Question

if tan∅+ 1/tan∅=2 then show that tan^2∅+1/tan^2∅=2

Answer 1

Let theta= x (so calculation is simpler)

$\tan(x) + \frac{1}{ \tan(x) } = 2$

Squarring both sides

${( \tan(x) + \frac{1}{ \tan(x) })}^{2} = 4$

${ \tan( x) }^{2} + \frac{1}{ { \tan(x) }^{2} } - 2 = 4$

${ \tan( x) }^{2} + \frac{1}{ { \tan(x) }^{2} } = 2$

Answer 2

$\red {\huge{ \underline{ \overline{ \mid{ \bold{ \purple{ \: \: SOLUTION \: \: \red{\mid}}}}}}}}$

$\underline{\bold{ \: \: GIVEN \: \: : }} \: \: \: \: \bold{tan \phi + \frac{1}{tan \phi} = 2 }\\ \\ \underline{ \bold{ \: \: TO \: \: SHOW\: : }} \: \: \: \bold{tan {}^{2} \phi \: + \: \frac{1}{tan {}^{2} \phi} = 2 }$

$\bold{tan \phi + \frac{1}{tan \phi } = 2 } \\ \\ \bold{ = > \frac{tan {}^{2} \phi + 1}{tan \phi} = 2} \\ \\ \bold{ = > tan \phi {}^{2} + 1 = 2 \: tan \phi} \\ \\ \bold{ = > tan {}^{2} \phi - 2 \: tan \phi + 1 = 0 } \\ \\ \bold{ = > ( tan \phi - 1) {}^{2} = 0} \\ \\ \bold{ = > tan\phi - 1 =0} \\ \\ \bold{ = > tan \phi= 1 \: \: - - - - - - > (1)}$

$\underline{ \bold{ \: \: Putting \: \: the \: \: value \: \: of \: \: \boxed{ \: tan \phi \: } \: \: in \: \: L.H.S. \: \: }} \\ \\ \bold{L.H.S. = tan {}^{2} \phi + \frac{1}{tan \phi} }\\ \\ \bold{ = (1) {}^{2} + \frac{1}{1} \: \: \: \: \: [ From\: \: (1) \:] } \\ \\ \bold{ = 1 + 1} \\ \\ \bold{ = 2} \\ \\ \bold{ = R.H.S.}$

$\boxed{\bold{\purple{tan {}^{2} \phi \: + \: \frac{1}{tan {}^{2} \phi} = 2 }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \bold{ \: \: Hence \: \: Proved. \: \: }}$