Is cot(θ)(sin(θ)+tan(θ))=cos(θ)+1 ?

The answer to whether cot(θ)(sin(θ)+tan(θ))=cos(θ)+1 is True

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prove cot(θ)(sin(θ)+tan(θ))=cos(θ)+1

prove

cot (θ) (sin (θ) + tan (θ)) = cos (θ) + 1

True

Solution steps

cot (θ) (sin (θ) + tan (θ)) = cos (θ) + 1

Manipulating left side

cot (θ) (sin (θ) + tan (θ))

Express with sin, cos

(sin (θ) + tan (θ)) cot (θ)

Use the basic trigonometric identity:

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

= (sin (θ) + \frac{sin ( θ )}{cos ( θ )}) cot (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

= \frac{cos ( θ ) ( sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )} )}{sin ( θ )}

Join

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

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{\frac{s i n ( θ ) c o s ( θ ) + s i n ( θ )}{c o s ( θ )} cos ( θ )}{sin ( θ )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= sin (θ) cos (θ) + sin (θ)

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

Factor out common term

sin (θ)

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

Cancel the common factor:

sin (θ)

= cos (θ) + 1

= cos (θ) + 1

= 1 + cos (θ)

= cos (θ) + 1

We showed that the two sides could take the same form

\Rightarrow True

(2 - sin^{2} (x)) csc^{2} (x) = cot^{2} (x)

\frac{1}{tan ( A )} + tan (A) = \frac{2}{sin ( 2 A )}

2 sin^{2} (x) + 4 cos^{2} (x) = 2 - 4 sin^{2} (x)

1 + sin (θ) = cos (θ)

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

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