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

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

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

prove

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

True

Solution steps

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{sin ( θ )}{csc ( θ ) - cot ( θ )} = 1 + cos (θ)

sec (θ) - tan (θ) sin (θ) = \frac{1}{sec ( θ )}

cot (- a) cos (- a) + sin (- a) = - csc (a)

sin (2 a) - tan (a) = tan (a) cos (2 a)

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