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

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

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

prove

cos (B) csc (B) tan (B) = 1

True

Solution steps

cos (B) csc (B) tan (B) = 1

Manipulating left side

cos (B) csc (B) tan (B)

Express with sin, cos

cos (B) csc (B) tan (B)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (B)

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

Cancel the common factor:

cos (B)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ( θ ) + cot ( θ )}{tan ( θ )} = csc^{2} (θ)

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

\frac{cos ( θ )}{1 + sin ( θ )} + tan (θ) = sec (θ)

cos (x) \cdot tan (x) = sin (x)

\frac{sin ( 2 θ )}{sin ( θ )} - \frac{cos ( 2 θ )}{cos ( θ )} = sec (θ)