Question #498a1

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Answer 1 · 2017-11-06T11:18:04

Proof

# (cotx + tanx) / cotx # = #sec^2(x)#

Convert tan(x) and cot(x) in terms of sin and cos

#(cos(x)/sin(x)+sin(x)/cos(x))/(cos(x)/sin(x))#

Take the LCD on the numerator and apply

#((sin^2(x)+cos^2(x))/(sin(x)cos(x)))/(cos(x)/sin(x))#

Apply algebra

#((sin^2(x)+cos^2(x))/(cancel(sin(x))cos(x)))#x #(cancel(sin(x))/cos(x))#

#((sin^2(x)+cos^2(x))/(cos^2(x)))#

Apply trigonometric identity

#color(red)(sin^2(x)+cos^2(x)=1#

#1/cos^2(x)# #root#=#sec^2(x)#

Answer 2 · 2017-11-06T11:27:55

Please refer to a Proof given in the Explanation.

We have,

#(cotx+tanx)/cotx,#

#=cotx/cotx+tanx/cotx,#

#=1+tanx*tanx..........[because, 1/cotx=tanx],#

#=1+tan^2x,#

#=sec^2x.#