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

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

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

prove

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

True

Solution steps

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

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

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

csc (A - B) = \frac{csc ( A ) csc ( B )}{cot ( B ) - cot ( A )}

\frac{cot ^{2} ( θ )}{csc ^{2} ( θ )} = cos^{2} (θ)

(csc (a) + 1) (csc (a) - 1) = cot^{2} (a)

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

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

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