Is (sin(y)+tan(y))/(1+sec(y))=sin(y) ?

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

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prove (sin(y)+tan(y))/(1+sec(y))=sin(y)

prove

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

True

Solution steps

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

Manipulating left side

\frac{sin ( y ) + tan ( y )}{1 + sec ( y )}

Express with sin, cos

\frac{sin ( y ) + tan ( y )}{1 + sec ( y )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (y) = cos (y)

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

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

Join

sin (y) + \frac{sin ( y )}{cos ( y )} : \frac{sin ( y ) cos ( y ) + sin ( y )}{cos ( y )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (y)

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

Factor out common term

sin (y)

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

Cancel the common factor:

cos (y) + 1

= sin (y)

= sin (y)

= sin (y)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e cos^{2} (2 x) - sin^{2} (2 x) = cos (4 x)

p ro v e \frac{sec ^{2} ( x ) - 1}{sin ( x )} = \frac{sin ( x )}{1 - sin ^{2} ( x )}

p ro v e sin (\frac{π}{4} + x) + sin (\frac{π}{4} - x) = 2 cos (x)

p ro v e sin (θ) sec (θ) cot (θ) = 1

p ro v e 1 + sec^{2} (θ) sin^{2} (θ) = sec^{2} (θ)

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