Is sec(θ)-(tan(θ))/(csc(θ))=cos(θ) ?

The answer to whether sec(θ)-(tan(θ))/(csc(θ))=cos(θ) is True

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prove sec(θ)-(tan(θ))/(csc(θ))=cos(θ)

prove

sec (θ) - \frac{tan ( θ )}{csc ( θ )} = cos (θ)

True

Solution steps

sec (θ) - \frac{tan ( θ )}{csc ( θ )} = cos (θ)

Manipulating left side

sec (θ) - \frac{tan ( θ )}{csc ( θ )}

Express with sin, cos

- \frac{tan ( θ )}{csc ( θ )} + sec (θ)

Use the basic trigonometric identity:

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

= - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{csc ( θ )} + sec (θ)

Use the basic trigonometric identity:

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

= - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}} + sec (θ)

Use the basic trigonometric identity:

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

= - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}} + \frac{1}{cos ( θ )}

Simplify

- \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}} + \frac{1}{cos ( θ )} : \frac{- sin ^{2} ( θ ) + 1}{cos ( θ )}

- \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}} + \frac{1}{cos ( θ )}

\frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}} = \frac{sin ^{2} ( θ )}{cos ( θ )}

\frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}}

Divide fractions:

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

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

Refine

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Apply rule

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

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

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

= \frac{1 - sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

cot^{2} (x) sec^{2} (x) = cot^{2} (x) + 1

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

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

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

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