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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

(sec (x) + tan (x))^{2}

Express with sin, cos

(sec (x) + tan (x))^{2}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - sin^{2} (x)

Factor out common term

- 1

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

Factor

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

sin^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (sin (x) + 1) (sin (x) - 1)

= - (sin (x) + 1) (sin (x) - 1)

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

Cancel the common factor:

sin (x) + 1

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

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

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

Manipulating right side

\frac{csc ( x ) + 1}{csc ( x ) - 1}

Express with sin, cos

\frac{1 + csc ( x )}{- 1 + csc ( x )}

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1 + \frac{1}{s i n ( x )}}{- 1 + \frac{1}{s i n ( x )}}

Simplify

\frac{1 + \frac{1}{s i n ( x )}}{- 1 + \frac{1}{s i n ( x )}} : \frac{sin ( x ) + 1}{- sin ( x ) + 1}

\frac{1 + \frac{1}{s i n ( x )}}{- 1 + \frac{1}{s i n ( x )}}

Join

- 1 + \frac{1}{sin ( x )} : \frac{- sin ( x ) + 1}{sin ( x )}

- 1 + \frac{1}{sin ( x )}

Convert element to fraction:

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

= - \frac{1 \cdot sin ( x )}{sin ( x )} + \frac{1}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot sin ( x ) + 1}{sin ( x )}

Multiply:

1 \cdot sin (x) = sin (x)

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

= \frac{1 + \frac{1}{s i n ( x )}}{\frac{- s i n ( x ) + 1}{s i n ( x )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

= \frac{\frac{s i n ( x ) + 1}{s i n ( x )}}{\frac{- s i n ( x ) + 1}{s i n ( x )}}

Divide fractions:

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

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

Cancel the common factor:

sin (x)

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

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

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

Multiply by

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

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

Factor

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

missing stepId = G e n er i c S im pl i f y T i tl e 0 Eq

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

Simplify

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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