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

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

Solution

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

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

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}}

Simplify

= 1

The two sides are not equal

\Rightarrow False

sin (θ - \frac{3 π}{2}) = cos (θ)

cos^{2} (x) + 2 sin (x) = 2

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

sin (27 0^{\circ} + x) = - cos (x)

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