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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (a) = cos (a)

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (a)

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

Factor out common term

sin (a)

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

Cancel the common factor:

cos (a) + 1

= \frac{1}{sin ( a )}

= \frac{1}{sin ( a )}

= \frac{1}{sin ( a )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( a )}}

Simplify

\frac{1}{\frac{1}{c s c ( a )}}

Apply the fraction rule:

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

= \frac{csc ( a )}{1}

Apply rule

\frac{a}{1} = a

= csc (a)

csc (a)

csc (a)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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