Question

pls help me to solve this problem ​

Answer 1

Answer:

$\sf\:If\:\dfrac{\cos^4x }{\cos^2y} + \dfrac{\sin^4x}{\sin^2y} = 1\sf, then\:find\:\dfrac{\cos^4y}{\cos^2x} + \dfrac{\sin^4y}{\sin^2x} = ?$

⠀

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf \dfrac{\cos^4x}{\cos^2y} + \dfrac{\sin^4x}{\sin^2y}=1\\\\\\:\implies\sf \dfrac{\cos^4x\sin^2y + \sin^4x\cos^2y}{\cos^2y\sin^2y}=1\\\\\\:\implies\sf \cos^4x\sin^2y + (\sin^2x)^2\cos^2y = \cos^2y\sin^2y\\\\\\:\implies\sf \cos^4x\sin^2y + (1 - \cos^2x)^2\cos^2y = \cos^2y\sin^2y\\\\\\:\implies\sf \cos^4x\sin^2y + (\cos^4x + 1 - 2\cos^2x)\cos^2y = \cos^2y\sin^2y\\\\\\:\implies\sf \cos^4x\sin^2y + \cos^4x\cos^2y + \cos^2y - 2\cos^2x\cos^2y = \cos^2y\sin^2y\\\\\\:\implies\sf \cos^4x(\sin^2y + \cos^2y) + \cos^2y - \cos^2y\sin^2y- 2\cos^2x\cos^2y = 0\\\\\\:\implies\sf \cos^4x(1) + \cos^2y(1 - \sin^2y) - 2\cos^2x\cos^2y = 0\\\\\\:\implies\sf \cos^4x + \cos^2y(\cos^2y) - 2\cos^2x\cos^2y = 0\\\\\\:\implies\sf \cos^4x + \cos^4y- 2\cos^2x\cos^2y = 0\\\\\\:\implies\sf (\cos^2x - \cos^2y)^2 = 0\\\\\\:\implies\sf \cos^2x - \cos^2y = 0\\\\\\:\implies\sf \cos^2x = \cos^2y\\\\\\:\implies\sf Likewise, \:we\:get\: :\sin^2x = \sin^2y$

$\rule{200px}{.3ex}$

$\dashrightarrow\sf\:\: \dfrac{\cos^4y}{\cos^2x} + \dfrac{\sin^4y}{\sin^2x}\\\\\\\dashrightarrow\sf\:\: \dfrac{\cos^4y}{\cos^2y} + \dfrac{\sin^4y}{\sin^2y}\\\\\\\dashrightarrow\sf\:\: \dfrac{(\cos^2y)^{2}}{\cos^2y} + \dfrac{(\sin^2y)^{2}}{\sin^2y}\\\\\\\dashrightarrow\sf\:\: \cos^2y +\sin^2y\\\\\\\dashrightarrow\sf\:\: 1$

⠀

$\therefore\:\underline{\textsf{Hence, correct option is b) \textbf{1}}}.$

Answer 2

AnswEr-:

• $\sf{ If\: \dfrac{cos^{4} x}{cos^{2}y} + \dfrac{sin^{4}x }{sin^{2}y}= \:1 , \:then \: \dfrac{cos^{4} y}{cos^{2}x} + \dfrac{sin^{4}y }{sin^{2}x} \: = \:??}$

$\mathrm {\bf{\dag{Solution \:of\:Question \:-:}}}$

• $\underline{\mathrm {\bf{\star{Now\:According \:to\:the\:Question \:-:}}}}$

• $:\implies {\sf{ \dfrac{cos^{4} x}{cos^{2}y} + \dfrac{sin^{4}x }{sin^{2}y}= \:1}}$

• $\mathrm {L.C.M\:of\: = \: cos^{2}y \:and\:sin^{2}y \:is \:cos^{2}y sin^{2}y}$

• $:\implies {\sf{ \dfrac{ cos^{4}x sin^{2}y + sin^{4}x cos^{2}y }{cos^{2}y sin^{2}y } = \:1}}$

