Is (tan^2(θ))/(sin^2(θ))-1=tan^2(θ) ?

The answer to whether (tan^2(θ))/(sin^2(θ))-1=tan^2(θ) is True

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prove (tan^2(θ))/(sin^2(θ))-1=tan^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= - 1 + \frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )}

Simplify

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

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

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin^{2} (θ)

= \frac{1}{cos ^{2} ( θ )}

= - 1 + \frac{1}{cos ^{2} ( θ )}

Convert element to fraction:

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

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

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} ( θ ) + 1}{cos ^{2} ( θ )}

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

\frac{sin ( θ ) tan ( θ )}{cos ( θ )}

= tan (θ) \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

tan (θ) tan (θ)

Simplify

tan (θ) tan (θ) : tan^{2} (θ)

tan (θ) tan (θ)

Apply exponent rule:

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

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= tan^{2} (θ)

tan^{2} (θ)

tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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)

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