How to solve #tan(x)=(cos(x)-sin(x))/(cos(x)+sin(x))#?

2 Answers
Ratnaker Mehta
Jun 9, 2018

# x in {(4k+1)pi/8 | k in ZZ}#.

Explanation:

In order that #tanx# be defined, #cosx!=0#.

#:.# Dividing the Nr. and Dr. of the R.H.S. by #cosx!=0#, we have,

#tanx={(cosx-sinx)/cosx}/{(cosx+sinx)/cosx}#.

#:. tanx={cosx/cosx-sinx/cosx}/{cosx/cosx+sinx/cosx}#.

#:. tanx=(1-tanx)/(1+tanx)={tan(pi/4)-tanx}/{1+tan(pi/4)tanx)#.

#:. tanx=tan(pi/4-x).................(star)#.

But, #tantheta=tanalpha rArr theta=kpi+alpha, k in ZZ#.

#:. (star) rArr x=kpi+(pi/4-x), or, 2x=(4k+1)pi/4, k in ZZ.#

#:. x in {(4k+1)pi/8 | k in ZZ}#.

Nghi N.
Jun 9, 2018

#x = pi/4 + kpi#

Explanation:

#tan x = (cos x - sin x)/(cos x + sin x)#
Reminder of identities:
#cos x - sin x = sqrt2cos (x + pi/4)#
#cos x + sin x = sqrt2sin (x + pi/4)#
Hence,
#tan x = cot (x + pi/4) = tan (pi/2 - x - pi/4) = tan (pi/4 - x)#
Unit circle and property of tan function -->
#x = pi/4 - x + kpi#
#2x = pi/4 + kpi#
#x = pi/8 + kpi#

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