Is sqrt((1-sin(x))/(1+sin(x)))=sec(x)-tan(x) ?

The answer to whether sqrt((1-sin(x))/(1+sin(x)))=sec(x)-tan(x) is False

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prove sqrt((1-sin(x))/(1+sin(x)))=sec(x)-tan(x)

prove

\frac{1 - sin ( x )}{1 + sin ( x )} = sec (x) - tan (x)

False

Solution steps

\frac{1 - sin ( x )}{1 + sin ( x )} = sec (x) - tan (x)

Show that the two sides are not equal

For

\frac{1 - sin ( x )}{1 + sin ( x )} = sec (x) - tan (x)

plug in

x = π

\frac{1 - sin ( π )}{1 + sin ( π )} = 1

\frac{1 - sin ( π )}{1 + sin ( π )}

Use the following trivial identity:

sin (π) = 0

sin (π)

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

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

Simplify

= 1

sec (π) - tan (π) = - 1

sec (π) - tan (π)

Use the following trivial identity:

sec (π) = (- 1)

sec (π)

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= (- 1)

Use the following trivial identity:

tan (π) = 0

tan (π)

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

= - 1 - 0

Simplify

= - 1

The two sides are not equal

\Rightarrow False

Is sqrt((1-sin(x))/(1+sin(x)))=sec(x)-tan(x) ?
The answer to whether sqrt((1-sin(x))/(1+sin(x)))=sec(x)-tan(x) is False