Question

Solve the below trigonometric function,
$\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} =$
A.$\sec \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)$
B.$\cos \left( {\dfrac{\pi }{8} - \dfrac{\alpha }{2}} \right)$
C.$\tan \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)$
D.$\cot \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{2}} \right)$

Answer 1

First, we shall analyze the given information so that we can able to solve the problem. Here we are asked to calculate the value of the given trigonometric expression. To find the required answer, we need to apply the appropriate formulae and trigonometric identities in the given expression. Here, we need to take out $\sqrt 2$ as a common term from both the numerator and the denominator. Next, we need to arrange the terms for our convenience. Then, we need to apply the required formulae to obtain the desired answer.
Formula to be used:
The required trigonometric identity and trigonometric formulae to be used are as follows.
a) $\sin \dfrac{\pi }{4} = \dfrac{1}{2}$
b) $\cos \dfrac{\pi }{4} = \dfrac{1}{2}$
c) $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$
d) $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$
e) $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
f) $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$

Complete answer:
Let us solve the given expression.
$\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} = \dfrac{{\left( {\sqrt 2 - \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\sin \alpha - \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\cos \alpha } \right)}}{{\left( {\dfrac{{\sqrt 2 }}{{\sqrt 2 }}\sin \alpha - \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\cos \alpha } \right)}}$
In the above step, we have multiplied and divided by $\sqrt 2$ in the coefficient $1$
$= \dfrac{{\sqrt 2 \left( {1 - \dfrac{1}{{\sqrt 2 }}\sin \alpha - \dfrac{1}{{\sqrt 2 }}\cos \alpha } \right)}}{{\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin \alpha - \dfrac{1}{{\sqrt 2 }}\cos \alpha } \right)}}$ (Here we have taken $\sqrt 2$ as the common term)
$= \dfrac{{1 - \dfrac{1}{{\sqrt 2 }}\sin \alpha - \dfrac{1}{{\sqrt 2 }}\cos \alpha }}{{\dfrac{1}{{\sqrt 2 }}\sin \alpha - \dfrac{1}{{\sqrt 2 }}\cos \alpha }}$
$= \dfrac{{1 - \left[ {\dfrac{1}{{\sqrt 2 }}\sin \alpha + \dfrac{1}{{\sqrt 2 }}\cos \alpha } \right]}}{{\dfrac{1}{{\sqrt 2 }}\sin \alpha - \dfrac{1}{{\sqrt 2 }}\cos \alpha }}$ (Here we have taken $1$ as a common term)
$= \dfrac{{1 - \left[ {\sin \dfrac{\pi }{4}\sin \alpha + \cos \dfrac{\pi }{4}\cos \alpha } \right]}}{{\cos \dfrac{\pi }{4}\sin \alpha - \sin \dfrac{\pi }{4}\cos \alpha }}$ (Here we applied $\sin \dfrac{\pi }{4} = \dfrac{1}{2}$ and $\cos \dfrac{\pi }{4} = \dfrac{1}{2}$ )
$= \dfrac{{1 - \left[ {\cos \alpha \cos \dfrac{\pi }{4} + \sin \alpha \sin \dfrac{\pi }{4}} \right]}}{{\sin \alpha \cos \dfrac{\pi }{4} - \cos \alpha \sin \dfrac{\pi }{4}}}$ (Here we rewritten the terms for our convenience)
Now, we need to apply the formulae $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ and $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ .
Also, we can note that a is equal to $\alpha$ and b is equal to $\dfrac{\pi }{4}$ . So, we need to just put $a = \alpha$ and $b = \dfrac{\pi }{4}$ in the above formulae.
Thus, we have
$\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} = \dfrac{{1 - \left[ {\cos \left( {\alpha - \dfrac{\pi }{4}} \right)} \right]}}{{\sin \left( {\alpha - \dfrac{\pi }{4}} \right)}}$
$= \dfrac{{2{{\sin }^2}\dfrac{{\left( {\alpha - \dfrac{\pi }{4}} \right)}}{2}}}{{\sin \left( {\alpha - \dfrac{\pi }{4}} \right)}}$ (Here we applied the trigonometric identity $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$ and $\theta = \alpha - \dfrac{\pi }{4}$ )
$= \dfrac{{2{{\sin }^2}\dfrac{{\left( {\alpha - \dfrac{\pi }{4}} \right)}}{2}}}{{2\sin \dfrac{{\left( {\alpha - \dfrac{\pi }{4}} \right)}}{2}\cos \dfrac{{\left( {\alpha - \dfrac{\pi }{4}} \right)}}{2}}}$ (Here we applied the identity $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ and $\theta = \alpha - \dfrac{\pi }{4}$ )
$= \dfrac{{2{{\sin }^2}\left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)}}{{2\sin \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)\cos \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)}}$
$= \dfrac{{\sin \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)}}{{\cos \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)}}$
$= \tan \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)$
Therefore, $\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} = \tan \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)$

Hence, option C) is the required answer.

Note:
In the mid part of the problem, we can also assume $\theta = \alpha - \dfrac{\pi }{4}$.
$\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} = \dfrac{{1 - \left[ {\cos \left( {\alpha - \dfrac{\pi }{4}} \right)} \right]}}{{\sin \left( {\alpha - \dfrac{\pi }{4}} \right)}}$
In the above step, we shall assume that $\theta = \alpha - \dfrac{\pi }{4}$
Thus, we have $\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} =$ $\dfrac{{1 - \left[ {\cos \theta } \right]}}{{\sin \theta }}$
$= \dfrac{{2{{\sin }^2}\dfrac{\theta }{2}}}{{\sin \theta }}$ (Here we applied the trigonometric identity $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$ )
$= \dfrac{{2{{\sin }^2}\dfrac{\theta }{2}}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}$ (Here we applied the trigonometric identity $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ )
$= \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}$
$= \tan \dfrac{\theta }{2}$
Now, we shall substitute our assumption $\theta = \alpha - \dfrac{\pi }{4}$
Thus, we have $\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} = \tan \left( {\dfrac{{\alpha - \dfrac{\pi }{4}}}{2}} \right)$
$= \tan \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)$
Therefore, $\dfrac{{\left( {\sqrt 2 - \sin \alpha - \cos \alpha } \right)}}{{\left( {\sin \alpha - \cos \alpha } \right)}} = \tan \left( {\dfrac{\alpha }{2} - \dfrac{\pi }{8}} \right)$
Hence, we got the same answer as we got earlier. It will be easy for us when we assumed $\theta = \alpha - \dfrac{\pi }{4}$