Is (1+cot^2(x))tan(x)=csc(x)sec(x) ?

The answer to whether (1+cot^2(x))tan(x)=csc(x)sec(x) is True

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prove (1+cot^2(x))tan(x)=csc(x)sec(x)

Solution

prove

(1 + cot^{2} (x)) tan (x) = csc (x) sec (x)

Solution

True

Solution steps

(1 + cot^{2} (x)) tan (x) = csc (x) sec (x)

Manipulating left side

(1 + cot^{2} (x)) tan (x)

Express with sin, cos

(1 + cot^{2} (x)) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

Multiply

\frac{1}{sec ( x )} \cdot \frac{1}{csc ( x )} : \frac{1}{sec ( x ) csc ( x )}

\frac{1}{sec ( x )} \cdot \frac{1}{csc ( x )}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (x) csc (x)

sec (x) csc (x)

sec (x) csc (x)

= csc (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 (1+cot^2(x))tan(x)=csc(x)sec(x) ?
The answer to whether (1+cot^2(x))tan(x)=csc(x)sec(x) is True