Verify the identity:
$$\dfrac{1+\sec \alpha + an \alpha }{1+\sec \alpha - an \alpha } = \dfrac{1+\sin \alpha }{\cos \alpha } $$

Question

Verify the identity:
$$\dfrac{1+\sec \alpha +	an \alpha  }{1+\sec \alpha -	an \alpha } =  \dfrac{1+\sin \alpha  }{\cos \alpha  }  $$

Uniteasy Admin · Accepted Answer

$$\because \sec \alpha =\dfrac{1}{\cos \alpha }  $$
$$	herefore  \dfrac{1+\sec \alpha +	an \alpha }{1+\sec \alpha -	an \alpha } $$
$$=\dfrac{1+\dfrac{1}{\cos \alpha }+\dfrac{\sin \alpha }{\cos \alpha }  }{1+\dfrac{1}{\cos \alpha }-\dfrac{\sin \alpha }{\cos \alpha }  } $$
$$=\dfrac{1+\cos \alpha+\sin \alpha  }{1+\cos \alpha -\sin \alpha  } $$
$$(1+\cos \alpha+\sin \alpha )\cos \alpha $$
$$=\cos \alpha+\cos^2 \alpha+ \sin \alpha\cos \alpha$$n
$$(1+\cos \alpha -\sin \alpha )(1+\sin \alpha)$$
$$=1+\sin \alpha+\cos \alpha +\sin \alpha  \cos \alpha-\sin \alpha -\sin^2\alpha   $$
$$=1+\cos \alpha+\sin \alpha  \cos \alpha -\sin^2\alpha  $$n
$$(\cos \alpha+\cos^2 \alpha+ \sin \alpha\cos \alpha)-(1+\cos \alpha+\sin \alpha  \cos \alpha -\sin^2\alpha)$$
$$=\cos^2 \alpha+\sin^2\alpha-1$$
$$=0$$
So we can reverse the process to verify the identity.

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