Question

Prove that:

Answer 1

Step-by-step explanation:

$\frac{ \cos( \alpha ) }{1 - \sin( \alpha ) } = \sec( \alpha ) + \tan( \alpha )$

$\frac{ \cos( \alpha ) (1 + \sin( \alpha ) }{(1 - \sin( \alpha ) )(1 + \sin( \alpha ) }$

$\frac{ \cos( \alpha ) + \cos( \alpha ) \sin( \alpha ) }{ \cos {}^{2} ( \alpha ) }$

$\frac{ \cos( \alpha ) }{ \cos {}^{2} ( \alpha ) } + \frac{ \cos( \alpha ) \sin( \alpha ) }{ \cos {}^{2} ( \alpha ) }$

$\sec( \alpha ) + \tan( \alpha )$

Answer 2

Answer:

See below for proof using trigonometric identities.

Step-by-step explanation:

Given expression:

$\dfrac{\cos \alpha}{1-\sin \alpha}$

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

$\implies \dfrac{\cos \alpha(1+\sin \alpha)}{(1-\sin \alpha)(1+\sin \alpha)}$

Expand the brackets:

$\implies \dfrac{\cos \alpha+\cos \alpha\sin \alpha}{1-\sin^2 \alpha}$

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

$\implies \dfrac{\cos \alpha+\cos \alpha\sin \alpha}{\cos^2 \alpha}$

$\textsf{Apply the addition fraction rule} \quad \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}:$

$\implies \dfrac{\cos \alpha}{\cos^2 \alpha}+\dfrac{\cos \alpha\sin \alpha}{\cos^2 \alpha}$

Cancel the common factor:

$\implies \dfrac{1}{\cos\alpha}+\dfrac{\sin \alpha}{\cos \alpha}$

$\textsf{Apply the trigonometric identities} \quad \dfrac{1}{\cos \theta} \equiv \sec \theta \:\:\textsf{ and }\:\: \dfrac{\sin \theta}{\cos \theta} \equiv \tan \theta :$

$\implies \sec \alpha + \tan \alpha$

Learn more about trigonometric identities here:

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