Question

√(sec tetha+1)(sec tetha-1)


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Answer 1

Answer:

$=TanA$

P.s. I will be Replacing theta with A

Step-by-step explanation:

$\sqrt{(SecA +1)( SecA -1)}$

$=\sqrt{Sec^{2} -1}$ ∵ $a^{2}-b^{2} = (a-b)(a+b)$

$=\sqrt{Tan^{2}A}$ ∵$Sec^{2}A -Tan^{2} = 1$

$=TanA$

Answer 2

Answer:

$\tan(theeta)$

Step-by-step explanation:

we know that

${ \sec(theeta) }^{2} - 1 = { \tan(theeta) }^{2}$

by using identities

${a}^{2} - {b}^{2} = (a + b)(a - b)$

${ \sec(theeta) }^{2} - 1 =( sec{theeta} - 1)(sectheeta + 1)$

$\sqrt[2]{ (\sec(theeta) + 1)( \sec(theeta) - 1) }= \sqrt{ { \tan(theeta) }^{2} }$