• $\mathrm {By\:Cross\:Multiplication}$

• $:\implies {\sf{ \dfrac{ cos^{4}x sin^{2}y + sin^{4}x cos^{2}y }{cos^{2}y sin^{2}y } = \:1}}$

• $:\implies {\sf{ cos^{4}x sin^{2}y + sin^{4}x cos^{2}y = \:cos^{2}y sin^{2} }}$

• $:\implies {\sf{ cos^{4}x sin^{2}y + (sin^{2}x)^{2} cos^{2}y = \:cos^{2}y sin^{2} }}$

• $\sf{ 1-cos^{2}x =sin^{2}x}$

• $:\implies {\sf{ cos^{4}x sin^{2}y + \purple{(sin^{2}x)}^{2} cos^{2}y = \:cos^{2}y sin^{2} }}$

• $:\implies {\sf{ cos^{4}x sin^{2}y + \purple{(1 - cos^{2}x)}^{2} cos^{2}y = \:cos^{2}y sin^{2} }}$

• $\mathrm {(a-b)^{2} = a^{2} + b^{2} - 2ab }$

• $:\implies {\sf{ cos^{4}x sin^{2}y + \purple{(1-cos^{2}x)^{2}} cos^{2}y = \:cos^{2}y sin^{2} }}$

• $:\implies {\sf{ cos^{4}x sin^{2}y + \purple{(cos^{4}x+1 - 2cos^{2}x)} cos^{2}y = \:cos^{2}y sin^{2} }}$

• $:\implies {\sf{ cos^{4}x sin^{2}y + cos^{4}xcos^{2}y + cos^{2} y - 2cos^{2}x cos^{2}y = \:cos^{2}y sin^{2} }}$

• $\sf{ cos^{2}x +sin^{2}x=1}$

• $:\implies {\sf{ cos^{4}x ( sin^{2}y + cos^{2}y)+ cos^{2}y - cos^{2} y sin^{2}y - 2cos^{2}x cos^{2}y =0 }}$

• $:\implies {\sf{ cos^{4}x ( 1 )+ cos^{2}y (1-sin^{2}y)- 2cos^{2}x cos^{2}y =0 }}$

• $:\implies {\sf{ cos^{4}x+ cos^{2}y (cos^{2}y)- 2cos^{2}x cos^{2}y =0 }}$

• $:\implies {\sf{ cos^{4}x+ cos^{4}y - 2cos^{2}x cos^{2}y =0 }}$

• $:\implies {\sf{ (cos^{2}x-cos^{2}y)^{2} =0 }}$

• $:\implies {\sf{ cos^{2}x-cos^{2}y =0 }}$

• $:\implies {\sf{ cos^{2}x=cos^{2}y }}$

• and , Like $:\implies {\sf{ cos^{2}x=cos^{2}y }}$ this ,

• $:\implies {\sf{ sin^{2}x=sin^{2}y }}$

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$\implies {\mathrm { \dfrac{cos^{4} y}{cos^{2}x} + \dfrac{sin^{4}y }{sin^{2}x}=??}}$

• As , We have finded that ,

• $:\implies {\sf{ cos^{2}x=cos^{2}y }}$

• and ,

• $:\implies {\sf{ sin^{2}x=sin^{2}y }}$

Now , Solve ,

• $\implies {\mathrm { \dfrac{cos^{4} y}{cos^{2}x} + \dfrac{sin^{4}y }{sin^{2}x}}}$

• $\implies {\mathrm { \dfrac{cos^{4} y}{cos^{2}y} + \dfrac{sin^{4}y }{sin^{2}y}}}$

• $\implies {\mathrm { \dfrac{(cos^{2} y)^{2}}{cos^{2}y} + \dfrac{(sin^{2}y)^{2} }{sin^{2}y}}}$

• $\implies {\mathrm { cos^{2}y + sin^{2}y }}$

• $\implies {\mathrm { 1 }}$

Hence ,

• Option B or (1) is Correct.

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