Is 1/(cos^2(x))-2=0 ?

The answer to whether 1/(cos^2(x))-2=0 is False

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prove 1/(cos^2(x))-2=0

prove

\frac{1}{cos ^{2} ( x )} - 2 = 0

False

Solution steps

\frac{1}{cos ^{2} ( x )} - 2 = 0

Show that the two sides are not equal

For

\frac{1}{cos ^{2} ( x )} - 2 = 0

plug in

x = 0

\frac{1}{cos ^{2} ( 0 )} - 2 = - 1

\frac{1}{cos ^{2} ( 0 )} - 2

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= \frac{1}{1 ^{2}} - 2

Simplify

= - 1

The two sides are not equal

\Rightarrow False

sin (θ) = cos^{2} (θ) + cos^{2} (θ) \cdot sin (θ)

sin (a) = tan (a) cos (a)

cos (x) sec (x) - sin^{2} (x) = cos^{2} (x)

\frac{sin ( 2 x )}{cos ( 2 x ) + sin ^{2} ( x )} = 2 tan (x)

tan (θ) sec (θ) (csc (θ) - sin (θ)) = 1