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Question Number 105366 by bemath last updated on 28/Jul/20

Answered by $@y@m last updated on 28/Jul/20

15. ((tan A(sec A+1)+tan A(sec A−1))/(sec^2  A−1))  =((2tan Asec A)/(tan^2 A))  =2cosec A

$$\mathrm{15}.\:\frac{\mathrm{tan}\:{A}\left(\mathrm{sec}\:{A}+\mathrm{1}\right)+\mathrm{tan}\:{A}\left(\mathrm{sec}\:{A}−\mathrm{1}\right)}{\mathrm{sec}^{\mathrm{2}} \:{A}−\mathrm{1}} \\ $$$$=\frac{\mathrm{2tan}\:{A}\mathrm{sec}\:{A}}{\mathrm{tan}\:^{\mathrm{2}} {A}} \\ $$$$=\mathrm{2cosec}\:{A} \\ $$

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