Is (tan(y))/(csc(y))= 1/(cos(y))-1/(sec(y)) ?

The answer to whether (tan(y))/(csc(y))= 1/(cos(y))-1/(sec(y)) is True

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prove (tan(y))/(csc(y))= 1/(cos(y))-1/(sec(y))

prove

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

True

Solution steps

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

Manipulating left side

\frac{tan ( y )}{csc ( y )}

Express with sin, cos

= \frac{sin ^{2} ( y )}{cos ( y )}

Rewrite using trig identities

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

Manipulating right side

\frac{1}{cos ( y )} - \frac{1}{sec ( y )}

Express with sin, cos

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin^{2} (x) (csc^{2} (x) - 1) = cos^{2} (x)

\frac{1}{sin ( x ) cot ( x )} = \frac{1}{cos ( x )}

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

Is (tan(y))/(csc(y))= 1/(cos(y))-1/(sec(y)) ?
The answer to whether (tan(y))/(csc(y))= 1/(cos(y))-1/(sec(y)) is True