If x= cosec A+ cos A and y cosec A- cos A then prove that2 बाय �

1.	If x= cosec A+ cos A and y cosec A- cos A then prove that2 बाय एक्स प्लस वाई का होल स्क्वायर प्लस एक्स माइनस वन बाय टू का होल स्क्वायर इक्वल टू वन
Answer» 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 ════════════════════════════ ➠ x + y = csc A + cos A + csc A - cos A ➠ x + y = 2 csc A $\to \sf \dfrac{2}{x+y}=\dfrac{2}{2\ csc\ A}\\\\\to \sf \dfrac{2}{x-y} =\dfrac{1}{csc\ A}\\\\\to \sf{\dfrac{2}{x+y}=sin\ A}\\\\\to \sf{\bf{\bigg(\dfrac{2}{x+y}\bigg)^2=sin^2A}}$ ════════════════════════════ ➠ x - y = csc A + cos A - csc A + cos A ➠ x - y = 2 cos A $\to \sf \dfrac{x-y}{2}=\dfrac{2\ cos\ A}{2}\\\\\to \sf \dfrac{x-y}{2}=cos\ A\\\\\to \sf{\bf \bigg(\dfrac{x-y}{2}\bigg)^2=cos^2A}$ ════════════════════════════ 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

If x= cosec A+ cos A and y cosec A- cos A then prove that2 बाय एक्स प्लस वाई का होल स्क्वायर प्लस एक्स माइनस वन बाय टू का होल स्क्वायर इक्वल टू वन

Answer»

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

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