Question #0a625

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Answer 1 · 2018-01-28T11:23:21

Please see the Explanation.

We replace #sec^2A" by "1+tan^2A#, and get,

#(cotA-1)/{2-(1+tan^2A)}#,

#=(cotA-1)/(1-tan^2A)#,

#=(cotA-1)/(1-1/cot^2A)#,

#=(cotA-1)/{(cot^2A-1)/cot^2A}#,

#={cancel((cotA-1))cot^2A}/{cancel((cotA-1))(cotA+1)}#,

#=cot^2A/(cotA+1)#,

#=cot^2A/(cotA+1)xxtanA/tanA#,

#={cotA(cotAtanA)}/{(cotAtanA+tanA)}#,

#=cotA/(1+tanA)," as desired!"#.