Is tan(θ)*csc(θ)= 1/(cos(θ)) ?

The answer to whether tan(θ)*csc(θ)= 1/(cos(θ)) is True

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prove tan(θ)*csc(θ)= 1/(cos(θ))

prove

tan (θ) \cdot csc (θ) = \frac{1}{cos ( θ )}

True

Solution steps

tan (θ) csc (θ) = \frac{1}{cos ( θ )}

Manipulating left side

tan (θ) csc (θ)

Express with sin, cos

csc (θ) tan (θ)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )} tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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

cos (2 θ) = \frac{csc ^{2} ( θ ) - 2}{csc ^{2} ( θ )}

tan (a) = \frac{1}{cot ( a )}

cos^{4} (x) + sin^{4} (x) = 1

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

Is tan(θ)*csc(θ)= 1/(cos(θ)) ?
The answer to whether tan(θ)*csc(θ)= 1/(cos(θ)) is True