Question

tan^2(4)-sec^2(4)=

Answer 1

Question 21019

$tan^{2}(4)-sec^{2}(4)=$ Answer.-1.

B is the midpoint of line $AC$. $AB=5x$ and $BC=2x+3$. Find $AB$, $BC$, and $AC$.

Solution.

Since

$sec^{2}(x)=\frac{1}{cos^{2}(4)}$, then using the equality $cos^{2}(4)+sin^{2}(4)=1$, we get

$sec^{2}(4)=\frac{cos^{2}(4)+sin^{2}(4)}{cos^{2}(4)}=1+tan^{2}(4).$

Finally,

$tan^{2}(4)-sec^{2}(4)=tan^{2}(4)-(1+tan^{2}(4))=-1$

Answer. -1.