If tan⁡θ+tan⁡ϕ=a,cot⁡θ+cot⁡ϕ=b,θ−ϕ=α(≠0) then

ab=tan⁡θtan⁡ϕ,tan2⁡α=tan2⁡(θ−φ)=tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ2 =a2−4(a/b)1+(a/b)2=ab(ab−4)(a+b)2Since tan2⁡a>0,ab>4

If tan⁡θ+tan⁡ϕ=a,cot⁡θ+cot⁡ϕ=b,θ−ϕ=α(≠0) then

ab=tan⁡θtan⁡ϕ,tan2⁡α=tan2⁡(θ−φ)=tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ2 =a2−4(a/b)1+(a/b)2=ab(ab−4)(a+b)2Since tan2⁡a>0,ab>4

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If $\tan θ + \tan ϕ = a, \cot θ + \cot ϕ = b, θ - ϕ = α (\neq 0)$ then

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$a b < 4$

$a b = 4$

$a b > 4$

$a b = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{a}{b} = \tan θ \tan ϕ, \tan^{2} α = \tan^{2} (θ - φ) \\ = {(\frac{\tan θ - \tan ϕ}{1 + \tan θ \tan ϕ})}^{2} \end{array}$

$= \frac{a^{2} - 4 (a / b)}{(1 + (a / b)^{2}} = \frac{a b (a b - 4)}{(a + b)^{2}}$

Since $\tan^{2} a > 0, a b > 4$

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If tan⁡θ+tan⁡ϕ=a,cot⁡θ+cot⁡ϕ=b,θ−ϕ=α(≠0) then

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If $\tan θ + \tan ϕ = a, \cot θ + \cot ϕ = b, θ - ϕ = α (\neq 0)$ then