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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

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

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

Simplify

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

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

Factor

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

sec^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (sec (x) + 1) (sec (x) - 1)

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

Cancel the common factor:

sec (x) - 1

= sec (x) + 1

= sec (x) + 1

= sec (x) + 1

Express with sin, cos

1 + sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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{cos ( x ) + 1}{cos ( x )}

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

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

cos (α + β) - cos (α - β) = - 2 sin (α) cos (β)

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

7 sec (θ) = \frac{7 sec ( θ ) - 7}{1 - cos ( θ )}

2 tan (x) sin (x) - 3 cos (x) = 2

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

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