Is tan(t)sin(t)=(sec^2(t)-1)/(sec(t)) ?

The answer to whether tan(t)sin(t)=(sec^2(t)-1)/(sec(t)) is True

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prove tan(t)sin(t)=(sec^2(t)-1)/(sec(t))

Solution

prove

tan (t) sin (t) = \frac{sec ^{2} ( t ) - 1}{sec ( t )}

Solution

True

Solution steps

tan (t) sin (t) = \frac{sec ^{2} ( t ) - 1}{sec ( t )}

Manipulating left side

tan (t) sin (t)

Express with sin, cos

sin (t) tan (t)

Use the basic trigonometric identity:

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

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

Simplify

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

sin (t) \frac{sin ( t )}{cos ( t )}

Multiply fractions:

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

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

sin (t) sin (t) = sin^{2} (t)

sin (t) sin (t)

Apply exponent rule:

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

sin (t) sin (t) = sin^{1 + 1} (t)

= sin^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= sin^{2} (t)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1 - ( \frac{1}{s e c ( t )} ) ^{2}}{\frac{1}{s e c ( t )}}

Simplify

\frac{1 - ( \frac{1}{s e c ( t )} ) ^{2}}{\frac{1}{s e c ( t )}}

Apply the fraction rule:

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

= \frac{( 1 - ( \frac{1}{s e c ( t )} ) ^{2} ) sec ( t )}{1}

(\frac{1}{sec ( t )})^{2} = \frac{1}{sec ^{2} ( t )}

(\frac{1}{sec ( t )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sec ^{2} ( t )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( t )}

= \frac{sec ( t ) ( - \frac{1}{s e c ^{2} ( t )} + 1 )}{1}

Join

1 - \frac{1}{sec ^{2} ( t )} : \frac{sec ^{2} ( t ) - 1}{sec ^{2} ( t )}

1 - \frac{1}{sec ^{2} ( t )}

Convert element to fraction:

1 = \frac{1 sec ^{2} ( t )}{sec ^{2} ( t )}

= \frac{1 \cdot sec ^{2} ( t )}{sec ^{2} ( t )} - \frac{1}{sec ^{2} ( t )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sec ^{2} ( t ) - 1}{sec ^{2} ( t )}

Multiply:

1 \cdot sec^{2} (t) = sec^{2} (t)

= \frac{sec ^{2} ( t ) - 1}{sec ^{2} ( t )}

= \frac{\frac{s e c ^{2} ( t ) - 1}{s e c ^{2} ( t )} sec ( t )}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{sec ^{2} ( t ) - 1}{sec ^{2} ( t )} sec (t)

Multiply fractions:

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

= \frac{( sec ^{2} ( t ) - 1 ) sec ( t )}{sec ^{2} ( t )}

Cancel the common factor:

sec (t)

= \frac{sec ^{2} ( t ) - 1}{sec ( t )}

\frac{sec ^{2} ( t ) - 1}{sec ( t )}

\frac{sec ^{2} ( t ) - 1}{sec ( t )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan(t)sin(t)=(sec^2(t)-1)/(sec(t)) ?
The answer to whether tan(t)sin(t)=(sec^2(t)-1)/(sec(t)) is True