Jainam Ravani
Last Activity: 4 Years ago
The answer is 28:
First convert the tan terms to cot using allied angles. You will get:
tan^2(π/16)+cot^2(π/16)+tan^2(3π/16)+cot^2(3π/16)
Taking π/16 as a, and converting to sin and cos, you get:
(Sin^4a+cos^4a)/sin^2acos^2a+(sin^4(3a)+cos^4(3a))/sin^2(3a)cos^2(3a)
=(1-sin^2acos^2a)/sin^2acos^2a+(1-sin^2(3a)cos^2(3a)/sin^2(3a)cos^2(3a)
Using the identity sin2a=2sinacosa:
=(4-2sin^2(2a))/sin^2(2a)+(4-2sin^2(6a))/sin^2(6a)
=4/sin^2(2a)+4/sin^2(2a)-4
=4(sin^(6a)+sin^(2a))/sin^2(2a)cos^2(2a)-4
sin^2(6a)=cos^2(2a). Using this and simplifying:
4/sin^2(2a)cos^2(2a)-4
Using the sin2a identity:
16/sin^2(4a)-4
=32-4
=28