Is sec(a)-cos(a)=sin(a)tan(a) ?

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

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prove sec(a)-cos(a)=sin(a)tan(a)

prove

sec (a) - cos (a) = sin (a) tan (a)

True

Solution steps

sec (a) - cos (a) = sin (a) tan (a)

Manipulating left side

sec (a) - cos (a)

Express with sin, cos

- cos (a) + sec (a)

Use the basic trigonometric identity:

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

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

Simplify

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

- cos (a) + \frac{1}{cos ( a )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (a) cos (a) + 1

cos (a) cos (a) = cos^{2} (a)

cos (a) cos (a)

Apply exponent rule:

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

cos (a) cos (a) = cos^{1 + 1} (a)

= cos^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= cos^{2} (a)

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

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

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

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

Manipulating right side

sin (a) tan (a)

Express with sin, cos

sin (a) tan (a)

Use the basic trigonometric identity:

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

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

Simplify

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

sin (a) \frac{sin ( a )}{cos ( a )}

Multiply fractions:

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

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

sin (a) sin (a) = sin^{2} (a)

sin (a) sin (a)

Apply exponent rule:

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

sin (a) sin (a) = sin^{1 + 1} (a)

= sin^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= sin^{2} (a)

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( a )}{cos ( a )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

(sec (x) + tan (x)) (1 - sin (x)) = cos (x)

sin^{7} (x) = (1 - cos^{2} (x))^{3} sin (x)

tan (A) + cot (A) = sec (A) csc (A)

\frac{sec ( θ )}{csc ( θ )} = tan (θ)

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

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