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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{tan ^{2} ( b )}{tan ( b )}

Cancel the common factor:

tan (b)

= tan (b)

= tan (b)

We showed that the two sides could take the same form

\Rightarrow True

cos (- θ + \frac{π}{2}) = sin (θ)

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

- 1 - cot^{2} (x) = - csc^{2} (x)

sin (2 x) + cos (x) = 0

cot (x) (tan (x) sin (x) + cos (x)) = csc (x)

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