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

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

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

prove

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

True

Solution steps

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

Manipulating right 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}{cot ^{2} ( x )} + sec (x) cos (x) = csc^{2} (x)

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

1 csc (θ) - (cos (θ)) (cot (θ)) = sin (θ)

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

sin^{2} (3 x) = 9 sin^{3} (x) cos^{3} (x)

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