Is 9/(sec(x)+1)-9/(sec(x)-1)=0 ?

The answer to whether 9/(sec(x)+1)-9/(sec(x)-1)=0 is False

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prove 9/(sec(x)+1)-9/(sec(x)-1)=0

prove

\frac{9}{sec ( x ) + 1} - \frac{9}{sec ( x ) - 1} = 0

False

Solution steps

\frac{9}{sec ( x ) + 1} - \frac{9}{sec ( x ) - 1} = 0

Show that the two sides are not equal

For

\frac{9}{sec ( x ) + 1} - \frac{9}{sec ( x ) - 1} = 0

plug in

x = 1

\frac{9}{sec ( 1 ) + 1} - \frac{9}{sec ( 1 ) - 1} = - 7.42109 \dots

\frac{9}{sec ( 1 ) + 1} - \frac{9}{sec ( 1 ) - 1}

Refine to a decimal form

= - 7.42109 \dots

The two sides are not equal

\Rightarrow False

sec^{2} (x) + csc^{2} (x) = (sec (x) csc (x))^{2}

tan (x) = \frac{sin ( x ) csc ( x )}{cot ( x )}

tan (x) sec (x) cos (x) = tan (x)

\frac{1 + sin ( x )}{sin ( x )} = 1 + csc (x)

cos (2 y) = \frac{1 - tan ^{2} ( y )}{1 + tan ^{2} ( y )}

Is 9/(sec(x)+1)-9/(sec(x)-1)=0 ?
The answer to whether 9/(sec(x)+1)-9/(sec(x)-1)=0 is False