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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

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 )}{1 + sec ( x )}

Cancel the common factor:

sec (x) + 1

= sec (x) - 1

= 1 + sec (x) - 1

Group like terms

= sec (x) + 1 - 1

1 - 1 = 0

= sec (x)

= sec (x)

= sec (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( 4 x )}{4} = \frac{sin ( x ) cos ( x )}{4}

csc^{2} (x) \cdot cos^{2} (x) = cot^{2} (x)

\frac{1}{sec ^{3} ( x ) cos ^{4} ( x )} = sec (x)

csc^{2} (θ) + 1 = cot^{2} (θ)

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

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