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#