Question

If sin A =1/2 cos B =root3/2 where A is in2nd quadrant and B is in 1st quadrant. find the value of tan (A+B)​

Answer 1

$\huge{ \bold{ \underline {\underline{Answer:}}}}$

$\tan(A+B) = \sqrt{3}$

${ \bold{ \underline {\underline{Step \: by \: step \: explanation:}}}}$

${ \bold{ \underline {\underline{Given \: that:}}}}$

$\sin(A) = \frac{1}{2} ; \cos(B)= \sqrt{ \frac{3}{2} } \\ { \bold{ \underline {\underline{We \: know \: that:}}}}\\ So,A=30° and \: B=30° \\ A \: is \: in \: 2nd \: quadrant \: and \: \\ B \: is \: in \: 1st \: quadrant.</p><p> \\ \bold{In \: 1st \: quadrant \: all\:} \\ \bold{are \: positive.In \: 2nd \: quadrant \:} \: \\ \bold{sin \: and \: cosec \: are \: positive}$

${ \bold{ \underline {\underline{To \: find:}}}}$

tan(A+B) and that equals to

$\tan(A+B) = \frac{tanA+tanB}{1 - tanAtanB}$

${ \bold{ \underline {\underline{Values \: are:}}}}$

$\tan(A) = \tan(30°) = \frac{1}{ \sqrt{3} }$

$\tan(B) = \tan(30°) = \frac{1}{ \sqrt{3}}$

Because A = B = 30°

${ \bold{ \underline {\underline{Solution:}}}}$

$\frac{ \tan(30) + \tan(30) }{1- \tan(30) \tan(30) } = \frac{ \frac{1}{ \sqrt{3} } + \frac{1}{ \sqrt{3} } }{1 - \frac{1}{ \sqrt{3} } \times \frac{1}{ \sqrt{3} } } \\ = \frac{ \frac{2}{ \sqrt{3} } }{1 - \frac{1}{ \sqrt{9} } } = \frac{ \frac{2}{ \sqrt{3} } }{1 - \frac{1}{3} } = \frac{ \frac{2}{ \sqrt{3} } }{ \frac{3 - 1}{3} } = \frac{ \frac{2}{ \sqrt{3} } }{ \frac{2}{3} } \\ \implies{} \frac{3}{ \sqrt{3} } = \sqrt{3} </p><p>$

(or) you can do directly like this:

$\tan(A + B) = \tan(30 + 30) \\ \implies{ \tan(60) } = \sqrt{3}$

Answer 2

Answer:

Answer for this question is square root of 3