Question

1-cos 2ø÷sin 2ø= tan ø. prove​

Answer 1

To Prove :-

$\rm :\longmapsto\: \:\: \:\dfrac{1 - cos2 \phi \: }{sin2\phi \:} = tan\phi \:$

$\large\underline{\sf{Solution-}}$

Consider LHS,

$\rm :\longmapsto\: \:\: \:\dfrac{1 - cos2 \phi \: }{sin2\phi \:}$

We know,

$\boxed{ \red{ \bf \: 1 - cos2x = {2sin}^{2} x}}$

and

$\boxed{ \red{ \bf \:sin2x = 2sinx \: cosx}}$

On substituting these values, we get

$\rm :\longmapsto\: \: = \: \:\dfrac{ {2sin}^{2}\phi \: }{2sin\phi \: \: cos\phi \:}$

$\rm :\longmapsto\: \: = \: \:\dfrac{sin\phi \:}{cos\phi \:}$

$\rm :\longmapsto\: \: = \: \:tan\phi \:$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \red{ \bf \: cos2x = {2cos}^{2} x - 1}}$

$\boxed{ \red{ \bf \: cos2x = 1 - {2sin}^{2} x }}$

$\boxed{ \red{ \bf \: cos2x = {cos}^{2} x - {sin}^{2} x }}$

$\boxed{ \red{ \bf \: tan2x = \dfrac{2tanx}{1 - {tan}^{2}x }}}$

$\boxed{ \red{ \bf \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x }}}$

$\boxed{ \red{ \bf \: 1 + cos2x = {2cos}^{2} x}}$

$\boxed{ \red{ \bf \: 1 - cos2x = {2sin}^{2} x}}$