Please solve q 18?

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2 Answers
May 2, 2018

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

May 3, 2018

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