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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

Factor

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

sec^{2} (A) - 1

Rewrite

1

1^{2}

= sec^{2} (A) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sec (A) + 1) (sec (A) - 1)

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

Cancel the common factor:

sec (A) + 1

= sec (A) - 1

= 1 + sec (A) - 1

Group like terms

= sec (A) + 1 - 1

1 - 1 = 0

= sec (A)

= sec (A)

= sec (A)

We showed that the two sides could take the same form

\Rightarrow True

sin (x - \frac{π}{6}) + cos (x) = sin (x + π)

tan^{2} (2 3^{\circ}) - sec^{2} (2 3^{\circ}) = - 1

(1 - cos^{2} (t)) (1 + cos^{2} (t)) = 1

cos (x) (sec^{2} (x)) = sec (x)

csc (x) + cot^{2} (x) csc (x) = csc^{3} (x)

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