Question

Please solve the question​

Answer 1

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

$\rm :\longmapsto\:0 < x < \dfrac{\pi}{4}$

and

$\rm :\longmapsto\:cosx + sinx = \dfrac{5}{4}$

and

$\rm :\longmapsto\:cosx - sinx = \dfrac{ \sqrt{k} }{m}$

where, k and m are relatively primes.

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{the \: value \: of \: k + m}\end{cases}\end{gathered}\end{gathered}$

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

Given that,

$\rm :\longmapsto\:cosx + sinx = \dfrac{5}{4}$

We know that,

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

So, Replace x by sinx and y by cosx, we get,

$\rm :\longmapsto\: {(cosx + sinx)}^{2} + {(cosx - sinx)}^{2} = 2( {sin}^{2}x + {cos}^{2}x)$

On substituting the value of sinx + cosx, we get

$\rm :\longmapsto\:\dfrac{25}{16} + {(cosx - sinx)}^{2} = 2 \times 1$

$\rm :\longmapsto\:\dfrac{25}{16} + {(cosx - sinx)}^{2} = 2$

$\rm :\longmapsto\: {(cosx - sinx)}^{2} = 2 - \dfrac{25}{16}$

$\rm :\longmapsto\: {(cosx - sinx)}^{2} = \dfrac{32 - 25}{16}$

$\rm :\longmapsto\: {(cosx - sinx)}^{2} = \dfrac{7}{16}$

$\rm :\longmapsto\: {cosx - sinx}= \pm \: \dfrac{ \sqrt{7} }{4} - - - (1)$

But it is given that,

$\rm :\longmapsto\:0 < x < \dfrac{\pi}{4}$

and we know that,

$\rm :\longmapsto\:0 < x < \dfrac{\pi}{4} \: \implies \: cosx > sinx$

So, equation (1) reduces to

$\rm :\longmapsto\: {cosx - sinx}= \: \dfrac{ \sqrt{7} }{4}$

But it is given that,

$\rm :\longmapsto\:cosx - sinx = \dfrac{ \sqrt{k} }{m}$

So, On comparing, we get

$\rm :\longmapsto\:k = 7 \: \: \: and \: \: \: m = 4$

And k and m are relatively prime as HCF (4, 7) = 1

Hence,

$\bf\implies \:k + m = 7 + 4 = 11$

Additional Information :-

$\red{ \boxed{ \rm{ \: sin(A + B) = sinAcosB + sinBcosA}}}$

$\red{ \boxed{ \rm{ \: sin(A - B) = sinAcosB - sinBcosA}}}$

$\red{ \boxed{ \rm{ \: cos(A + B) = cosAcosB - sinAsinB}}}$

$\red{ \boxed{ \rm{ \: cos(A - B) = cosAcosB + sinAsinB}}}$

$\red{ \boxed{ \rm{ \: tan(A + B) = \frac{tanA + tanB}{1 - tanAtanB}}}}$

$\red{ \boxed{ \rm{ \: tan(A - B) = \frac{tanA - tanB}{1 + tanAtanB}}}}$