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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Factor

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

- 1 + sec^{2} (x)

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 )}{- 1 + sec ( x )}

Cancel the common factor:

sec (x) - 1

= sec (x) + 1

= 1 + sec (x)

= sec (x) + 1

We showed that the two sides could take the same form

\Rightarrow True

sin (θ) - cos^{2} (θ) = sin^{3} (θ)

1 - tan^{2} (a) = cos (2 a) sec^{2} (a)

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

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

cos (0) tan (0) = sin (0)

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