If x=cosecA+cosA and y=cosecA−cosA then prove that (x+y2)2+(2x−y)2−1=0.

Given:-⇒x=cosecA+cosA⇒y=cosecA−cosATo prove:-⇒(x+y2)2+(2x−y)2−1=0∴x+y=cosecA+cosA+cosecA−cosA=2cosecAx−y=cosecA+cosA−cosecA+cosA=2cosA∴(x+y2)2+(2x−y)2−1=(2cosecA2)2+(22cosA)2−1⇒ (sinA)2+(cosA)2−1⇒ sin2A+cos2A−1⇒ 1−1⇒ 0Hence proved.

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If x= {cosec}A + A and y= {cosec}A- A then prove that ({2}{x

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If $x = cosec A + cos A$ and $y = cosec A - cos A$ then prove that $(\frac{2}{x + y})^{2} + (\frac{x - y}{2})^{2} - 1 = 0$ .

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Emma Discussed

x = cosec A + cos A

and

y = cosec A - cos A

then prove that

(\frac{2}{x + y})^{2} + (\frac{x - y}{2})^{2} - 1 = 0

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Given:-

\Rightarrow

x = cosec A + cos A

\Rightarrow

y = cosec A - cos A

To prove:-

\Rightarrow

(\frac{2}{x + y})^{2} + (\frac{x - y}{2})^{2} - 1 = 0

∴ x + y = cosec A + cos A + cosec A - cos A = 2 cosec A

x - y = cosec A + cos A - cosec A + cos A = 2 cos A

∴ (\frac{2}{x + y})^{2} + (\frac{x - y}{2})^{2} - 1 = (\frac{2}{2 cosec A})^{2} + (\frac{2 cos A}{2})^{2} - 1

\Rightarrow (sin A)^{2} + (cos A)^{2} - 1

\Rightarrow sin^{2} A + cos^{2} A - 1

\Rightarrow 1 - 1

\Rightarrow 0

Hence proved.

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Question Text	If $x = cosec A + cos A$ and $y = cosec A - cos A$ then prove that $(\frac{2}{x + y})^{2} + (\frac{x - y}{2})^{2} - 1 = 0$ .
Answer Type	Text solution:1
Upvotes	150

If x=cosecA+cosA and y=cosecA−cosA then prove that (x+y2​)2+(2x−y​)2−1=0.

Text solutionVerified

Practice questions from similar books

If $x = cosec A + cos A$ and $y = cosec A - cos A$ then prove that $(\frac{2}{x + y})^{2} + (\frac{x - y}{2})^{2} - 1 = 0$ .