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

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

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prove 1/(csc^2(x))+cos^2(x)=1

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= cos^{2} (x) + \frac{1}{( \frac{1}{s i n ( x )} ) ^{2}}

Simplify

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

cos^{2} (x) + \frac{1}{( \frac{1}{s i n ( x )} ) ^{2}}

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} = sin^{2} (x)

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{s i n ^{2} ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

2 cos^{2} (θ) - 1 = 1 - 2 sin^{2} (θ)

cot (x) = \frac{csc ( x )}{sec ( x )}

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

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

sec (2 x) = \frac{sec ^{2} ( x )}{1 - tan ^{2} ( x )}

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