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AP EAPCET 06-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 63 of 160

Marks: +1, -0

If

x \sin (\alpha+y)=\sin y

y=\;\frac{m}{x^{2}+2 n x+1}

m^{2}=

$1-n^{2}$
$1+n$
$1-n$
$n^{2}-1$

Solution: 👈: Video Solution

Given,

x \sin \;(\alpha+y)= \sin \;y

and

y=\;\frac{m}{x^2+2 n x+1}

,

x \sin \;(\alpha+y)= \sin \;y

\;\Rightarrow \;\; \;\frac{ \sin\;(\alpha+y)}{ \sin\; y}=\;\frac{1}{x}

\;\Rightarrow \;\; \;\frac{ \sin\;\alpha \cos y+\cos\alpha \sin\; y}{ \sin\; y}=\;\frac{1}{x}

\;\Rightarrow \;\; \sin\;\alpha \cot y+\cos\alpha=\;\frac{1}{x}

\;\Rightarrow \;\; \cot y=\;\frac{1-x \cos\alpha}{x \sin\;\alpha}

\;\Rightarrow \;\; \tan y=\;\frac{x \sin\;\alpha}{1-x \cos\alpha}

\;\Rightarrow \;\; y=\tan ^{-1} \left[\;\frac{x \sin\;\alpha}{1-x \cos\alpha}\right]

y^{'}=\;\;\frac{1}{1+ \left(\;\frac{x \sin\;\alpha}{1-x \cos\alpha}\right) ^2}

\;\left[\;\frac{(1-x \cos\alpha) \sin\;\alpha-x \sin\;\alpha(-\cos\alpha)}{(1-x \cos\alpha)^2}\right]

\Rightarrow y^{'}=\;\;\frac{(1-x \cos\alpha)^2}{(1-x \cos\alpha)^2+(x \sin\;\alpha)^2}

\;\left[\;\frac{ \sin\;\alpha-x \cos\alpha \sin\;\alpha+x \cos\alpha \sin\;\alpha}{(1-x \cos\alpha)^2}\right]

\;\Rightarrow y^{'}=\;\frac{ \sin\alpha}{1+x^2 \cos^2\alpha-2 x \cos\alpha+x^2 \sin^2\alpha}

\;\Rightarrow y^{'}=\;\frac{ \sin\;\alpha}{x^2-2 x \cos\alpha+1}

Here,

m= \sin\;\alpha

and

n=-\cos\alpha

m^2 = \sin^2\alpha

\;=1-\cos ^2\alpha=1-n^2

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