If tan(α+β)=1,cos(α−β)=32 then cot(3α+β).cot(α+3β)=

tan(α+β)=1⇒α+β=450cos(α−β)=32⇒α−β=300 Now 3α+β=1200,α+3β=600 cot(3α+β).cot(α+3β)=−13×13=−13

If tan(α+β)=1,cos(α−β)=32 then cot(3α+β).cot(α+3β)=

tan(α+β)=1⇒α+β=450cos(α−β)=32⇒α−β=300 Now 3α+β=1200,α+3β=600 cot(3α+β).cot(α+3β)=−13×13=−13

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If $\tan (α + β) = 1, \cos (α - β) = \frac{\sqrt{3}}{2}$ then $\cot (3 α + β) . \cot (α + 3 β) =$

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$\frac{1}{3}$

$- \frac{1}{3}$

–3

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Detailed Solution

$\begin{matrix} \tan (α + β) = 1 \Rightarrow α + β = 45^{0} \\ \cos (α - β) = \frac{\sqrt{3}}{2} \Rightarrow α - β = 30^{0} \end{matrix}$

Now $3 α + β = 120^{0}, α + 3 β = 60^{0}$

$\cot (3 α + β) . \cot (α + 3 β) = - \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = - \frac{1}{3}$

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If tan(α+β)=1,cos(α−β)=32 then cot(3α+β).cot(α+3β)=

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If $\tan (α + β) = 1, \cos (α - β) = \frac{\sqrt{3}}{2}$ then $\cot (3 α + β) . \cot (α + 3 β) =$