How do you simplify #(tanx)/(1+secx) + (1+secx)/tanx = 2cscx#?

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Answer 1 · 2018-03-20T12:05:10

#LHS=(tanx)/(1+secx) + (1+secx)/tanx#

#=(tanx(secx-1))/((secx+1)(secx-1) )+ (1+secx)/tanx#

#=(tanx(secx-1))/(sec^2x-1)+ (1+secx)/tanx#

#=(tanx(secx-1))/tan^2x+ (1+secx)/tanx#

#=(secx-1)/tanx+ (1+secx)/tanx#

#=secx/tanx-1/tanx+ 1/tanx+secx/tanx#

#=(2secx)/tanx#

#=(2/cosx)/(sinx/cosx)#

#=2/sinx#

# = 2cscx=RHS#