Is (cos(b))/(sec(b))+(sin(b))/(csc(b))=csc^2(b)-cot^2(b) ?

The answer to whether (cos(b))/(sec(b))+(sin(b))/(csc(b))=csc^2(b)-cot^2(b) is True

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prove (cos(b))/(sec(b))+(sin(b))/(csc(b))=csc^2(b)-cot^2(b)

prove

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

True

Solution steps

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

Manipulating left side

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )}

Express with sin, cos

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )}

Use the basic trigonometric identity:

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

= \frac{cos ( b )}{\frac{1}{c o s ( b )}} + \frac{sin ( b )}{csc ( b )}

Use the basic trigonometric identity:

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

= \frac{cos ( b )}{\frac{1}{c o s ( b )}} + \frac{sin ( b )}{\frac{1}{s i n ( b )}}

Simplify

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

\frac{cos ( b )}{\frac{1}{c o s ( b )}} + \frac{sin ( b )}{\frac{1}{s i n ( b )}}

\frac{cos ( b )}{\frac{1}{c o s ( b )}} = cos^{2} (b)

\frac{cos ( b )}{\frac{1}{c o s ( b )}}

Apply the fraction rule:

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

= \frac{cos ( b ) cos ( b )}{1}

Apply rule

\frac{a}{1} = a

= cos (b) cos (b)

Apply exponent rule:

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

cos (b) cos (b) = cos^{1 + 1} (b)

= cos^{1 + 1} (b)

Add the numbers:

1 + 1 = 2

= cos^{2} (b)

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

\frac{sin ( b )}{\frac{1}{s i n ( b )}}

Apply the fraction rule:

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

= \frac{sin ( b ) sin ( b )}{1}

Apply rule

\frac{a}{1} = a

= sin (b) sin (b)

Apply exponent rule:

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

sin (b) sin (b) = sin^{1 + 1} (b)

= sin^{1 + 1} (b)

Add the numbers:

1 + 1 = 2

= sin^{2} (b)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

1 = csc^{2} (x) - cot^{2} (x)

= csc^{2} (b) - cot^{2} (b)

= csc^{2} (b) - cot^{2} (b)

We showed that the two sides could take the same form

\Rightarrow True

2 cos^{2} (\frac{θ}{2}) - cos (θ) - 1 = 0

\frac{1}{csc ( θ ) - cot ( θ )} = csc (θ) + cot (θ)

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

cos (θ + \frac{π}{4}) = \frac{2}{2} (cos (θ) - sin (θ))

5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

Is (cos(b))/(sec(b))+(sin(b))/(csc(b))=csc^2(b)-cot^2(b) ?
The answer to whether (cos(b))/(sec(b))+(sin(b))/(csc(b))=csc^2(b)-cot^2(b) is True