Question

Please Solve its urgent

Answer 1

Step-by-step explanation:

replace θ by A

(i)tan(Π/4+A) tan(Π/4-A)

=(tanΠ/4+tanA/1-tanΠ/4tanA)(tanΠ/4 -tanA/1+tanΠ/4tanA)

=(1+tanA/1-tanA)(1-tanA/1+tanA)

=1

(ii)tan(Π/4+A)-tanΠ/4-A)

=tan(45+A)-tan(45-A)

=(tan45+tanA/1+tan45tanA)- (tan45-tanA/1+tan45tanA)

=(1+tanA/1-tanA)-(1-tanA/1+tanA)

=(1+tanA)²-(1-tanA)²/1-tan²A

=1+2tanA+tan²A-1+2tanA-tan²A/1-tan²A

=4tanA/1-tan²A

=2.2tanA/1-tan²A

=2tan2A