Is (tan^2(θ))/(sec(θ)+1)=sec(θ)-1 ?

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

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prove (tan^2(θ))/(sec(θ)+1)=sec(θ)-1

prove

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

True

Solution steps

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

Manipulating left side

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1}

Rewrite using trig identities

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

sec^{2} (θ) - 1

Rewrite

1

1^{2}

= sec^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sec (θ) + 1) (sec (θ) - 1)

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

Cancel the common factor:

sec (θ) + 1

= sec (θ) - 1

= sec (θ) - 1

= sec (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

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

cot (x) - 1 = \frac{csc ( x ) - sec ( x )}{sec ( x )}

tan (θ) = 2

sec (x) = sec (- x)

csc^{2} (θ) - 1 = cot^{2} (θ)

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