Is 1+1/(tan^2(α))= 1/(sin^2(α)) ?

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

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prove 1+1/(tan^2(α))= 1/(sin^2(α))

prove

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

True

Solution steps

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

Manipulating left side

1 + \frac{1}{tan ^{2} ( α )}

Express with sin, cos

1 + \frac{1}{tan ^{2} ( α )}

Use the basic trigonometric identity:

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

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

Simplify

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

1 + \frac{1}{( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

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

\frac{1}{( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

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

(\frac{sin ( α )}{cos ( α )})^{2}

Apply exponent rule:

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

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

= \frac{1}{\frac{s i n ^{2} ( α )}{c o s ^{2} ( α )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin^{2} (α) = sin^{2} (α)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ^{2} ( α )}

= \frac{1}{sin ^{2} ( α )}

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{sin ( t ) cos ( t )}{tan ( t )} = cos^{2} (t)

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

sec (θ) (sec (θ) - cos (θ)) = tan^{2} (θ)

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