Is (csc(x))/(sin(x))-(cot(x))/(tan(x))=1 ?

The answer to whether (csc(x))/(sin(x))-(cot(x))/(tan(x))=1 is True

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prove (csc(x))/(sin(x))-(cot(x))/(tan(x))=1

prove

\frac{csc ( x )}{sin ( x )} - \frac{cot ( x )}{tan ( x )} = 1

True

Solution steps

\frac{csc ( x )}{sin ( x )} - \frac{cot ( x )}{tan ( x )} = 1

Manipulating left side

\frac{csc ( x )}{sin ( x )} - \frac{cot ( x )}{tan ( x )}

Express with sin, cos

- \frac{cot ( x )}{tan ( x )} + \frac{csc ( x )}{sin ( x )}

Use the basic trigonometric identity:

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

= - \frac{\frac{c o s ( x )}{s i n ( x )}}{tan ( x )} + \frac{csc ( x )}{sin ( x )}

Use the basic trigonometric identity:

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

= - \frac{\frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x )}{c o s ( x )}} + \frac{csc ( x )}{sin ( x )}

Use the basic trigonometric identity:

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

= - \frac{\frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x )}{c o s ( x )}} + \frac{\frac{1}{s i n ( x )}}{sin ( x )}

Simplify

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

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

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

\frac{\frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x )}{c o s ( x )}}

Divide fractions:

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

\frac{\frac{1}{s i n ( x )}}{sin ( x )} = \frac{1}{sin ^{2} ( x )}

\frac{\frac{1}{s i n ( x )}}{sin ( x )}

Apply the fraction rule:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{sin ^{2} ( x )}{sin ^{2} ( x )}

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x )}{1 + cos ( x )} = csc (x) - cot (x)

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

(tan (x) + cot (x))^{4} = csc^{4} (x) sec^{4} (x)

sin (x) tan (x) = sec (x) - cos (x)

tan^{2} (x) - sin^{2} (x) = sin^{4} (x) sec^{2} (x)

Is (csc(x))/(sin(x))-(cot(x))/(tan(x))=1 ?
The answer to whether (csc(x))/(sin(x))-(cot(x))/(tan(x))=1 is True