Cot B + cos B divided by sec B - cos B = csc B + 1 divided by tan^2 B can it be verified?

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Answer 1 · 2018-03-13T04:26:54

Yes... see below

$\frac{cot B + cos B}{sec B - cos B} = \frac{csc B + 1}{{tan}^{2} B}$ $(Cot B + cos B)/(sec B - cos B) = (csc B + 1)/ tan^2 B$

$\frac{cot B + cos B}{\frac{1}{cos B} - \frac{{cos}^{2} B}{cos B}} =$ $(CotB+cosB)/(1/cosB - cos^2 B/cosB)=$

$\frac{\frac{cos B}{sin B} + \frac{cos B sin B}{sin B}}{\frac{{sin}^{2} B}{cos B}} =$ $(CosB/SinB+(cosBsinB)/sinB)/(sin^2B/cosB)=$

$(\frac{cos B + cos B sin B}{sin B}) \cdot \frac{cos B}{{sin}^{2} B} =$ $((CosB+cosBsinB)/sinB)*(cosB)/(sin^2B)=$

$(\frac{{cos}^{2} B + {cos}^{2} B sin B}{{sin}^{3} B}) =$ $((Cos^2B+cos^2BsinB)/sin^3B)=$

$\frac{{cos}^{2} B}{{sin}^{3} B} + \frac{{cos}^{2} B sin B}{{sin}^{3} B} =$ $Cos^2B/sin^3B+(cos^2BsinB)/sin^3B=$

$\frac{{cos}^{2} B}{{sin}^{2} B} \cdot \frac{1}{sin B} + \frac{{cos}^{2} B}{{sin}^{2} B} =$ $Cos^2B/sin^2B*1/sinB+cos^2B/sin^2B=$

${cot}^{2} B \cdot csc B + {cot}^{2} B =$ $cot^2B*cscB+cot^2B=$

$\frac{csc B}{{tan}^{2} B} + \frac{1}{{tan}^{2} B} =$ $cscB/tan^2B+1/tan^2B=$

$\frac{csc B + 1}{{tan}^{2} B}$ $(cscB+1)/tan^2B$