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Question

If f(x)=(tanx2)(1+secx)(1+sec2x)(1+sec22x)×(1+sec23x) then the value of f(π32) is

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Solution

1+secx=cosx+1cosx=2cos2x2cosx

1+sec2x=cos2x+1cos2x=2cos2x1+1cos2x=2cos2xcos2x

1+sec22x=cos4x+1cos4x=2cos22x1+1cos4x=2cos22xcos4x

1+sec23x=2cos24xcos 8x

f(x)=sinx2cosx2×2cos2x2cosx×2cos2xcos2x×2cos22xcos4x×2cos24xcos8x

=sinx2×2cosx2×2cosx×2cos2x×2cos4xcos8x

=sinx×2cosx×2cos2x×2cos4xcos8x

=sin8xcos8x=tan8x

f(x)=tan 8x
f(π32)=tan(8×π32)=tanπ4=1

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