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8cos-x-8sin-x-3-55tan-x-55-tan-x-




Question Number 168664 by cortano1 last updated on 15/Apr/22
      { ((8cos x−8sin x=3)),((55tan x+((55)/(tan x)) =?)) :}
$$\:\:\:\:\:\begin{cases}{\mathrm{8cos}\:{x}−\mathrm{8sin}\:{x}=\mathrm{3}}\\{\mathrm{55tan}\:{x}+\frac{\mathrm{55}}{\mathrm{tan}\:{x}}\:=?}\end{cases} \\ $$
Answered by greogoury55 last updated on 15/Apr/22
 (8cos x−8sin x)^2  = 9   64−64sin 2x = 9  ⇒55 = 64sin 2x  ⇒sin 2x= ((55)/(64))   ⇒55(((sin x)/(cos x)) +((cos x)/(sin x)))=  ⇒55((2/(sin 2x)))= ((55×2)/((((55)/(64))))) = 128
$$\:\left(\mathrm{8cos}\:{x}−\mathrm{8sin}\:{x}\right)^{\mathrm{2}} \:=\:\mathrm{9} \\ $$$$\:\mathrm{64}−\mathrm{64sin}\:\mathrm{2}{x}\:=\:\mathrm{9} \\ $$$$\Rightarrow\mathrm{55}\:=\:\mathrm{64sin}\:\mathrm{2}{x} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}=\:\frac{\mathrm{55}}{\mathrm{64}} \\ $$$$\:\Rightarrow\mathrm{55}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\:+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right)= \\ $$$$\Rightarrow\mathrm{55}\left(\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}{x}}\right)=\:\frac{\mathrm{55}×\mathrm{2}}{\left(\frac{\mathrm{55}}{\mathrm{64}}\right)}\:=\:\mathrm{128}\: \\ $$

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