Is (csc(θ)+1)/(1+sin(θ))=csc(θ) ?

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

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prove (csc(θ)+1)/(1+sin(θ))=csc(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ( θ ) + 1}{1 + sin ( θ )}

Express with sin, cos

\frac{1 + csc ( θ )}{1 + sin ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 + \frac{1}{sin ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ ) + 1}{sin ( θ )}

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin (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

sec (A) csc (A) = csc (A) tan (A)

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

sin (x) \cdot tan (x) + cos (- x) = sec (x)

cos (2 α) = cos^{2} (α) - sin^{2} (α)

\frac{tan ( x )}{1 + tan ^{2} ( x )} = cos (x) sin (x)

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