R3→R3−(cotx)R1
f(x)=∣∣
∣
∣
∣∣tanxsinxcosxtanxsecx+cosxcos2xcos2xcosec2x00cos2x−secx−cos2xsinx∣∣
∣
∣
∣∣
⇒f(x)=(sinxcosx−cos3xsinx)(cos2x−secx−cos2xsinx)
We know that,
(sinxcosx−cos3xsinx)(cos2x−secx) is an even function so,
−π∫+πf(x) dx=−π∫+π−cos3x−cos5x dx⇒I=−π∫+π−cos3x−cos5x dx⇒I=+π∫−πcos3x+cos5x dx⇒I=2+π∫0cos3x+cos5x dx
Substituting x→π−x
⇒I=2+π∫0−cos3x−cos5x dx⇒I=0
⇒k=0