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

The answer to whether 1/(1+sin(t))=(sec(t)-tan(t))sec(t) is True

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prove 1/(1+sin(t))=(sec(t)-tan(t))sec(t)

prove

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

True

Solution steps

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

Manipulating right side

(sec (t) - tan (t)) sec (t)

Express with sin, cos

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

Rewrite using trig identities

= \frac{1}{sin ( t ) + 1}

= \frac{1}{1 + sin ( t )}

We showed that the two sides could take the same form

\Rightarrow True

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

sin (\frac{π}{3}) = \frac{3}{2}

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

cos (t) sec (\frac{π}{2} - t) = cot (t)

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

Is 1/(1+sin(t))=(sec(t)-tan(t))sec(t) ?
The answer to whether 1/(1+sin(t))=(sec(t)-tan(t))sec(t) is True