Respuesta :

culveryolanda46

Step-by-step explanation:

[tex] \frac{ \cos( \alpha ) }{1 - \sin( \alpha ) } = \sec( \alpha ) + \tan( \alpha ) [/tex]

[tex] \frac{ \cos( \alpha ) (1 + \sin( \alpha ) }{(1 - \sin( \alpha ) )(1 + \sin( \alpha ) } [/tex]

[tex] \frac{ \cos( \alpha ) + \cos( \alpha ) \sin( \alpha ) }{ \cos {}^{2} ( \alpha ) } [/tex]

[tex] \frac{ \cos( \alpha ) }{ \cos {}^{2} ( \alpha ) } + \frac{ \cos( \alpha ) \sin( \alpha ) }{ \cos {}^{2} ( \alpha ) } [/tex]

[tex] \sec( \alpha ) + \tan( \alpha ) [/tex]

semsee45

Answer:

See below for proof using trigonometric identities.

Step-by-step explanation:

Given expression:

[tex]\dfrac{\cos \alpha}{1-\sin \alpha}[/tex]

Multiply the numerator and denominator by the conjugate of the denominator:

[tex]\implies \dfrac{\cos \alpha(1+\sin \alpha)}{(1-\sin \alpha)(1+\sin \alpha)}[/tex]

Expand the brackets:

[tex]\implies \dfrac{\cos \alpha+\cos \alpha\sin \alpha}{1-\sin^2 \alpha}[/tex]

Apply the trigonometric identity  sin²θ + cos²θ ≡ 1  to the denominator:

[tex]\implies \dfrac{\cos \alpha+\cos \alpha\sin \alpha}{\cos^2 \alpha}[/tex]

[tex]\textsf{Apply the addition fraction rule} \quad \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}:[/tex]

[tex]\implies \dfrac{\cos \alpha}{\cos^2 \alpha}+\dfrac{\cos \alpha\sin \alpha}{\cos^2 \alpha}[/tex]

Cancel the common factor:

[tex]\implies \dfrac{1}{\cos\alpha}+\dfrac{\sin \alpha}{\cos \alpha}[/tex]

[tex]\textsf{Apply the trigonometric identities} \quad \dfrac{1}{\cos \theta} \equiv \sec \theta \:\:\textsf{ and }\:\: \dfrac{\sin \theta}{\cos \theta} \equiv \tan \theta :[/tex]

[tex]\implies \sec \alpha + \tan \alpha[/tex]

Learn more about trigonometric identities here:

https://brainly.com/question/27938536

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