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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 + tan^{2} (\frac{π}{2} - x)

Rewrite using trig identities

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Simplify

1 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )} : \frac{sin ^{2} ( x ) + cos ^{2} ( x )}{sin ^{2} ( x )}

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

Rewrite using trig identities

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

Manipulating right side

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

Rewrite using trig identities

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

Remove parentheses:

(a) = a

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

cos (θ) - sin (θ) sin (2 θ) = cos (θ) cos (2 θ)

1 + (cot (A))^{2} = \cdot (csc (A))^{2}

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