Is ((tan^2(x)))/(sec(x)+1)=((1-cos(x)))/(cos(x)) ?

The answer to whether ((tan^2(x)))/(sec(x)+1)=((1-cos(x)))/(cos(x)) is True

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Popular Trigonometry >

prove ((tan^2(x)))/(sec(x)+1)=((1-cos(x)))/(cos(x))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{tan ^{2} ( x )}{sec ( x ) + 1}

Express with sin, cos

\frac{tan ^{2} ( x )}{1 + sec ( x )}

Use the basic trigonometric identity:

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

= \frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{1 + sec ( x )}

Use the basic trigonometric identity:

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

= \frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{1 + \frac{1}{c o s ( x )}}

Simplify

\frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{1 + \frac{1}{c o s ( x )}} : \frac{sin ^{2} ( x )}{cos ( x ) ( cos ( x ) + 1 )}

\frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{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{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{\frac{c o s ( x ) + 1}{c o s ( x )}}

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

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

= \frac{\frac{s i n ^{2} ( x )}{c o s ^{2} ( x )}}{\frac{c o s ( x ) + 1}{c o s ( x )}}

Divide fractions:

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

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

Cancel the common factor:

cos (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

1 - cos^{2} (x) : - (cos (x) + 1) (cos (x) - 1)

1 - cos^{2} (x)

Factor out common term

- 1

= - (cos^{2} (x) - 1)

Factor

cos^{2} (x) - 1 : (cos (x) + 1) (cos (x) - 1)

cos^{2} (x) - 1

Rewrite

1

1^{2}

= cos^{2} (x) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

Cancel the common factor:

cos (x) + 1

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

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

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

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

Simplify

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is ((tan^2(x)))/(sec(x)+1)=((1-cos(x)))/(cos(x)) ?
The answer to whether ((tan^2(x)))/(sec(x)+1)=((1-cos(x)))/(cos(x)) is True