Is (tan(x))/(csc(x))+cos(x)=sec(x) ?

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

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prove (tan(x))/(csc(x))+cos(x)=sec(x)

Solution

prove

\frac{tan ( x )}{csc ( x )} + cos (x) = sec (x)

Solution

True

Solution steps

\frac{tan ( x )}{csc ( x )} + cos (x) = sec (x)

Manipulating left side

\frac{tan ( x )}{csc ( x )} + cos (x)

Express with sin, cos

cos (x) + \frac{tan ( x )}{csc ( x )}

Use the basic trigonometric identity:

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

= cos (x) + \frac{\frac{s i n ( x )}{c o s ( x )}}{csc ( x )}

Use the basic trigonometric identity:

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

= cos (x) + \frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}}

Simplify

cos (x) + \frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}} : \frac{cos ^{2} ( x ) + sin ^{2} ( x )}{cos ( x )}

cos (x) + \frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}}

\frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}} = \frac{sin ^{2} ( x )}{cos ( x )}

\frac{\frac{s i n ( x )}{c o s ( x )}}{\frac{1}{s i n ( x )}}

Divide fractions:

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

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

Refine

= \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

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Frequently Asked Questions (FAQ)

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