Question

if tanA=12/13 find the value of 2sinAcosA/sinA²-cos²A​

Answer 1

Solution:-

Given:-

$\rm \implies \: \tan A \: = \dfrac{12}{13}$

To find:-

$\implies \rm\dfrac{2 \sin A \cos A}{ \sin {}^{2} A - \cos^{2} A}$

Now we know

$\rm \implies \: \tan A \: = \dfrac{12}{13} = \dfrac{p}{b}$

Now using pythagoras theorem

$\rm \: \implies \: {h}^{2} = {b}^{2} + {p}^{2}$

We have

$\rm \: p = 12,b = 13 \: and \: h = x$

Now

$\rm \implies {x}^{2} = (13 ){}^{2} + (12) ^{2}$

$\rm \implies \: {x}^{2} = 169 + 144$

$\rm \implies {x}^{2} = 313$

$\rm \implies \: x = \sqrt{313}$

So

$\rm \implies\: h = \sqrt{313}$

we get

$\rm \to \: \sin \: A = \dfrac{p}{h} = \dfrac{12}{ \sqrt{313} }$

$\rm \to \cos A = \dfrac{b}{h} = \dfrac{13}{ \sqrt{313} }$

Put the value on

$\implies \rm\dfrac{2 \sin A \cos A}{ \sin {}^{2} A - \cos^{2} A}$

$\rm \implies \dfrac{2 \times \dfrac{12}{ \sqrt{313} } \times \dfrac{13}{ \sqrt{313} } }{ \bigg( \dfrac{12}{ \sqrt{313} } \bigg)^{2} - \bigg(\dfrac{13}{ \sqrt{313} } \bigg)^{2} }$

$\rm \implies \dfrac{ \dfrac{312}{313} }{ \dfrac{144}{313} - \dfrac{169}{313} }$

$\rm \implies \dfrac{ \dfrac{312}{313} }{ \dfrac{ - 25}{313} }$

$\rm \implies \dfrac{312}{313} \times \dfrac{ - 313}{25}$

$\rm \implies \dfrac{312}{ \cancel{313}} \times \dfrac{ \cancel{ - 313}}{25}$

Answer

$\rm \implies \: - \dfrac{312}{25}$