Question #835cb

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Answer 1 · 2017-02-12T03:05:14

Given

$\frac{tan a}{tan b} = \sqrt{3}$ $tana/tanb=sqrt3$

$\Rightarrow \frac{cot b}{cot a} = \sqrt{3}$ $=>cotb/cota=sqrt3$

$=>cotb=sqrt3cota......[1]$

Again

$sina/sinb=sqrt2$

$=>cscb/csca=sqrt2$

$=>csc^2b/csc^2a=2$

$=>(1+cot^2b)/(1+cot^2a)=2$

Putting $cotb=sqrt3cota$

$=>(1+3cot^2a)/(1+cot^2a)=2$

$=>1+3cot^2a=2+2cot^2a$

$=>3cot^2a-2cot^2a=2-1=1$

$=>cota=1=cot(pi/4)=cot(pi+pi/4)$

$=>a=pi/4 and (5pi)/4$

Inserting $a =pi/4 or (5pi)/4$ in [1]

$=>cotb=sqrt3cota$

$=>cotb=sqrt3xx1=cot(pi/6)=cot(pi+pi/6)$

$=>b=pi/6 and (7pi)/6$