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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (a) + sin (a) tan (a)

Express with sin, cos

cos (a) + sin (a) tan (a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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 )}

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

Convert element to fraction:

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

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

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

cos (a) cos (a) + sin^{2} (a)

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) + sin^{2} (a)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( a )}}

Simplify

\frac{1}{\frac{1}{s e c ( a )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( a )}{1}

Apply rule

\frac{a}{1} = a

= sec (a)

sec (a)

sec (a)

We showed that the two sides could take the same form

\Rightarrow True

(csc (θ) + cot (θ)) (1 - cos (θ)) = sin (θ)

\frac{1 - cos ( x )}{sin ( x )} = csc (x) - cot (x)

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

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

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

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