If #tantheta+sectheta=((x-2)/(x+2))# find value of #costheta#?

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Answer 1 · 2018-02-22T11:38:10

#cosA=(x^2-4)/(x^2+4)#

Given that #tanA+secA=(x-2)/(x+2)#.....[1]
We know that,
#rarrsec^2A-tan^2A=1#

#rarr(secA+tanA)(secA-tanA)=1#

#rarrsecA-tanA=1/(secA+tanA)=1/((x-2)/(x+2))=(x+2)/(x-2)#.....[2]

Adding equation [1] and [2], we get

#rarrtanA+secA+secA-tanA=(x-2)/(x+2)+(x+2)/(x-2)#

#rarr2secA=((x-2)^2+(x+2)^2)/((x+2)(x-2))#

#rarr2/cosA=(2(x^2+2^2))/(x^2-2^2)#

#rarrcosA=(x^2-4)/(x^2+4)#