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Question Number 170752 by balirampatel last updated on 30/May/22
Solve:− ((sinx + cosx)/(sinx − cosx)) = (((√(3 )) − 1)/( (√(3 )) + 1))
$${Solve}:−\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$
Answered by Rasheed.Sindhi last updated on 30/May/22
((sinx + cosx)/(sinx − cosx)) = (((√(3 )) − 1)/( (√(3 )) + 1)) determinant ((((a/b)=(c/d)⇒((a+b)/(a−b))=((c+d)/(c−d))))) (((sinx + cosx)+(sinx − cosx))/((sinx + cosx)−(sinx − cosx))) = ((((√(3 )) − 1)+((√(3 )) + 1))/( ((√(3 )) − 1)−((√(3 )) + 1))) ((2sinx)/(2cosx))=((2(√3))/(−2)) tanx=−(√3) x=−(π/3)+nπ
$$\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$$$\begin{array}{|c|}{\frac{{a}}{{b}}=\frac{{c}}{{d}}\Rightarrow\frac{{a}+{b}}{{a}−{b}}=\frac{{c}+{d}}{{c}−{d}}}\\\hline\end{array} \\ $$$$\:\:\frac{\left({sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}\right)+\left({sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}\right)}{\left({sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}\right)−\left({sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}\right)+\left(\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}\right)}{\:\left(\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}\right)−\left(\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}\right)}\: \\ $$$$\frac{\mathrm{2}{sinx}}{\mathrm{2}{cosx}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{−\mathrm{2}} \\ $$$${tanx}=−\sqrt{\mathrm{3}} \\ $$$${x}=−\frac{\pi}{\mathrm{3}}+{n}\pi \\ $$
Answered by Rasheed.Sindhi last updated on 30/May/22
((sinx + cosx)/(sinx − cosx)) = (((√(3 ))+(−1))/( (√(3 )) −(−1))) determinant (((((a+b)/(a−b))=((c+d)/(c−d)) ⇒(a/b)=(c/d)))) ((sinx)/(cosx))=((√3)/(−1)) tanx=−(√3) x=−(π/3)+nπ
$$\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}+\left(−\mathrm{1}\right)}{\:\sqrt{\mathrm{3}\:}\:−\left(−\mathrm{1}\right)}\: \\ $$$$\begin{array}{|c|}{\frac{{a}+{b}}{{a}−{b}}=\frac{{c}+{d}}{{c}−{d}}\:\Rightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}}}\\\hline\end{array} \\ $$$$\frac{{sinx}}{{cosx}}=\frac{\sqrt{\mathrm{3}}}{−\mathrm{1}} \\ $$$${tanx}=−\sqrt{\mathrm{3}} \\ $$$${x}=−\frac{\pi}{\mathrm{3}}+{n}\pi \\ $$

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