Is csc(x)sec(x)= 1/(sin(x)cos(x)) ?

The answer to whether csc(x)sec(x)= 1/(sin(x)cos(x)) is True

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prove csc(x)sec(x)= 1/(sin(x)cos(x))

prove

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

True

Solution steps

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

Manipulating left side

csc (x) sec (x)

Express with sin, cos

csc (x) sec (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( x )} sec (x)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} : \frac{1}{sin ( x ) cos ( x )}

\frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

csc (Θ) + sin (- Θ) = \frac{cos ^{2} ( Θ )}{sin ( Θ )}

csc (A) = \frac{cos ( 2 A )}{sin ( A )} + \frac{sin ( 2 A )}{cos ( A )}

sin^{2} (θ) = 1 + cos^{2} (θ)

sin (\frac{π}{2}) = \frac{4}{2}

Is csc(x)sec(x)= 1/(sin(x)cos(x)) ?
The answer to whether csc(x)sec(x)= 1/(sin(x)cos(x)) is True