Is (cot(y))/(sec(y))=csc(y)-sin(y) ?

The answer to whether (cot(y))/(sec(y))=csc(y)-sin(y) is True

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prove (cot(y))/(sec(y))=csc(y)-sin(y)

prove

\frac{cot ( y )}{sec ( y )} = csc (y) - sin (y)

True

Solution steps

\frac{cot ( y )}{sec ( y )} = csc (y) - sin (y)

Manipulating left side

\frac{cot ( y )}{sec ( y )}

Express with sin, cos

\frac{cot ( y )}{sec ( y )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( y )}{s i n ( y )}}{sec ( y )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( y )}{s i n ( y )}}{\frac{1}{c o s ( y )}}

Simplify

\frac{\frac{c o s ( y )}{s i n ( y )}}{\frac{1}{c o s ( y )}} : \frac{cos ^{2} ( y )}{sin ( y )}

\frac{\frac{c o s ( y )}{s i n ( y )}}{\frac{1}{c o s ( y )}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{cos ( y ) cos ( y )}{sin ( y ) \cdot 1}

Refine

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

cos (y) cos (y) = cos^{2} (y)

cos (y) cos (y)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (y) cos (y) = cos^{1 + 1} (y)

= cos^{1 + 1} (y)

Add the numbers:

1 + 1 = 2

= cos^{2} (y)

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

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

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

Manipulating right side

csc (y) - sin (y)

Express with sin, cos

csc (y) - sin (y)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( y )} - sin (y)

Simplify

\frac{1}{sin ( y )} - sin (y) : \frac{1 - sin ^{2} ( y )}{sin ( y )}

\frac{1}{sin ( y )} - sin (y)

Convert element to fraction:

sin (y) = \frac{sin ( y ) sin ( y )}{sin ( y )}

= \frac{1}{sin ( y )} - \frac{sin ( y ) sin ( y )}{sin ( y )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - sin ( y ) sin ( y )}{sin ( y )}

1 - sin (y) sin (y) = 1 - sin^{2} (y)

1 - sin (y) sin (y)

sin (y) sin (y) = sin^{2} (y)

sin (y) sin (y)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (y) sin (y) = sin^{1 + 1} (y)

= sin^{1 + 1} (y)

Add the numbers:

1 + 1 = 2

= sin^{2} (y)

= 1 - sin^{2} (y)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( y )}{sin ( y )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

csc (θ) = cos (θ) cot (θ) + sin (θ)

cos^{4} (x) + sin^{2} (x) cos^{2} (x) = cos^{2} (x)

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

\frac{( sec ^{2} ( x ) )}{tan ( x )} = sec (x) csc (x)

Is (cot(y))/(sec(y))=csc(y)-sin(y) ?
The answer to whether (cot(y))/(sec(y))=csc(y)-sin(y) is True