Let θ ∈(0, (π/4)) and t1=( tan θ) tan θ, t2=( tan θ) cot θ, t3= ( cot θ) tan θ and t4=( cot θ) cot θ, then

Question

Let θ ∈(0, (π/4)) and t1=( tan θ) tan θ, t2=( tan θ) cot θ, t3= ( cot θ) tan θ and t4=( cot θ) cot θ, then (A) t1>t2>t3>t4 (B) t4>t3>t1>t2 (C) t3>t1>t2>t4 (D) t2>t3>t1>t4

Tardigrade Admin · Accepted Answer

We have θ ∈(0, (π/4)) t1 =( tan θ) tan θ ; t2=( tan θ) cot θ ; t3=( cot θ) tan θ t4 =( cot θ) cot θ - For θ ∈(0, (π/4)): tan θ ∈(0,1) and cot θ(1, ∞) This implies that t4=(>1)>t, t3=(>1)<1, t2=(<1)>1, t1=(<1)<1 Therefore, t4>t3>t1>t2

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

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

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

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

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

Q. Let θ∈(0,4π​) and t1​=(tanθ)tanθ,t2​=(tanθ)cotθ,t3​= (cotθ)tanθ and t4​=(cotθ)cotθ, then

t1​>t2​>t3​>t4​

t4​>t3​>t1​>t2​

t3​>t1​>t2​>t4​

t2​>t3​>t1​>t4​

We have θ∈(0,4π​) t1​=(tanθ)tanθ;t2​=(tanθ)cotθ;t3​=(cotθ)tanθ t4​=(cotθ)cotθ - For θ∈(0,4π​):tanθ∈(0,1) and cotθ(1,∞) This implies that t4​=(>1)>t,t3​=(>1)<1,t2​=(<1)>1,t1​=(<1)<1 Therefore, t4​>t3​>t1​>t2​

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

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

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

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

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