Question

Sin²α(1+sin^-1α+ ctgα)(1-sin^-1α+ctgα) упростите 

Answer 1

Sin²α·(1+1/sinα + ctgα)(1-1/sinα+ctgα) =

= Sin²α·(1+ ctgα +1/sinα )(1 +ctgα-1/sin²α) =

= Sin²α·[(1 +ctgα)²-1/sin²α] =

= Sin²α·[(1 +ctgα)²·sin²α - 1]/sin²α =

= (1 +ctgα)²/(1 + ctg²α) - 1 =

= (1 + ctg²α + 2ctgα)/(1 + ctg²α) - 1 =

= 1 + 2ctgα/(1 + ctg²α) -1

= 2ctgα/(1 + ctg²α) =

= 2ctgα/(1 + ctg²α) =

= 2ctgα·sin²α =

= 2cosα·sin²α /sinα =

= 2cosα·sinα =

= sin2α

Answer 2

$sin^{2}x(1+ctgx+frac{1}{sinx})(1+ctgx-frac{1}{sinx})=$$sin^{2}x(1+2ctgx+ctg^{2}x-frac{1}{sin^{2}x})=sin^{2}x*2ctgx=sin2x$