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Sagot :
Recall the angle sum identities for cosine,
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
cos(x - y) = cos(x) cos(y) + sin(x) sin(y)
Now, by definition of tangent, we have
tan(x) = sin(x) / cos(x)
⇒ tan(44°) tan(46°) = sin(44°) sin(46°) / (cos(44°) cos(46°))
From the identities above, we can show that
cos(x) cos(y) = 1/2 (cos(x - y) + cos(x + y))
sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))
so that
sin(44°) sin(46°) = 1/2 (cos(44° - 46°) - cos(44° + 46°))
⇒ sin(44°) sin(46°) = 1/2 (cos(-2°) - cos(90°))
⇒ sin(44°) sin(46°) = 1/2 cos(2°)
and
cos(44°) cos(46°) = 1/2 (cos(44° - 46°) + cos(44° + 46°))
⇒ cos(44°) cos(46°) = 1/2 (cos(-2°) + cos(90°))
⇒ cos(44°) cos(46°) = 1/2 cos(2°)
and in turn, it follows that
tan(44°) tan(46°) = (1/2 cos(2°)) / (1/2 cos(2°))
⇒ tan(44°) tan(46°) = 1
We have to prove that tan (44°) tan (46°) = 1 . So , for this we will use the sum identity for the tangent function i.e.
- [tex]{\boxed{\bf{\tan (x+y)=\dfrac{\tan (x)+\tan (y)}{1-\tan (x)\tan (y)}}}}[/tex]
So , here , we can easily get the answer by just putting x = 44° and y = 46° but let's first re-write it and then simplify to get the term tan(x) tan (y) , so that the answer will be just equal to what the product of tan (x) and tan (y) equal to
[tex]{:\implies \quad \sf \tan (x+y)\{1-\tan (x)\tan (y)\}=\tan (x)+\tan (y)}[/tex]
[tex]{:\implies \quad \sf 1-\tan (x)\tan (y)=\dfrac{\tan (x)+\tan (y)}{\tan (x+y)}}[/tex]
[tex]{:\implies \quad \sf \tan (x)\tan (y)=\dfrac{\tan (x)+\tan (y)}{\tan (x+y)}+1}[/tex]
Now , put x = 44° and y = 46°
[tex]{:\implies \quad \sf \tan (44^{\circ})\tan (46^{\circ})=\dfrac{\tan (44^{\circ})+\tan (46^{\circ})}{\tan (44^{\circ}+46^{\circ})}+1}[/tex]
[tex]{:\implies \quad \sf \tan (44^{\circ})\tan (46^{\circ})=\dfrac{\tan (44^{\circ})+\tan (46^{\circ})}{\tan (90^{\circ})}+1}[/tex]
[tex]{:\implies \quad \sf \tan (44^{\circ})\tan (46^{\circ})=\dfrac{\tan (44^{\circ})+\tan (46^{\circ})}{\infty}+1\quad \qquad \{\because \tan (90^{\circ})=\infty\}}[/tex]
[tex]{:\implies \quad \sf \tan (44^{\circ})\tan (46^{\circ})=0+1}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{\tan (44^{\circ})\tan (46^{\circ})=1}}}[/tex]
Hence , Proved
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