The value of (1+tanx)(1-tanx)+sec²x is 0 and the value of [tex]\frac{sec^{2}-4 }{sec+2}[/tex] is [tex]sec\theta \ -2[/tex].
Given that, (1+tanx)(1-tanx)+sec²x.
Now, (1-tan²x)+sec²x
=1-tan²x+sec²x
=1+sec²x-tan²x
=1-1=0
Now, evaluate [tex]\frac{sec^{2}-4 }{sec+2}[/tex]
[tex]=\frac{sec^{2} \theta \ -2^{2} }{sec\theta \ +2} =\frac{(sec \theta \ +2)(sec \theta \ -2)}{sec \theta \ +2}[/tex] (∵a²-b²=(a+b)(a-b))
[tex]=sec \theta \ -2[/tex]
Hence, the value of (1+tanx)(1-tanx)+sec²x is 0 and the value of [tex]\frac{sec^{2}-4 }{sec+2}[/tex] is [tex]sec\theta \ -2[/tex].
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