Is sec(β)-cos(β)=tan(β)sin(β) ?

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

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prove sec(β)-cos(β)=tan(β)sin(β)

prove

sec (β) - cos (β) = tan (β) sin (β)

True

Solution steps

sec (β) - cos (β) = tan (β) sin (β)

Manipulating left side

sec (β) - cos (β)

Express with sin, cos

- cos (β) + sec (β)

Use the basic trigonometric identity:

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

= - cos (β) + \frac{1}{cos ( β )}

Simplify

- cos (β) + \frac{1}{cos ( β )} : \frac{- cos ^{2} ( β ) + 1}{cos ( β )}

- cos (β) + \frac{1}{cos ( β )}

Convert element to fraction:

cos (β) = \frac{cos ( β ) cos ( β )}{cos ( β )}

= - \frac{cos ( β ) cos ( β )}{cos ( β )} + \frac{1}{cos ( β )}

Since the denominators are equal, combine the fractions:

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

= \frac{- cos ( β ) cos ( β ) + 1}{cos ( β )}

- cos (β) cos (β) + 1 = - cos^{2} (β) + 1

- cos (β) cos (β) + 1

cos (β) cos (β) = cos^{2} (β)

cos (β) cos (β)

Apply exponent rule:

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

cos (β) cos (β) = cos^{1 + 1} (β)

= cos^{1 + 1} (β)

Add the numbers:

1 + 1 = 2

= cos^{2} (β)

= - cos^{2} (β) + 1

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

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

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

Manipulating right side

tan (β) sin (β)

Express with sin, cos

sin (β) tan (β)

Use the basic trigonometric identity:

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

= sin (β) \frac{sin ( β )}{cos ( β )}

Simplify

sin (β) \frac{sin ( β )}{cos ( β )} : \frac{sin ^{2} ( β )}{cos ( β )}

sin (β) \frac{sin ( β )}{cos ( β )}

Multiply fractions:

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

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

sin (β) sin (β) = sin^{2} (β)

sin (β) sin (β)

Apply exponent rule:

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

sin (β) sin (β) = sin^{1 + 1} (β)

= sin^{1 + 1} (β)

Add the numbers:

1 + 1 = 2

= sin^{2} (β)

= \frac{sin ^{2} ( β )}{cos ( β )}

= \frac{sin ^{2} ( β )}{cos ( β )}

= \frac{sin ^{2} ( β )}{cos ( β )}

Rewrite using trig identities

\frac{sin ^{2} ( β )}{cos ( β )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( y ^{2} )}{cos ( y )} = sec (y) - cos (y)

sin^{2} (\frac{π}{4} - 2 π) = cos^{2} (\frac{π}{4} - 2 π)

sec (θ) = tan (θ) csc (θ)

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

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

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