Is (1-cos^2(t))(1+cos^2(t))=1 ?

The answer to whether (1-cos^2(t))(1+cos^2(t))=1 is False

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prove (1-cos^2(t))(1+cos^2(t))=1

prove

(1 - cos^{2} (t)) (1 + cos^{2} (t)) = 1

False

Solution steps

(1 - cos^{2} (t)) (1 + cos^{2} (t)) = 1

Show that the two sides are not equal

For

(1 - cos^{2} (t)) (1 + cos^{2} (t)) = 1

plug in

t = 0

(1 - cos^{2} (0)) (1 + cos^{2} (0)) = 0

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

= (1 - 1^{2}) (1 + 1^{2})

Simplify

= 0

The two sides are not equal

\Rightarrow False

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

csc (x) + cot^{2} (x) csc (x) = csc^{3} (x)

cot (x) = cos (x) arcsin (x)

\frac{1}{sin ( x ) sec ( x )} = cot (x)

3 + sin^{2} (x) = 4 - cos^{2} (x)

Is (1-cos^2(t))(1+cos^2(t))=1 ?
The answer to whether (1-cos^2(t))(1+cos^2(t))=1 is False