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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

For

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

plug in

b = 1

sec (1) + tan (1) = 3.40822 \dots

sec (1) + tan (1)

Refine to a decimal form

= 3.40822 \dots

\frac{cos ( 1 )}{1 - sin ( 0 )} = 0.54030 \dots

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

Refine to a decimal form

= 0.54030 \dots

The two sides are not equal

\Rightarrow False

2 sinh (ln (x)) + \frac{1 ^{2}}{2} - \frac{4}{3} = 2

2 cot^{2} (x) sin^{2} (x) = 1 + cos (2 x)

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

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

tan (u) + cos (2 u) \cdot tan (u) = sin (2 u)

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