Question

if tan x = 3 cot X then X is a 45 degree b 30° c 60 degree and 90 degree​

Answer 1

Given,

$\tan( \alpha ) = 3 \cot( \alpha )$

To finD :

$\alpha = \: ?$

From the given relation, we write :

$\dfrac{ \tan( \alpha ) }{ \cot( \alpha) } = 3 \\ \\ \longmapsto \: \dfrac{ \dfrac{ \sin( \alpha ) }{ \cos( \alpha ) } }{ \dfrac{ \cos( \alpha ) }{ \sin( \alpha ) } } = 3 \\ \\ \longmapsto \: \dfrac{ { \sin( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} } = 3 \: \\ \\ \longmapsto \: \tan {}^{2} ( \alpha) = \tan {}^{2} (60) \\ \\ \longmapsto \: \boxed{ \boxed{ \alpha = {60}^{ \circ}}}$

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\huge {\tt GIVEN :-}$

$\tt tan x = 3\:cot x$

$\huge {\tt TO\:FIND :-}$

$\tt x = ?$

$\huge\underline {\tt ANSWER :-}$

$\leadsto{\tt tan x = 3\:cotx}$

$\leadsto {\tt \dfrac{tanx}{cotx}=3}$

$\leadsto {\tt \dfrac {sinx}{cosx }\times \dfrac {sinx}{cosx}= 3 }$

$\leadsto {\tt \dfrac {{sin}^{2}x}{{cos}^{2}x}=3}$

$\leadsto {\tt {tan}^{2}x=3}$

$\leadsto {\tt {tan}^{2}x= {\sqrt {3}}^{2}}$

$\leadsto {\tt {tan}^{2}x={tan}^{2}60}$

Because Tan60 = √3 → tan²60 = (√3)² = 3

$\leadsto \huge {\boxed{\tt x=60}}$

$\rule{200}2$