Is sec(θ)-sin(θ)tan(θ)=cos(θ) ?

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

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prove sec(θ)-sin(θ)tan(θ)=cos(θ)

prove

sec (θ) - sin (θ) tan (θ) = cos (θ)

True

Solution steps

sec (θ) - sin (θ) tan (θ) = cos (θ)

Manipulating left side

sec (θ) - sin (θ) tan (θ)

Express with sin, cos

sec (θ) - sin (θ) tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{1}{cos ( θ )} - sin (θ) \frac{sin ( θ )}{cos ( θ )}

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

sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

Apply rule

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

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

= \frac{1 - sin ^{2} ( θ )}{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

sin (π - θ) = sin (θ)

tan (\frac{x}{2}) = csc (x) - cot (x)

tan (x) cot (x) = 1

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

\frac{1}{2} (cot (x) + tan (x)) = csc (2 x)

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