Is 1/(1+sin(x))+1/(1+csc(x))=1 ?

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

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prove 1/(1+sin(x))+1/(1+csc(x))=1

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

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

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

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

\frac{1}{1 + \frac{1}{s i n ( x )}}

Join

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

1 + \frac{1}{sin ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Apply the fraction rule:

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

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

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

Apply rule

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (θ) cos^{2} (θ) = 1

sec^{2} (x) cos^{2} (x) + cot^{2} (x) = csc^{2} (x)

\frac{1 - cos ( θ ) - 1 + cos ( θ )}{sin ( θ )} = 0

- (sin (x) + 2) (sin (x) - 2) = 4 - sin^{2} (x)

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

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