Is tan(θ)+1=(sin(θ))/(cos(θ))+1 ?

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

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prove tan(θ)+1=(sin(θ))/(cos(θ))+1

prove

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

True

Solution steps

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

Manipulating left side

tan (θ) + 1

Express with sin, cos

1 + tan (θ)

Use the basic trigonometric identity:

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

= 1 + \frac{sin ( θ )}{cos ( θ )}

Simplify

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

1 + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Multiply:

1 \cdot cos (θ) = cos (θ)

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

\frac{cos ( θ )}{cos ( θ )} = 1

= 1 + \frac{sin ( θ )}{cos ( θ )}

= 1 + \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )} + 1

We showed that the two sides could take the same form

\Rightarrow True

(1 + tan^{2} (α)) cos^{2} (α) = 1

\frac{1}{sin ^{2} ( x )} = \frac{1}{csc ^{2} ( x )}

sin^{2} (θ) = (1 - cos^{2} (θ))

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

3 + 3 cos (x) = 3 csc (x)

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