Is (cos^2(A))/(cos(A)-sin(A))+(sin(A))/(1-cot(A))=sin(A)+cos(A) ?

The answer to whether (cos^2(A))/(cos(A)-sin(A))+(sin(A))/(1-cot(A))=sin(A)+cos(A) is True

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Popular Trigonometry >

prove (cos^2(A))/(cos(A)-sin(A))+(sin(A))/(1-cot(A))=sin(A)+cos(A)

Solution

prove

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - cot ( A )} = sin (A) + cos (A)

Solution

True

Solution steps

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - cot ( A )} = sin (A) + cos (A)

Manipulating left side

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - cot ( A )}

Express with sin, cos

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - cot ( A )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}}

Join

1 - \frac{cos ( A )}{sin ( A )} : \frac{sin ( A ) - cos ( A )}{sin ( A )}

1 - \frac{cos ( A )}{sin ( A )}

Convert element to fraction:

1 = \frac{1 sin ( A )}{sin ( A )}

= \frac{1 \cdot sin ( A )}{sin ( A )} - \frac{cos ( A )}{sin ( A )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sin ( A ) - cos ( A )}{sin ( A )}

Multiply:

1 \cdot sin (A) = sin (A)

= \frac{sin ( A ) - cos ( A )}{sin ( A )}

= \frac{sin ( A )}{\frac{s i n ( A ) - c o s ( A )}{s i n ( A )}}

Apply the fraction rule:

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

= \frac{sin ( A ) sin ( A )}{sin ( A ) - cos ( A )}

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

sin (A) sin (A)

Apply exponent rule:

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

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

= sin^{1 + 1} (A)

Add the numbers:

1 + 1 = 2

= sin^{2} (A)

= \frac{sin ^{2} ( A )}{sin ( A ) - cos ( A )}

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

Least Common Multiplier of

cos (A) - sin (A), sin (A) - cos (A) : - (cos (A) - sin (A))

cos (A) - sin (A), sin (A) - cos (A)

Lowest Common Multiplier (LCM)

sin (A) - cos (A) = - (cos (A) - sin (A))

= cos (A) - sin (A), - (cos (A) - sin (A))

Compute an expression comprised of factors that appear either in

cos (A) - sin (A)

sin (A) - cos (A)

= - (cos (A) - sin (A))

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

- (cos (A) - sin (A))

For

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} :

multiply the denominator and numerator by

- 1

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} = \frac{cos ^{2} ( A ) ( - 1 )}{( cos ( A ) - sin ( A ) ) ( - 1 )} = \frac{- cos ^{2} ( A )}{- ( cos ( A ) - sin ( A ) )}

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

Since the denominators are equal, combine the fractions:

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

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

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

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

Cancel

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

\frac{- cos ^{2} ( A ) + sin ^{2} ( A )}{( cos ( A ) - sin ( A ) )}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

sin^{2} (A) - cos^{2} (A) = (sin (A) + cos (A)) (sin (A) - cos (A))

= \frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{( cos ( A ) - sin ( A ) )}

Cancel

\frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{( cos ( A ) - sin ( A ) )} : - (sin (A) + cos (A))

\frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{( cos ( A ) - sin ( A ) )}

cos (A) - sin (A) = - (sin (A) - cos (A))

= \frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{- ( sin ( A ) - cos ( A ) )}

Refine

= - \frac{( sin ( A ) + cos ( A ) ) ( sin ( A ) - cos ( A ) )}{sin ( A ) - cos ( A )}

Cancel the common factor:

sin (A) - cos (A)

= - (sin (A) + cos (A))

= - (sin (A) + cos (A))

Negate

- (sin (A) + cos (A)) = - sin (A) - cos (A)

= - sin (A) - cos (A)

= - (- sin (A) - cos (A))

Distribute parentheses

= - (- sin (A)) - (- cos (A))

Apply minus-plus rules

- (- a) = a

= sin (A) + cos (A)

= sin (A) + cos (A)

= cos (A) + sin (A)

= sin (A) + cos (A)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (cos^2(A))/(cos(A)-sin(A))+(sin(A))/(1-cot(A))=sin(A)+cos(A) ?
The answer to whether (cos^2(A))/(cos(A)-sin(A))+(sin(A))/(1-cot(A))=sin(A)+cos(A) is True