Is csc(θ)cos^2(θ)+sin(θ)=csc(θ) ?

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

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prove csc(θ)cos^2(θ)+sin(θ)=csc(θ)

prove

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

True

Solution steps

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

Manipulating left side

csc (θ) cos^{2} (θ) + sin (θ)

Express with sin, cos

sin (θ) + cos^{2} (θ) csc (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos^{2} (θ) \frac{1}{sin ( θ )}

Multiply fractions:

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (θ) sin (θ) + cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

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

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

sec (θ + \frac{π}{2}) = csc (θ)

\frac{1 + csc ( θ )}{sec ( θ )} - cot (θ) = cos (θ)

sin (2 θ) = \frac{2 cot ( θ )}{1 + cot ^{2} ( θ )}

cos (A) csc (A) = cot (A)

tan (θ) = \frac{1 - cos ( 2 θ )}{sin ( 2 θ )}

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