1/(cot^2 (x)) - 1/(cos^2 (x))

Mathematics

2 answers:

Airida [17]3 years ago

5 0

Step-by-step explanation:

1 is your answer its law of trig identises

tia_tia [17]3 years ago

5 0

$\\ \sf\longmapsto \dfrac{1}{cot^2x}-\dfrac{1}{cos^2x}$

$\\ \sf\longmapsto tan^2x-sec^2x$

$\\ \sf\longmapsto 1$

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The midpoint of AB is at (3,7) and A is at (0,-5). Where is B located?

olga nikolaevna [1]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
% (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\qquad thus
\\
----------------------------\\$ $\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 0}}\quad ,&{{ -5}})\quad 
% (c,d)
B&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
(3,7)\impliedby midpoint\qquad thus
\\ \quad \\
\left(\cfrac{{{ x_2 }} + {{ 0}}}{2}=3\quad ,\quad \cfrac{{{ y_2 }} + {{( -5)}}}{2}=7 \right)\to 
\begin{cases}
\cfrac{{{ x_2 }} + {{ 0}}}{2}=3
\\ \quad \\
\cfrac{{{ y_2 }} + {{ -5}}}{2}=7
\end{cases}
\\ \quad \\
solve\ for\ x_2\ and\ y_2$