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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{sec ( t ) - 1}{tan ( t )}

Express with sin, cos

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

Rewrite using trig identities

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

Manipulating right side

\frac{tan ( t )}{sec ( t ) + 1}

Express with sin, cos

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

We showed that the two sides could take the same form

\Rightarrow True

3 cos^{2} (x) - 3 sin^{2} (x) = 3 - 6 sin^{2} (x)

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

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

sec^{4} (x) = \frac{1}{cos ^{4} ( x )}

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

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