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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (θ)

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

Factor out common term

sin (θ)

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

Cancel the common factor:

cos (θ) + 1

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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