Question

tanQ+tanQ(90-Q)=secQsec(90-Q)​

Answer 1

Solution -

tanA+tan(90°-A)

=tanA+cotA [∵, tan(90°-A)=cotA]

=tanA+1/tanA

=(tan²A+1)/tanA

=sec²A/tanA [∵, sec²A-tan²A=1]

=(1/cos²A)/(sinA/cosA)

=1/cos²A×cosA/sinA

=1/(cosAsinA)

=(1/cosA)(1/sinA)

=secAcosecA

=secAsec(90°-A) [∵, sec(90°-A)=cosecA]

( proved )