Is tan(θ)(cos(θ)+cot(θ))=sin(θ)+1 ?

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

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prove tan(θ)(cos(θ)+cot(θ))=sin(θ)+1

prove

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

True

Solution steps

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

Manipulating left side

tan (θ) (cos (θ) + cot (θ))

Express with sin, cos

(cos (θ) + cot (θ)) tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Join

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

cos (θ) + \frac{cos ( θ )}{sin ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= cos (θ) sin (θ) + cos (θ)

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

Factor out common term

cos (θ)

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

Cancel the common factor:

cos (θ)

= sin (θ) + 1

= sin (θ) + 1

= 1 + sin (θ)

= sin (θ) + 1

We showed that the two sides could take the same form

\Rightarrow True

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

tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

\frac{1 - sin ( 2 x )}{cos ( 2 x )} = \frac{cos ( 2 x )}{1 + sin ( 2 x )}

sec (2 π) = 1 - tan^{2} (2 π)

tan^{2} (θ) cos^{2} (θ) = 1 - cos^{2} (θ)

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