Is prove-tan(t)-(cos(t))/(sin(t)-1)=sec(x) ?

The answer to whether prove-tan(t)-(cos(t))/(sin(t)-1)=sec(x) is False

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prove-tan(t)-(cos(t))/(sin(t)-1)=sec(x)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

t = 0

For

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

plug in

x = 1

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

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

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

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

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

Simplify

= 1

sec (1) = 1.85081 \dots

sec (1)

Refine to a decimal form

= 1.85081 \dots

The two sides are not equal

\Rightarrow False

\frac{cot ( x )}{tan ( x ) sec ( x )} = csc (x) - 1

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

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

tan^{2} (A) + 1 = sec^{2} (A)

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

Is prove-tan(t)-(cos(t))/(sin(t)-1)=sec(x) ?
The answer to whether prove-tan(t)-(cos(t))/(sin(t)-1)=sec(x) is False