Question

Please solve q 18?

Answer 1

Given that `A+B=90^@` then `A=90-B^@`

`rarr(tanAtanB+tanAcotB)/(sinAsecB)-(sin^2B)/(cos^2A)`

`=(tanA[tanB+cotB])/(sinAsecB)-(sin^2B)/(cos^2(90^@-B)`

`=((cancel(sinA)/cosA)[sinB/cosB+cosB/sinB])/(cancel(sinA)/cosB)-(sin^2B)/(sin^2B)`

`=((1/cosA)[(sin^2B+cos^2B)/(sinB*cancel(cosB))])/(1/cancel(cosB))-1`

`=1/(cos(90^@-B)sinB)-1`

`=1/sin^2B-1=(1-sin^2B)/sin^2B=(cos^2B)/(sin^2B)=cot^2B`

Answer 2

Given that `A+B=90^@` then A=90-B^@#
Now

`(tanAtanB+tanAcotB)/(sinAsecB)-(sin^2B)/(cos^2A)`

`(tan(90^@-B)tanB+tan(90^@-B)cotB)/(sin(90^@-B)secB)-(sin^2B)/(cos^2(90^@-B))`

`=(cotBtanB+cotBcotB)/(cosBsecB)-(sin^2B)/(sin^2B)`

`=1+cot^2B-1=cot^2B`