Let -0- -4 and t 1-tan-tan- t 2tan- t 3-tan- and t4-tan- thenA- t1-t2-t3-t4B- t4-t3-t1-t2C- t3-t1-t2-t4D- t2-t3-t1-t4

The correct option is B A when θ∈( 0 , π/4) , tan θ lt; cot θ nbsp; Sin tan θ lt; 1 and cot θ gt; 1 nbsp; ∴ (tan θ)^tan θ lt; 1 and (cot θ)^ta ...

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Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

Let θ∈0, π/4 ...

Question

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

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Solution

The correct option is B

A when $θ \in (0, \frac{π}{4}), t a n θ < c o t θ$

Sin $t a n θ < 1 a n d c o t θ > 1$

$∴$ $(t a n θ)^{t a n θ} < 1 a n d (c o t θ)^{t a n θ} > 1$

$∴$ $t_{4} > t_{1}$ which only holds in (b).

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Let θ∈(0,π4) and t1=(tanθ)tanθ,t2(tanθ)cott3=(cotθ)tanθ and t4(cotθ)tanθ , then

The correct option is B A when θ∈(0,π4),tanθ<cotθ Sin tanθ<1 and cotθ>1 ∴ (tanθ)tanθ<1 and (cotθ)tanθ>1 ∴ t4>t1 which only holds in (b).

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

The correct option is B

A when $θ \in (0, \frac{π}{4}), t a n θ < c o t θ$

Sin $t a n θ < 1 a n d c o t θ > 1$

$∴$ $(t a n θ)^{t a n θ} < 1 a n d (c o t θ)^{t a n θ} > 1$

$∴$ $t_{4} > t_{1}$ which only holds in (b).