Prove that #(1+cotx)/cscx-secx/(tanx+cotx)=cosx#?

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Answer 1 · 2017-01-23T05:59:45

Please see below.

#(1+cotx)/cscx-secx/(tanx+cotx)#

= #(1+cosx/sinx)/(1/sinx)-(1/cosx)/(sinx/cosx+cosx/sinx)#

= #(sinx+cosx)-(1/cosx)/((sin^2x+cos^2x)/(sinxcosx)#

mutiplying numerator and denominator in first term by #sinx#

= #(sinx+cosx)-(1/cosx)/(1/(sinxcosx)#

= #(sinx+cosx)-1/cosx xxsinxcosx#

= #sinx+cosx-sinx#

= #cosx#