Is tan(x)(cos(x)+cot(x))=sin(x)+1 ?

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

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prove tan(x)(cos(x)+cot(x))=sin(x)+1

prove

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

True

Solution steps

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

Manipulating left side

tan (x) (cos (x) + cot (x))

Express with sin, cos

(cos (x) + cot (x)) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= cos (x) sin (x) + cos (x)

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

Factor out common term

cos (x)

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

Cancel the common factor:

cos (x)

= sin (x) + 1

= sin (x) + 1

= 1 + sin (x)

= sin (x) + 1

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (x) - sin^{2} (x) tan^{2} (x) = sin^{2} (x)

(1 - cos (x)) (1 + cos (x)) (1 + cot^{2} (x)) = 1

1 + tan^{2} (6 0^{\circ}) = sec^{2} (6 0^{\circ})

1 + cos (θ) = sin (θ)

sin (x) = sin (x + 2 π)

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