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Question

If f(x)=∣ ∣ ∣tan xsin xcos xtan xsec x+cos xcos2xcos2xcosec2x1cos2xcos2x∣ ∣ ∣ Then π+πf(x)dx=2kπ. Therefore, the value of k is

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Solution

R3R3(cotx)R1
f(x)=∣ ∣ ∣ ∣tanxsinxcosxtanxsecx+cosxcos2xcos2xcosec2x00cos2xsecxcos2xsinx∣ ∣ ∣ ∣
f(x)=(sinxcosxcos3xsinx)(cos2xsecxcos2xsinx)
We know that,
(sinxcosxcos3xsinx)(cos2xsecx) is an even function so,
π+πf(x) dx=π+πcos3xcos5x dxI=π+πcos3xcos5x dxI=+ππcos3x+cos5x dxI=2+π0cos3x+cos5x dx
Substituting xπx
I=2+π0cos3xcos5x dxI=0
k=0

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