Show that $$\dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x} }{\sqrt{1-\sin x}+\sqrt{1-\cos x} } =\sqrt{2}+1 $$

Question

Show that $$\dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x}    }{\sqrt{1-\sin x}+\sqrt{1-\cos x}    } =\sqrt{2}+1   $$

Uniteasy Admin · Accepted Answer

The first step is to show that $$\dfrac{\cos x+\cos\Big(  \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) } =\sqrt{2}+1  $$
Using Sum to product identities
$$\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}$$
$$\sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}$$
Here
$$ \alpha = x$$, $$ \beta = \dfrac{\pi}{4}-x $$
$$\dfrac{\alpha +\beta }{2} = \dfrac{\pi}{8} $$, $$\dfrac{\alpha - \beta }{2} = x-\dfrac{\pi}{8} $$
$$\dfrac{\cos x+\cos\Big(  \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) }$$
$$=\dfrac{2\cos\dfrac{\pi}{8}\cos(x-\dfrac{\pi}{8} ) }{2\sin\dfrac{\pi}{8}\cos(x-\dfrac{\pi}{8} ) } = \cot\dfrac{\pi}{8} $$ 
Using half angle identity for tangent function to find the value of $$	an\dfrac{\pi}{8}$$ ,
$$	an \dfrac{\alpha }{2}=\sqrt{\dfrac{1-\cos \alpha }{1+\cos \alpha} } $$
$$	an\dfrac{\pi}{8}=\sqrt{\dfrac{1-\cos\dfrac{\pi}{4} }{1+\cos\dfrac{\pi}{4} } } $$ 
$$=\sqrt{\dfrac{1-\dfrac{\sqrt{2} }{2} }{1+\dfrac{\sqrt{2} }{2} } }  $$ 
$$=\sqrt{\dfrac{2-\sqrt{2} }{2+\sqrt{2} } } $$ 
$$=\dfrac{2-\sqrt{2}  }{\sqrt{2} } = \sqrt{2}-1 $$
Then, the value of $$\cot\dfrac{\pi}{8} $$ is the reciprocal of $$	an\dfrac{\pi}{8}$$ 
$$\cot\dfrac{\pi}{8}  = \dfrac{1}{	an\dfrac{\pi}{8} }$$
$$=\dfrac{1}{\sqrt{2} -1 } =\sqrt{2}+1 $$
Therefore,
$$\dfrac{\cos x+\cos\Big(  \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) }=\sqrt{2}+1 $$ 
Next step is to show that 
which is silver ratio.
$$\dfrac{\cos x+\cos\Big(  \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) } =\dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x}  }{\sqrt{1-\sin x}+\sqrt{1-\cos x}  } $$
Use the half angle identities
$$\sin \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1-\cos \alpha }{2} } $$  
$$\cos \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1+\cos \alpha }{2} }$$
$$\dfrac{\cos x+\cos\Big(  \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) }$$
$$=\dfrac{\sqrt{\dfrac{1+\cos x }{2} }+ \sqrt{\dfrac{1+\cos (\dfrac{\pi}{2} - x) }{2}}}{\sqrt{\dfrac{1-\cos x }{2} }+ \sqrt{\dfrac{1-\cos (\dfrac{\pi}{2} - x) }{2}}} $$ 
$$=\dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x}  }{\sqrt{1-\sin x}+\sqrt{1-\cos x}  } $$
Therefore
$$\dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x}  }{\sqrt{1-\sin x}+\sqrt{1-\cos x}  } =\sqrt{2}+1$$

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