Is csc(t)cot(t)=(1+cot^2(t))/(sec(t)) ?

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

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Popular Trigonometry >

prove csc(t)cot(t)=(1+cot^2(t))/(sec(t))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

csc (t) cot (t)

Express with sin, cos

cot (t) csc (t)

Use the basic trigonometric identity:

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

= \frac{cos ( t )}{sin ( t )} csc (t)

Use the basic trigonometric identity:

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

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

Simplify

\frac{cos ( t )}{sin ( t )} \cdot \frac{1}{sin ( t )} : \frac{cos ( t )}{sin ^{2} ( t )}

\frac{cos ( t )}{sin ( t )} \cdot \frac{1}{sin ( t )}

Multiply fractions:

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

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

Multiply:

cos (t) \cdot 1 = cos (t)

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

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

sin (t) sin (t)

Apply exponent rule:

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

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

= sin^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= sin^{2} (t)

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

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

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

Manipulating right side

\frac{1 + cot ^{2} ( t )}{sec ( t )}

Express with sin, cos

\frac{1 + cot ^{2} ( t )}{sec ( t )}

Use the basic trigonometric identity:

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

= \frac{1 + ( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{sec ( t )}

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{1 + ( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{\frac{1}{c o s ( t )}}

Apply the fraction rule:

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is csc(t)cot(t)=(1+cot^2(t))/(sec(t)) ?
The answer to whether csc(t)cot(t)=(1+cot^2(t))/(sec(t)) is True