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Question

lf x=cosnθ,y=sinnθ then d2ydx2=

A
nn1.cos2n1θsinn3θ
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B
n2n.tann3θcosn+1θsinθ
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C
n1.tann2θ.sec2θ
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D
nsinn1θcosn2θ
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Solution

The correct option is B n2n.tann3θcosn+1θsinθ
dydx=dydθdxdθ=nsinn1θcosθncosn1θsinθ

=tann2θ

d2ydx2=ddθ(dydx)dθdx

=((n2)tann3sinθ)ncosn1θsinθ

=(n2)ntann3θcosn+1θsinθ


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