Question

Tan²a + cot²a + 2 = sec²a × cosec²a

Answer 1

Answer...

tan²A + cot²A + 2 = sec²A × cosec²Atan²A + cot²A = sec²A × cosec²A - 2LHS = tan²A + cot²A= (sec²A - 1 ) + (cosec²A - 1)= sec²A + cosec²A - 2

tan²A + cot²A + 2 = sec²A × cosec²Atan²A + cot²A = sec²A × cosec²A - 2LHS = tan²A + cot²A= (sec²A - 1 ) + (cosec²A - 1)= sec²A + cosec²A - 2=\frac{1}{{cos}^{2}\alpha} + \frac{1}{{sin}^{2}\alpha} - 2 \\ \\ = \frac{{cos}^{2}\alpha + {sin}^{2}\alpha}{{cos}^{2}\alpha× {sin}^{2}\alpha}- 2 \\ \\ \huge\boxed{{sin}^{2}\theta + {cos}^{2}\theta = 1} \\ \\ = \frac{1}{{cos}^{2}\alpha\: {cosec}^{2}\alpha} -2\\ \\= {sec}^{2}\alpha \:{cosec}^{2}\alpha - 2 \\ \\

Hence proved...