Question

Prove that sin^-6 theta - cot^6 theta = 1+3cot^2 theta sin^-2 theta

Answer 1

ANSWER :-

IN MY PHONE THETA IS NOT AVAILABLE SO I LET IT WITH ALPHA

${ sin }^{ - 6 } \alpha - {cot}^{6} \alpha = 1 + 3 {cot}^{2} \alpha \: {sin}^{2} \alpha$

$left \: hand \: side \\ 1 \div {sin}^{6} \alpha - {cos}^{6} \alpha \div {sin}^{6} \alpha \\ 1 - {cos}^{6 } \alpha \div {sin}^{6 } \alpha$

${sin}^{6} \alpha + 3 {sin}^{2} \alpha {cos}^{2} \alpha ( {sin }^{2} \alpha + {cos}^{2} \alpha ) \div {sin}^{6} \alpha \\ {sin}^{4} \alpha ( {sin }^{2} \alpha + 3 {cos}^{2} \alpha \: {sin}^{2} \alpha + 3 {cos}^{4} \div {sin}^{6} \\ {sin}^{2} \alpha (1 + 3 {cos \alpha }^{2} ) + 3 {cos}^{4 } \alpha \div {sin \alpha }^{2}$