Question

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

Answer 1

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,......`