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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin^{2} (x)

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

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

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

Manipulating right side

1 + tan^{2} (x)

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

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

sec (y) + tan (y) = \frac{cos ( y )}{1 - sin ( y )}

(csc (x) + cot (x)) (sec (x) - 1) = tan (x)

cos (x - \frac{π}{2}) = cos (x) tan (x)

csc (- θ) = - csc (θ)

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