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

Θ∈(0, π4) ⇒tanθ lt; 1 and θ gt; 1. Let tanθ = 1 x and θ = 1 + y. where x, y gt; 0 and are very small, then nbsp; t_1 = (1 x)^1 x, t_2 = (1 x)^1+y,t_3 ...

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

Byju's Answer

Standard XIII

Mathematics

Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

If θ∈0, π4 an...

Question

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

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_{2}

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

Open in App

Solution

The correct option is B $t_{4} > t_{3} > t_{1} > t_{2}$
$θ \in (0, \frac{π}{4}) \Rightarrow tan θ < 1$ and $cot θ > 1.$
Let $tan θ = 1 - x$ and $cot θ = 1 + y .$ where $x, y > 0$ and are very small, then
$t_{1} = (1 - x)^{1 - x}, t_{2} = (1 - x)^{1 + y}, t_{3} = (1 + y)^{1 - x}, t_{4} = (1 + y)^{1 + y}$
Clearly, $t_{4} > t_{3}$ and $t_{1} > t_{2} .$ Also $t_{3} > t_{1} .$
Thus, $t_{4} > t_{3} > t_{1} > t_{2} .$

Suggest Corrections

Similar questions

Q. 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

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

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:

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

Join BYJU'S Learning Program

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

The correct option is B t4>t3>t1>t2θ∈(0,π4)⇒tanθ<1 and cotθ>1. Let tanθ=1−x and cotθ=1+y. where x,y>0 and are very small, then t1=(1−x)1−x,t2=(1−x)1+y,t3=(1+y)1−x,t4=(1+y)1+y Clearly, t4>t3 and t1>t2. Also t3>t1. Thus, t4>t3>t1>t2.

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