Not trigo again

Geometry Level 3

sec ⁡ 4 α tan ⁡ 2 β + sec ⁡ 4 β tan ⁡ 2 α \large{\frac{\sec^{4} \alpha}{\tan^{2} \beta}+\frac{\sec^{4} \beta}{\tan^{2} \alpha}}

If the above expression is defined, then the minimum value it can take

4 8 10 6

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Neehar Dingari
Jul 16, 2015

tan \alpha = a, tan\beta = b. The expression is given as ((1+a^2)^2)/b^2 + ((1+b^2)^2)/a^2 = (a^2((1+a^2)^2) + b^2((1+b^2)^2))/(a^2 b^2).

1+a^2 >=2a and 1+b^2>=2b.

Using this the above expressions is greater than or equal to 8.

Possible answers are 8 and 10.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

@Tanishq Varshney , this person said that there's 2 answer. Can you verify/disprove him/her? Thanks

Pi Han Goh - 5 years, 7 months ago

Can you please explain how you got the inequality 1+a^2>=2a

Adithya S - 4 years, 4 months ago

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