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