Is (cos(x))/(1-sin(x))=tan(x/2-pi/4) ?

The answer to whether (cos(x))/(1-sin(x))=tan(x/2-pi/4) is False

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prove (cos(x))/(1-sin(x))=tan(x/2-pi/4)

prove

\frac{cos ( x )}{1 - sin ( x )} = tan (\frac{x}{2} - \frac{π}{4})

False

Solution steps

\frac{cos ( x )}{1 - sin ( x )} = tan (\frac{x}{2} - \frac{π}{4})

Show that the two sides are not equal

For

\frac{cos ( x )}{1 - sin ( x )} = tan (\frac{x}{2} - \frac{π}{4})

plug in

x = 0

\frac{cos ( 0 )}{1 - sin ( 0 )} = 1

\frac{cos ( 0 )}{1 - sin ( 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

Use the following trivial identity:

sin (0) = 0

sin (0)

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}{1 - 0}

Simplify

= 1

tan (\frac{0}{2} - \frac{π}{4}) = - 1

tan (\frac{0}{2} - \frac{π}{4})

Simplify

= tan (- \frac{π}{4})

Use the following property:

tan (- x) = - tan (x)

tan (- \frac{π}{4}) = - tan (\frac{π}{4})

= - tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

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}

= 1

= - 1

The two sides are not equal

\Rightarrow False

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

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

cot (x - y) = \frac{cot ( x ) cot ( y ) + 1}{cot ( y ) - cot ( x )}

\frac{cot ( a )}{cos ( a )} = csc (a)

cos (2 θ) = \frac{2 - sec ^{2} ( θ )}{sec ^{2} ( θ )}

Is (cos(x))/(1-sin(x))=tan(x/2-pi/4) ?
The answer to whether (cos(x))/(1-sin(x))=tan(x/2-pi/4) is False