- Let -0- -4 and t1-tan-tan-t2-tan- t3-tan-and t4- then

The correct option is A t_4>t_3>t_1>t_2 Given θ∈(0, π/4) , then tanθ lt;1 and θ gt;1 .Let tanθ=1 h and θ=1+h where h ...

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Standard XII

Mathematics

Property 2

. Let θ∈0, π...

Question

. Let $θ \in (0, \frac{π}{4})$ and $t_{1} = (tan θ)^{tan θ}, t_{2} = (tan θ)^{cot θ}, t_{3} = (cot θ)^{tan θ}$ and $t_{4} = (cot θ)^{cot θ}$ , then

t_{1} > t_{2} > t_{3} > t_{4}

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t_{4} > t_{3} > t_{1} > t_{2}

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t_{3} > t_{1} > t_{2} > t_{4}

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t_{2} > t_{3} > t_{1} > t_{4}

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Solution

The correct option is A $t_{4} > t_{3} > t_{1} > t_{2}$
Given $θ \in (0, \frac{π}{4})$ , then $tan θ < 1$ and $cot θ > 1$ .
Let $tan θ = 1 - h$ and $cot θ = 1 + h$ where $h$ is very small and positive.
then $t_{1} = (1 - h)^{1 - h}, t_{2} = (1 - h)^{1 + h},$ $t_{3} = (1 + h)^{1 - h}$
and $t_{4} = (1 + h)^{1 + h}$
Hence $t_{4} > t_{3} > t_{1} > t_{2}$ .

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. Let θ∈(0,π4) and t1=(tanθ)tanθ, t2=(tanθ)cotθ, t3=(cotθ)tanθ and t4=(cotθ)cotθ, then

The correct option is A t4>t3>t1>t2Given θ∈(0,π4), then tanθ<1 and cotθ>1.Let tanθ=1−h and cotθ=1+h where h is very small and positive. then t1=(1−h)1−h, t2=(1−h)1+h, t3=(1+h)1−hand t4=(1+h)1+hHence t4>t3>t1>t2.

. Let $θ \in (0, \frac{π}{4})$ and $t_{1} = (tan θ)^{tan θ}, t_{2} = (tan θ)^{cot θ}, t_{3} = (cot θ)^{tan θ}$ and $t_{4} = (cot θ)^{cot θ}$ , then