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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

tan (x) sin (x)

Express with sin, cos

sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

Rewrite using trig identities

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( x ) - 1}{sec ^{2} ( x )}

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sec (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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