How do you solve #(1+tanA)^2 + (1+cotA)^2 = (secA+cosecA)^2#?

1 Answer
Shiva Prakash M V
May 3, 2018

Hence, the values ofA satisfying the equation

#(1+tanA)^2+(1+cotA)^2=(secA+cscA)^2# are,

#A=......,(-7pi)/2,-3pi,(-5pi)/2,-2pi,(-3pi)/2,-pi,-pi/2,0,pi/2,pi,(3pi)/2,2pi,(5pi)/2,3pi,(7pi)/2,......#

Explanation:

#(1+tanA)^2+(1+cotA)^2=(secA+cscA)^2#

#tanA=sinA/cosA#

#cotA=cosA/sinA#

#secA=1/cosA#

#cscA=1/sinA#

#(1+sinA/cosA)^2+(1+cosA/sinA)^2=(1/cosA+1/sinA)^2#

#(cosA+sinA)^2/cos^2A+(sinA+cosA)^2/sin^2A=(sinA+cosA)^2/(sinAcosA)#

#(cosA+sinA)^2(1/cos^2A+1/sin^2A)=(cosA+sinA)^2(1/(sinAcosA))#

#(sin^2A+cos^2A)/(sin^2Acos^2A)=1/(sinAcosA)#

#1/(sinAcosA)^2=1/(sinAcosA)#

#(sinAcosA)^2=sinAcosA#

#(sinAcosA)^2-sinAcosA=0#

#sinAcosA(sinAcosA-1)=0#

#sinA=0, cosA=0, sinAcosA-1=0#

#sinA=0->A=0,pi,2pi,3pi,...#
#cosA=0->A=pi/2,(3pi)/2,(5pi)/2,(7pi)/2,......#
#sinAcosA=0->1/2sin2A=0->sin2A=0#
#2A=0,pi,2pi,3pi,...#
#A=0,pi/2,pi,(3pi)/2,2pi,(5pi)/2,3pi,(7pi)/2,......#

Hence, the values ofA satisfying the equation

#(1+tanA)^2+(1+cotA)^2=(secA+cscA)^2# are,

#A=......,(-7pi)/2,-3pi,(-5pi)/2,-2pi,(-3pi)/2,-pi,-pi/2,0,pi/2,pi,(3pi)/2,2pi,(5pi)/2,3pi,(7pi)/2,......#

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