²∅ = 1

Then,

$\frac{1}{ x } = \frac{ \sec(A) + \tan(A) }{1}$

1/x = SecA + tanA

Hence, proved

I hope this will help you

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