First slide

Trigonometric Identities

Question

Which of the following is/are correct?

Moderate

A

$(\tan x)^{\ln (\sin x)} > (\cot x)^{\ln (\sin x)}, \forall x \in (0, π / 4)$
B

$4^{\ln cosec x} < 5^{\ln cosec x}, \forall x \in (0, π / 2)$
C

$(1 / 2)^{\ln (\cos x)} < (1 / 3)^{\ln (\cos x)}, \forall x \in (0, π / 2)$
D

$2^{\ln (\tan x)} > 2^{\ln (\sin x)}, \forall x \in (0, π / 2)$

Solution

a.
$\begin{array}{l} For, x \in (0, \frac{π}{4}), \tan x < \cot x \\ Also, \ln (\sin x) < 0 \end{array}$
$\Rightarrow (\tan x)^{\ln (\sin x)} > (\cot x)^{\ln (\sin x)}$
b. For $x \in (0, \frac{π}{2}), cosec x \geq 1$
$\begin{array}{l} \Rightarrow \ln (cosec x) \geq 0 \\ \Rightarrow 4^{\ln (cosec x)} < 5^{\ln (cosec x)} \end{array}$
c.
$\begin{array}{l} x \in (0, \frac{π}{2}) \Rightarrow \cos x \in (0, 1) \\ \Rightarrow \ln (\cos x) < 0 \\ Also, \frac{1}{2} > \frac{1}{3} \\ \Rightarrow {(\frac{1}{2})}^{\ln (\cos x)} < {(\frac{1}{3})}^{\ln (\cos x)} \end{array}$
d.
For $x \in (0, \frac{π}{2})$
Since sin x < tan x, we get
ln(sin x) < ln(tan x)
$\Rightarrow 2^{\ln (\sin x)} < 2^{\ln (\tan x)}$

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