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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

Refine

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

sec (y) - cos (y)

Express with sin, cos

- cos (y) + sec (y)

Use the basic trigonometric identity:

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

= - cos (y) + \frac{1}{cos ( y )}

Simplify

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

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

Convert element to fraction:

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

= - \frac{cos ( y ) cos ( y )}{cos ( y )} + \frac{1}{cos ( y )}

Since the denominators are equal, combine the fractions:

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

= \frac{- cos ( y ) cos ( y ) + 1}{cos ( y )}

- cos (y) cos (y) + 1 = - cos^{2} (y) + 1

- cos (y) cos (y) + 1

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)

= - cos^{2} (y) + 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

cos (3 x) = cos (2 x + x)

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

2 - csc^{2} (x) = 1 - cot^{2} (x)

1 - sin^{2} (θ) = cos^{2} (θ)

sin (8 x) = 2 sin (4 x) cos (4 x)

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