Question

In a right angel triangle ABC , right angel is at B ,if Ta na =√3 then find the value of 1) sin A cosC + Cosa sinC 2) Cos A cos C -sin A sin C

Answer 1

Answer:

Step-by-step explanation:

tan =$\frac{ \sqrt{3} }{1}$ =$\frac{p}{b}$

$h^{2} =p^{2} +b^{2}$

$h^{2}= \sqrt{3} ^{2} +1^{2}$

$h^{2} = 3+1$

$h=\sqrt{4}$

$h=2$

1) sinAcosC + cosAsinC

$\frac{p}{h} *\frac{b'}{h'} +\frac{b}{h} *\frac{p'}{h'} \\\\\frac{\sqrt{3}k }{1k} \frac{\sqrt{3k} }{1k} +\frac{1k}{2k}\frac{1k}{2k}\\\\\frac{3k^{2} }{1k^{2} } +\frac{1k^{2} }{4k^{2} }\\\\3 +\frac{1}{4}\\\frac{12 + 1}{4} \\\\\frac{13}{4}$

1) cosAcosC - sinAsinC

$\frac{b}{h} *\frac{b'}{h'} -\frac{p}{h} *\frac{p'}{h'} \\\\\frac{1k }{2k} *\frac{\sqrt{3k} }{2k} -\frac{\sqrt{3} k}{2k}*\frac{1k}{2k}\\\\\frac{\sqrt{3} k^{2} }{4k^{2} } -\frac{\sqrt{3} k^{2} }{4k^{2} }\\\\\frac{\sqrt{3} }{4} -\frac{\sqrt{3} }{4}\\\frac{\sqrt{3} - \sqrt{3} }{4} \\\\\frac{0}{4}\\\\=0$