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

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

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

prove

sin (x) \cdot tan (x) + cos (- x) = sec (x)

True

Solution steps

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

Manipulating left side

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

Use the negative angle identity:

cos (- x) = cos (x)

= 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

cos (2 α) = cos^{2} (α) - sin^{2} (α)

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

2 csc (2 x) = \frac{1}{sin ( x ) cos ( x )}

sin (u) (csc (u) - sin (u)) = cos^{2} (u)

cos^{2} (θ) - sin^{2} (θ) = 1

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