Is (tan(x))/(sin(x)cos(x))=tan^2(x)+1 ?

The answer to whether (tan(x))/(sin(x)cos(x))=tan^2(x)+1 is True

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prove (tan(x))/(sin(x)cos(x))=tan^2(x)+1

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin (x)

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

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

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

Manipulating right side

tan^{2} (x) + 1

Express with sin, cos

1 + tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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cos (4 x) = 1 - 8 cos^{2} (x) + 8 cos^{4} (x)

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

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Is (tan(x))/(sin(x)cos(x))=tan^2(x)+1 ?
The answer to whether (tan(x))/(sin(x)cos(x))=tan^2(x)+1 is True