Question

in triangle ABC <A=90°, tanB= root3.find the value of sinB cosC+cosB sinC​

Answer 1

Here is your answer

Answer 2

As we know,
$\tan(a) = \frac{opposite}{adjecent}$
For
$\tan \: B \: = \sqrt{3}$
Using this relation,
we have the relation in the second diagram,


$\sin(a) = \frac{oppsite}{hypotaneous}$
$\cos(a) = \frac{adjacent}{opposite}$


SinBcosC+ cosBsinC=
$= ( \frac{ \sqrt{3}x }{2x} \times \frac{ \sqrt{3} x}{2x} ) + ( \frac{x}{2x} \times \frac{ x }{2x} ) \\ = \frac{3 {x}^{2} }{4 {x}^{2} } + \frac{ {x}^{2} }{4 {x}^{2} } \\ = \frac{4 {x}^{2} }{4 {x}^{2} } = 1$