Is cot(pi/2-x)csc(x)=sec(x) ?

The answer to whether cot(pi/2-x)csc(x)=sec(x) is True

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prove cot(pi/2-x)csc(x)=sec(x)

prove

cot (\frac{π}{2} - x) csc (x) = sec (x)

True

Solution steps

cot (\frac{π}{2} - x) csc (x) = sec (x)

Manipulating left side

cot (\frac{π}{2} - x) csc (x)

Rewrite using trig identities

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Express with sin, cos

= \frac{1}{cos ( x )}

Rewrite using trig identities

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

csc^{2} (x) + sec^{2} (x) = csc^{2} (x) sec^{2} (x)

cos (6 0^{\circ}) = cos^{2} (3 0^{\circ}) - sin^{2} (3 0^{\circ})

sec^{2} (θ) cot (θ) = tan (θ) + cot (θ)

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

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

Is cot(pi/2-x)csc(x)=sec(x) ?
The answer to whether cot(pi/2-x)csc(x)=sec(x) is True