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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

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

Join

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

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (θ)

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

Factor out common term

sin (θ)

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

Cancel the common factor:

cos (θ) + 1

= sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

sec (2 θ) = \frac{sec ^{2} ( θ )}{1 - tan ^{2} ( θ )}

cos (2 a) = cos^{2} (a) - sin^{2} (a)

tan (θ) + cot (θ) = sec (θ) \cdot csc (θ)

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

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

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