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

The correct option is D t_4>t_3>t_1>t_2 When θ∈(0,π4) tanθ lt;θ and tanθ lt;1 and θ gt;1 . ∴ (tanθ)^θ lt;1 and (θ)^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Property 2

Let θϵ 0,π/...

Question

Let $θ ϵ (0, \frac{π}{4})$ 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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $t_{4} > t_{3} > t_{1} > t_{2}$
When $θ \in (0, \frac{π}{4})$

$tan θ < cot θ$

and $tan θ < 1$ and $cot θ > 1$ .

$∴ (tan θ)^{cot θ} < 1$ and $(cot θ)^{tan θ} > 1$

$∴ t_{3} > t_{2}$

$(cot θ)^{cot θ} > 1$ and it is also greater than

$(cot θ)^{tan θ}$ as $tan θ < 1$

$∴ t_{4} > t_{3}$

Between $t_{2}$ and $t_{1}$ ,

as $(tan θ)^{cot θ} = \frac{1}{(cot θ)^{cot θ}}$ in the power of denominators.

$cot θ > tan θ$

$cot θ^{cot θ} > tan θ^{cot θ}$

So, $\frac{1}{(cot θ)^{cot θ}} < \frac{1}{(cot θ)^{tan θ}}$

$∴ t_{2} < t_{1}$

$∴ t_{2} < t_{1}$

$∴ t_{4} > t_{3} > t_{1} > t_{2}$

Suggest Corrections

Similar questions

Let $θ \in (0, \frac{π}{4}) a n d t_{1} = (t a n θ)^{t a n θ}, t_{2} (t a n θ)^{c o t} t_{3} = (c o t θ)^{t a n θ} a n d t_{4} (c o t θ)^{t a n θ}$ , then

Q. If

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

A block of mass 'm' is suspended by different springs of spring constants shown in the figure.

Let time period of oscillation in these four positions be T_1,T₂, T₃ and T₄. Then

Q. If with standard notations

t_{1}, t_{2}, t_{3}, t_{4}

are four co–normal points on the hyperbola

x y = c^{2}

then the orthocentre of the triangle formed by joining the points

t_{1}, t_{2}, t_{3}

is given by

Q. If

t_{1} = (tan x)^{cot x}, t_{2} = (cot x)^{cot x}, t_{3} = (tan x)^{tan x}, t_{4} = (cot x)^{tan x}, 0 < x < \frac{π}{4}

, then:

Join BYJU'S Learning Program

Let θϵ(0,π4)and t1=(tanθ)tanθ,t2=(tanθ)cotθ,t3=(cotθ)tanθand t4=(cotθ)cotθthen

Let $θ ϵ (0, \frac{π}{4})$ 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