Is (1-tan^2(x))/(1+tan^2(x))=tan(x) ?

The answer to whether (1-tan^2(x))/(1+tan^2(x))=tan(x) is False

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prove (1-tan^2(x))/(1+tan^2(x))=tan(x)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

\frac{1 - tan ^{2} ( 0 )}{1 + tan ^{2} ( 0 )}

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= \frac{1 - 0 ^{2}}{1 + 0 ^{2}}

Simplify

= 1

tan (0) = 0

tan (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= 0

The two sides are not equal

\Rightarrow False

\frac{1 - cos ( 2 x )}{tan ( x )} = sin (2 x)

\frac{cos ( θ )}{sec ( θ ) - tan ( θ )} = 1 sin (θ)

cot (2 x) + csc (2 x) = cot (x)

tan (x) + 2 cos (x) \cdot csc (x) = sec (x) \cdot csc (x) + cot (x)

- \frac{cos ( b )}{1 - sin ( b )} + \frac{cos ( b )}{1 + sin ( b )} = - 2 tan (b)

Is (1-tan^2(x))/(1+tan^2(x))=tan(x) ?
The answer to whether (1-tan^2(x))/(1+tan^2(x))=tan(x) is False