Is csc(y)cos^2(y)+sin(y)=csc(y) ?

The answer to whether csc(y)cos^2(y)+sin(y)=csc(y) is True

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prove csc(y)cos^2(y)+sin(y)=csc(y)

prove

csc (y) cos^{2} (y) + sin (y) = csc (y)

True

Solution steps

csc (y) cos^{2} (y) + sin (y) = csc (y)

Manipulating left side

csc (y) cos^{2} (y) + sin (y)

Express with sin, cos

sin (y) + cos^{2} (y) csc (y)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos^{2} (y) \frac{1}{sin ( y )}

Multiply fractions:

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

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

Multiply:

1 \cdot cos^{2} (y) = cos^{2} (y)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

sin (y) sin (y)

Apply exponent rule:

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

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

= sin^{1 + 1} (y)

Add the numbers:

1 + 1 = 2

= sin^{2} (y)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc (y)

csc (y)

csc (y)

We showed that the two sides could take the same form

\Rightarrow True

2 cos^{2} (x) = 2 - 2 sin^{2} (x)

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

cos (\frac{3 π}{2} + x) = sin (x)

csc (2 A) + cot (2 A) = cot (A)

1 - sin (θ) = cos (θ)

Is csc(y)cos^2(y)+sin(y)=csc(y) ?
The answer to whether csc(y)cos^2(y)+sin(y)=csc(y) is True