Question

Lim x tends to π 1+sec^3 x/tan^2 x plzzz solve it......

Answer 1

We have,limx→π1+sec3xtan2x=limx→π13+(secx)3sec2x−1=limx→π(1+secx)(12−1×secx+sec2x)(secx+1)(secx−1)=limx→π(secx+1)(1−secx+sec2x)(secx+1)(secx−1)=limx→π(1−secx+sec2x)(secx−1)=1−secπ+sec2πsecπ−1=1−(−1)+(−1)2−1−1=1+1+1−2=−32

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