Is 1/(sec(x)+1)+1/(cos(x)+1)=1 ?

The answer to whether 1/(sec(x)+1)+1/(cos(x)+1)=1 is True

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prove 1/(sec(x)+1)+1/(cos(x)+1)=1

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{1}{1 + cos ( x )} + \frac{1}{1 + \frac{1}{c o s ( x )}}

\frac{1}{1 + cos ( x )} + \frac{1}{1 + \frac{1}{c o s ( x )}} = 1

\frac{1}{1 + cos ( x )} + \frac{1}{1 + \frac{1}{c o s ( x )}}

\frac{1}{1 + \frac{1}{c o s ( x )}} = \frac{cos ( x )}{cos ( x ) + 1}

\frac{1}{1 + \frac{1}{c o s ( x )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

= \frac{1}{\frac{c o s ( x ) + 1}{c o s ( x )}}

Apply the fraction rule:

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

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

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

Apply rule

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 - sin ^{4} ( x )}{cos ^{2} ( x )} = 2 - cos^{2} (x)

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

tan (x) cos (x) = sin (x)

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

sin (a + b) + sin (a - b) = 2 sin (a) cos (b)

Is 1/(sec(x)+1)+1/(cos(x)+1)=1 ?
The answer to whether 1/(sec(x)+1)+1/(cos(x)+1)=1 is True