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Question

If cosx2+cosx22+cosx23++cosx2n=f(x) , then 12tanx2+122tanx22++12ntanx2n=

A
f(x)f(x)
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B
f(x)f(x)
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C
f(x)f(x)
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D
0
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Solution

The correct option is C f(x)f(x)
cosx2cosx22cosx23....cosx2n=f(x)
LEt 12tanx2+122tanx22+....12ntanx2n==g(x)
cosx2=f1; cosx22=f2; cosx23=f3;.......cosx2n=fn
Differentiating all f1 till fn
ddx(cosx2)=12sinx2
.........................................
ddx(cosx2n)=12nsinx2n=fn
f1f1=(12sinx2)cos(x2)=12tanx2=g1
--------------------------------------
fnfn=(12nsinx2n)cos(x2n)=12ntanx2n=gn
g1.....gn=f1f1+....fnfn
gn=f(x)f(x)

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