Is tan(θ)cot(θ)-1=1-sec(θ)cos(θ) ?

The answer to whether tan(θ)cot(θ)-1=1-sec(θ)cos(θ) is True

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prove tan(θ)cot(θ)-1=1-sec(θ)cos(θ)

prove

tan (θ) cot (θ) - 1 = 1 - sec (θ) cos (θ)

True

Solution steps

tan (θ) cot (θ) - 1 = 1 - sec (θ) cos (θ)

Manipulating left side

tan (θ) cot (θ) - 1

Express with sin, cos

- 1 + cot (θ) tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

- 1 + \frac{cos ( θ )}{sin ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )} = 0

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

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

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

Cancel the common factor:

cos (θ)

= \frac{sin ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= 1

= - 1 + 1

Simplify

= 0

= 0

= 0

Manipulating right side

1 - sec (θ) cos (θ)

Express with sin, cos

1 - cos (θ) sec (θ)

Use the basic trigonometric identity:

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

= 1 - cos (θ) \frac{1}{cos ( θ )}

1 - cos (θ) \frac{1}{cos ( θ )} = 0

1 - cos (θ) \frac{1}{cos ( θ )}

cos (θ) \frac{1}{cos ( θ )} = 1

cos (θ) \frac{1}{cos ( θ )}

Multiply fractions:

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

= \frac{1 \cdot cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= 1

= 1 - 1

Simplify

= 0

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{cot ( 2 x )}{cos ( 2 x )} = csc (2 x)

p ro v e tan (\frac{7 π}{6}) = \frac{3}{3}

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p ro v e cos^{2} (A) tan^{2} (A) = sin^{2} (A)

p ro v e csc (y) cos^{2} (y) + sin (y) = csc (y)

Is tan(θ)cot(θ)-1=1-sec(θ)cos(θ) ?
The answer to whether tan(θ)cot(θ)-1=1-sec(θ)cos(θ) is True