Answer:
Proved
Step-by-step explanation:
[tex](1-cosB)(1+cosB)=\frac{1}{cosec^2B} \\\\Use\\\\ (a+b)(a-b)=a^2-b^2 ,\\so,\\=(1)^2-(cosB)^2\\=1-cos^2B\\\\Now\ , we\ know\ \\sin^2B+cos^2B=1\\\\so,\\sin^2B=1-cos^2B\\so we just replace,\\\\=1-cos^2B=>sin^2B\\and\ since\ sinB=\frac{1}{cosecB} \\\\=sin^2B=>\frac{1}{cosec^2B}\\[/tex]
