Question

Tg(5n/4+x)+tg(5n/4-x)​

Answer 1

$tg(\frac{5\pi }{4}+x)+tg(\frac{5\pi }{4}-x)=\frac{sin\frac{5\pi }{2}}{cos(\frac{5\pi }{4}+x)\cdot cos(\frac{5\pi }{4}-x)}=\\\\=\frac{1}{\frac{1}{2}\cdot (cos\frac{5\pi }{2}+cos2x)}}=\frac{2}{1+cos2x}=\frac{2}{2cos^2x}=\frac{1}{cos^2x}=1+tg^2x\\\\\\\\P.S.\; \; \; \; tg(\alpha +\beta )=\frac{sin(\alpha +\beta )}{cos\alpha \cdot cos\beta }\; \; \; ,\; \; \; cos^2\alpha =\frac{1+cos2\alpha }{2}\; \; ,\\\\1+tg^2\alpha =\frac{1}{cos^2\alpha }$

Answer 2

The expression you provided seems to be a combination of trigonometric functions. However, there is a small mistake: the correct notation for the tangent function is "tan," not "tg." Assuming you meant "tan" instead of "tg," the expression would be:

tan(5n/4 + x) + tan(5n/4 - x)

where "tan" represents the tangent function and "n" and "x" are variables.

Unfortunately, I cannot simplify this expression further without knowing the specific values of "n" and "x." If you have specific values for these variables or any other constraints, I could help you simplify the expression or provide a numerical result. Otherwise, this is the simplified form of the expression with the given trigonometric functions.