IIT JEE , AIEEE Forum - Mathematics , Trignometry - Trigonometry-please explain

nknikhilesh1

If a and b are positive quantities and a

b.Find the minimum positive values of asec

-btan

.

sunip_the_mini_mastermind

whats the answer

nknikhilesh1

root under a²-b² but please explain

sunip_the_mini_mastermind

yes...that's the answer....just apply condition of minima and u will get sin@=b/a....now find sec@ ,tan@ and put the values in the given expression to get the desired result

nknikhilesh1

Can u please solve this question for me,please...

ramkumar_november

$LET\;\;y=asec\theta-btan\theta$

$\frac{dy}{d\theta}\;=\;asec\theta tan\theta-bsec^2\theta$

$asec\theta tan\theta-bsec^2\theta\;=\;0$

$sin\theta\;=\;\frac{b}{a}$

$sec\theta\;=\;\frac{a}{\sqrt{a^2-b^2}}$

$tan\theta\;=\;\frac{b}{\sqrt{a^2-b^2}}$

the second derivative will be positive.....

$y\;=\;a\bigg(\frac{a}{\sqrt{a^2-b^2}}\bigg)-b\bigg(\frac{b}{\sqrt{a^2-b^2}}\bigg)$

$y=\sqrt{a^2-b^2}$

since the second derivative is positive.... this must be the minimum value ....

Milo_Breakin_free

edison

Differentiate w.r.t to theta and equate it to zero, to obtain the results

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