Is cos(θ)*tan(θ)*csc(θ)=1 ?

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

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prove cos(θ)tan(θ)csc(θ)=1

prove

cos (θ) \cdot tan (θ) \cdot csc (θ) = 1

True

Solution steps

cos (θ) tan (θ) csc (θ) = 1

Manipulating left side

cos (θ) tan (θ) csc (θ)

Express with sin, cos

cos (θ) csc (θ) tan (θ)

Use the basic trigonometric identity:

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

= cos (θ) \frac{1}{sin ( θ )} tan (θ)

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= \frac{1 \cdot cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{1}{2} cot (\frac{x}{2}) - \frac{1}{2} tan (\frac{x}{2}) = cot (x)

p ro v e \frac{- cot ( x ) + 1}{tan ( x ) - 1} = cot (x)

p ro v e 4 cos^{2} (a) + 4 sin^{2} (a) = 4

p ro v e \frac{sin ( a )}{tan ( a )} = cos (a)

p ro v e sin^{2} (3 x) + cos^{2} (3 x) = 1

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