Question

The 8. is just the number of the question.

Answer 1

The value of (1+tanx)(1-tanx)+sec²x is 0 and the value of $\frac{sec^{2}-4 }{sec+2}$ is $sec\theta \ -2$.

Given that, (1+tanx)(1-tanx)+sec²x.

Now, (1-tan²x)+sec²x

=1-tan²x+sec²x

=1+sec²x-tan²x

=1-1=0

Now, evaluate $\frac{sec^{2}-4 }{sec+2}$

$=\frac{sec^{2} \theta \ -2^{2} }{sec\theta \ +2} =\frac{(sec \theta \ +2)(sec \theta \ -2)}{sec \theta \ +2}$ (∵a²-b²=(a+b)(a-b))

$=sec \theta \ -2$

Hence, the value of (1+tanx)(1-tanx)+sec²x is 0 and the value of $\frac{sec^{2}-4 }{sec+2}$ is $sec\theta \ -2$.

To learn more about trigonometric identities visit:

brainly.com/question/12537661.

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