Is (csc^2(A))/(1+tan^2(A))=cot^2(A) ?

The answer to whether (csc^2(A))/(1+tan^2(A))=cot^2(A) is True

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prove (csc^2(A))/(1+tan^2(A))=cot^2(A)

prove

\frac{csc ^{2} ( A )}{1 + tan ^{2} ( A )} = cot^{2} (A)

True

Solution steps

\frac{csc ^{2} ( A )}{1 + tan ^{2} ( A )} = cot^{2} (A)

Manipulating left side

\frac{csc ^{2} ( A )}{1 + tan ^{2} ( A )}

Rewrite using trig identities

\frac{csc ^{2} ( A )}{1 + tan ^{2} ( A )}

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

= \frac{csc ^{2} ( A )}{sec ^{2} ( A )}

= \frac{csc ^{2} ( A )}{sec ^{2} ( A )}

Express with sin, cos

\frac{csc ^{2} ( A )}{sec ^{2} ( A )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

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

(\frac{1}{cos ( A )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Divide fractions:

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

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

Refine

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

\frac{cos ( A ) cot ( A )}{sin ( A )}

= cot (A) \frac{cos ( A )}{sin ( A )}

Use the basic trigonometric identity:

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

cot (A) cot (A)

Simplify

cot (A) cot (A) : cot^{2} (A)

cot (A) cot (A)

Apply exponent rule:

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

cot (A) cot (A) = cot^{1 + 1} (A)

= cot^{1 + 1} (A)

Add the numbers:

1 + 1 = 2

= cot^{2} (A)

cot^{2} (A)

cot^{2} (A)

We showed that the two sides could take the same form

\Rightarrow True

sin (3 a) = 3 sin (a) - 4 sin^{3} (a)

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

\frac{1}{tan ( θ ) csc ( θ )} = cos (θ)

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

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

Is (csc^2(A))/(1+tan^2(A))=cot^2(A) ?
The answer to whether (csc^2(A))/(1+tan^2(A))=cot^2(A) is True