Is tan(y)+cot(y)=sec(y)csc(y) ?

The answer to whether tan(y)+cot(y)=sec(y)csc(y) is True

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prove tan(y)+cot(y)=sec(y)csc(y)

Solution

prove

tan (y) + cot (y) = sec (y) csc (y)

Solution

True

Solution steps

tan (y) + cot (y) = sec (y) csc (y)

Manipulating left side

tan (y) + cot (y)

Express with sin, cos

cot (y) + tan (y)

Use the basic trigonometric identity:

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

= \frac{cos ( y )}{sin ( y )} + tan (y)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Least Common Multiplier of

sin (y), cos (y) : sin (y) cos (y)

sin (y), cos (y)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (y)

cos (y)

= sin (y) cos (y)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (y) cos (y)

For

\frac{cos ( y )}{sin ( y )} :

multiply the denominator and numerator by

cos (y)

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

For

\frac{sin ( y )}{cos ( y )} :

multiply the denominator and numerator by

sin (y)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

Multiply

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (y) csc (y)

sec (y) csc (y)

sec (y) csc (y)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove 1/(sin(x))=csc(x)prove

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

prove (csc(θ))/(sec(θ))=cot(θ)prove

\frac{csc ( θ )}{sec ( θ )} = cot (θ)

prove sec(x)csc(x)=tan(x)+cot(x)prove

sec (x) csc (x) = tan (x) + cot (x)

prove sin(x)*sec(x)=tan(x)prove

sin (x) \cdot sec (x) = tan (x)

prove cot(B)sin(B)=cos(B)prove

cot (B) sin (B) = cos (B)

Frequently Asked Questions (FAQ)

Is tan(y)+cot(y)=sec(y)csc(y) ?
The answer to whether tan(y)+cot(y)=sec(y)csc(y) is True