Question

if m=cosA-sinA and n=cosA+sinA, then prove that m/n - n/m = (-4sinAcosA)/(cos²A-sin²A) = -4/(cotA-tanA).
whoever will answer correctly, i'll mark him/her as brainliest ​

Answer 1

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

Given that

$\rm :\longmapsto\:m = cosA - sinA$

and

$\rm :\longmapsto\:n = cosA + sinA$

Now, Consider

$\rm :\longmapsto\:\dfrac{m}{n} - \dfrac{n}{m}$

$\rm \:  =  \:  \:\dfrac{ {m}^{2} - {n}^{2} }{mn}$

$\rm \:  =  \:  \: \dfrac{(m + n)(m - n)}{mn}$

$\rm \:  =  \:  \: \dfrac{(cosA - sinA + cosA + sinA)(cosA - sinA - cosA - sinA)}{(cosA + sinA)(cosA - sinA)}$

$\rm \:  =  \:  \: \dfrac{(2cosA)( - 2sinA)}{ {cos}^{2}A - {sin}^{2}A}$

$\rm \:  =  \:  \: \dfrac{ - 4 \: cosA \: sinA}{ {cos}^{2}A - {sin}^{2}A}$

Divide numerator and denominator by sinA cosA

$\rm \:  =  \:  \: \dfrac{ - 4}{\dfrac{ {cos}^{2}A - {sin}^{2}A}{cosA \: sinA} }$

$\rm \:  =  \:  \: \dfrac{ - 4}{\dfrac{ {cos}^{2}A }{cosAsinA} - \dfrac{ {sin}^{2} A}{cosAsinA} }$

$\rm \:  =  \:  \: \dfrac{ - 4}{\dfrac{cosA}{sinA} - \dfrac{sinA}{cosA} }$

$\rm \:  =  \:  \: \dfrac{ - 4}{cotA - tanA}$

Hence, proved

Identities Used :-

$\boxed{ \sf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

$\boxed{ \sf{ \: \frac{sinA}{cosA} = tanA}}$

$\boxed{ \sf{ \: \frac{cosA}{sinA} = cotA}}$

More to know

$\boxed{ \sf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

$\boxed{ \sf{ \: {sec}^{2}x - {tan}^{2}x = 1}}$

$\boxed{ \sf{ \: {cosec}^{2}x - {cot}^{2}x = 1}}$

Answer 2

$\large\underline{\sf{Solution-}}</p><p>$

Given that

$\rm :\longmapsto\:m = cosA - sinA \\ \\ and \\ \\ \rm :\longmapsto\:n = cosA + sinA</p><p>$

Now, Consider

$\rm :\longmapsto\:\dfrac{m}{n} - \dfrac{n}{m} \\ \\ \rm \: = \: \:\dfrac{ {m}^{2} - {n}^{2} }{mn} \\ \\ \rm \: = \: \: \dfrac{(m + n)(m - n)}{mn} \\ \\ \rm \: = \: \: \dfrac{(cosA - sinA + cosA + sinA)(cosA - sinA - cosA - sinA)}{(cosA + sinA)(cosA - sinA)} \\ \\ \rm \: = \: \: \dfrac{(2cosA)( - 2sinA)}{ {cos}^{2}A - {sin}^{2}A} </p><p> \\ \\ \rm \: = \: \: \dfrac{ - 4 \: cosA \: sinA}{ {cos}^{2}A - {sin}^{2}A} </p><p> </p><p>$

Divide numerator and denominator by sinA cosA

$\rm \: = \: \: \dfrac{ - 4}{\dfrac{ {cos}^{2}A - {sin}^{2}A}{cosA \: sinA} } \\ \\ \rm \: = \: \: \dfrac{ - 4}{\dfrac{ {cos}^{2}A }{cosAsinA} - \dfrac{ {sin}^{2} A}{cosAsinA} } \\ \\ \rm \: = \: \: \dfrac{ - 4}{\dfrac{cosA}{sinA} - \dfrac{sinA}{cosA} } \\ \\ \rm \: = \: \: \dfrac{ - 4}{cotA - tanA} \\ \\ Hence, \: \: proved \\ \\ Identities \: \: Used :- \\ \\ \boxed{ \sf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}} \\ \\ \boxed{ \sf{ \: \frac{sinA}{cosA} = tanA}}$

$\boxed{ \sf{ \: \frac{cosA}{sinA} = cotA}}$