Is tan(x)cos(x)csc(x)=1 ?

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

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prove tan(x)cos(x)csc(x)=1

prove

tan (x) cos (x) csc (x) = 1

True

Solution steps

tan (x) cos (x) csc (x) = 1

Manipulating left side

tan (x) cos (x) csc (x)

Express with sin, cos

cos (x) csc (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

Cancel the common factor:

cos (x)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{csc ^{4} ( x ) - 1}{1 + csc ^{2} ( x )} = cot^{2} (x)

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

p ro v e cos^{2} (x) = \frac{1}{2} (cos (2 x) + 1)

p ro v e (csc (x) - 1) (csc (x) + 1) = cot^{2} (x)

p ro v e tan^{2} (y) (csc^{2} (y) - 1) = 1