Question

If x= cosec A+ cos A and y cosec A- cos A then prove that



2 बाय एक्स प्लस वाई का होल स्क्वायर प्लस एक्स माइनस वन बाय टू का होल स्क्वायर इक्वल टू वन ​

Answer 1

Correct Question :

If x = csc A +cos A and y = csc A - cos A then prove that

• (2/x+y)²+ (x-y/2)²- 1 =0

Solution :

x = csc A + cos A

y = csc A - cos A

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➠ x + y = csc A + cos A + csc A - cos A

➠ x + y = 2 csc A

$\to \sf \dfrac{2}{x+y}=\dfrac{2}{2\ csc\ A}\\\\\to \sf \dfrac{2}{x-y} =\dfrac{1}{csc\ A}\\\\\to \sf{\dfrac{2}{x+y}=sin\ A}\\\\\to \sf{\bf{\bigg(\dfrac{2}{x+y}\bigg)^2=sin^2A}}$

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➠ x - y = csc A + cos A - csc A + cos A

➠ x - y = 2 cos A

$\to \sf \dfrac{x-y}{2}=\dfrac{2\ cos\ A}{2}\\\\\to \sf \dfrac{x-y}{2}=cos\ A\\\\\to \sf{\bf \bigg(\dfrac{x-y}{2}\bigg)^2=cos^2A}$

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LHS

$\\ \to \sf \bigg(\dfrac{2}{x+y}\bigg)^2+\bigg(\dfrac{x-y}{2}\bigg)^2-1$

$\\ \to \sf \sf{\bf{sin^2A+cos^2A}}-1$

• From Trigonometric identities ,

• $\sf{\bf{sin^2x+cos^2x=1}}$

where ,

• x is any function

$\\ \to \sf 1 - 1$

$\\ \to \sf\ \; 0$

RHS

Hence Proved

Answer 2

$\pink\bigstar$QUESTION:-

If x=cosec A+cos A and y=cosec A -cos A then prove that

________

$\sf \bigg( \dfrac{2}{x + y} \bigg)^{2}+ \bigg(\dfrac{x - y}{2} \bigg)^{2} = 1$

$\blue\bigstar$SOLUTION:-

We have

⠀⠀⠀

$\sf LHS = {\bigg(\dfrac{2}{x + y} \bigg)}^{2} + {\bigg( \dfrac{x - y}{2} \bigg)}^{2}$

Substituting the value of x and y in above we get

$\sf {\bigg(\dfrac{2}{x + y} \bigg)}^{2} = \bigg \{\dfrac{2}{(cosecA + cosA) + (cosecA - cosA)} \bigg \}^{2}$

$\sf = \bigg( \dfrac{ \cancel{2}}{ \cancel{2}cosecA} \bigg)^{2}$

$\sf = \bigg(\dfrac{1}{cosecA}\bigg)^{2}$

$\sf \pink{ = sin^{2} A}.....(1) \: \: \: \: \bigg| \longrightarrow\dfrac{1}{cosecA= sinA} \bigg|$

And

⠀⠀

$\sf \bigg(\dfrac{x - y }{2} \bigg)^{2} = \bigg \{ \dfrac{(cosecA + cosA) -( cosecA - cosA)}{2} \bigg \}^{2}$

$\sf = \bigg(\dfrac{ \cancel{2}cosA}{ \cancel{2}} \bigg)^{2}$

$\sf \blue{ = {cos}^{2} A} ....(2)$

⠀⠀⠀

Now LHS

⠀⠀⠀

$\sf = {\bigg(\dfrac{2}{x + y} \bigg)}^{2} + {\bigg( \dfrac{x - y}{2} \bigg)}^{2}$

$\sf = sin^{2}A + cos^{2}A \: \: \: (from \: 1 \: and \: 2)$

$= 1$

Hence,Proved

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