Is (tan(θ)sin(θ))/(sec(θ)-1)=1+cos(θ) ?

The answer to whether (tan(θ)sin(θ))/(sec(θ)-1)=1+cos(θ) is True

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prove (tan(θ)sin(θ))/(sec(θ)-1)=1+cos(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{tan ( θ ) sin ( θ )}{sec ( θ ) - 1}

Express with sin, cos

\frac{sin ( θ ) tan ( θ )}{- 1 + sec ( θ )}

Use the basic trigonometric identity:

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

= \frac{sin ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{- 1 + sec ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

- 1 + \frac{1}{cos ( θ )} : \frac{- cos ( θ ) + 1}{cos ( θ )}

- 1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{- cos ( θ ) + 1}{cos ( θ )}

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

Multiply

sin (θ) \frac{sin ( θ )}{cos ( θ )} : \frac{sin ^{2} ( θ )}{cos ( θ )}

sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) sin ( θ )}{cos ( θ )}

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Divide fractions:

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

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

Cancel the common factor:

cos (θ)

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{1 - cos ( θ )}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (θ)

Factor out common term

- 1

= - (cos^{2} (θ) - 1)

Factor

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

cos^{2} (θ) - 1

Rewrite

1

1^{2}

= cos^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (cos (θ) + 1) (cos (θ) - 1)

= - (cos (θ) + 1) (cos (θ) - 1)

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

Cancel

- \frac{( cos ( θ ) + 1 ) ( cos ( θ ) - 1 )}{1 - cos ( θ )} : cos (θ) + 1

- \frac{( cos ( θ ) + 1 ) ( cos ( θ ) - 1 )}{1 - cos ( θ )}

- cos (θ) + 1 = - (cos (θ) - 1)

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

Refine

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

Cancel the common factor:

cos (θ) - 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= 1 + cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (tan(θ)sin(θ))/(sec(θ)-1)=1+cos(θ) ?
The answer to whether (tan(θ)sin(θ))/(sec(θ)-1)=1+cos(θ) is True