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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (x) + sin (x) tan (x)

Express with sin, cos

cos (x) + sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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

(cos (2 x) - sin (2 x))^{2} - 1 = sin (- 4 x)

cot^{2} (x) + \frac{1}{cos ( x ) sec ( x )} = csc^{2} (x)

\frac{1 + cos ( θ )}{sin ( θ )} = csc (θ) + cot (θ)

\frac{cos ( x + y ) + cos ( x - y )}{sin ( x - y ) + sin ( x + y )} = cot (x)

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