If t1-tan x x- t2- x x- t3-tan xtan x- t4- xtan x- 0 - x - -4- then-

The correct option is D t_2 > t_4 > t_3 > t_1 Since 0 lt; x lt; π4 ∴tan x lt; 1 and x gt; 1 ..... (1) ∴ Choose ...

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Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If t1=tan x...

Question

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:

t_{1} < t_{2} < t_{3} < t_{4}

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

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

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

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Solution

The correct option is D $t_{2} > t_{4} > t_{3} > t_{1}$
Since $0 < x < \frac{π}{4}$

$∴ tan x < 1$ and $cot x > 1$ ..... $(1)$

$∴$ Choose $tan x = 1 - k_{1}$ and $cot x = 1 + k_{2}$ , where

$k_{1}$ and $k_{2}$ are very small $+$ ve quantities.

$∴ t_{3} = (1 - k_{1})^{1 - k_{1}}, t_{1} = (1 - k_{1})^{1 + k_{2}}$ .... $(2)$

$t_{4} = (1 + k_{2})^{1 - k_{1}}, t_{2} = (1 + k_{2})^{1 + k_{2}}$ ..... $(3)$

$∴ t_{4} > t_{3}$ by $(3)$ and $(2)$ , $t_{2} > t_{4}$ by $(2)$ and $(3)$

$∴ t_{2} > t_{4} > t_{3}$ . Also $t_{3} > t_{1}$

$∴ t_{2} > t_{4} > t_{3} > t_{1}$ .

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