Is cos(b)+sin(b)tan(b)=sec(b) ?

The answer to whether cos(b)+sin(b)tan(b)=sec(b) is True

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prove cos(b)+sin(b)tan(b)=sec(b)

prove

cos (b) + sin (b) tan (b) = sec (b)

True

Solution steps

cos (b) + sin (b) tan (b) = sec (b)

Manipulating left side

cos (b) + sin (b) tan (b)

Express with sin, cos

cos (b) + sin (b) tan (b)

Use the basic trigonometric identity:

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

= cos (b) + sin (b) \frac{sin ( b )}{cos ( b )}

Simplify

cos (b) + sin (b) \frac{sin ( b )}{cos ( b )} : \frac{cos ^{2} ( b ) + sin ^{2} ( b )}{cos ( b )}

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

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

sin (b) \frac{sin ( b )}{cos ( b )}

Multiply fractions:

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

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

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

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)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( b )}}

Simplify

\frac{1}{\frac{1}{s e c ( b )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (b)

sec (b)

sec (b)

We showed that the two sides could take the same form

\Rightarrow True

2 sin (2 θ) (1 - 2 sin^{2} (θ)) = sin (4 θ)

cos^{2} (x) csc^{2} (x) = cot^{2} (x)

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

\frac{tan ^{2} ( θ )}{sec ^{2} ( θ )} = sin (+ cos (x))

4 cos^{2} (x) - 2 = 2 - 4 sin^{2} (x)

Is cos(b)+sin(b)tan(b)=sec(b) ?
The answer to whether cos(b)+sin(b)tan(b)=sec(b) is True