In a ∆ABC, right-angled at C, if tan⁡A=1/√3, find the value of sin⁡A cos⁡B+cos⁡A sin⁡B.

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Solution

Ans. tanA = $\frac{1}{\sqrt 3}$
By applying pythagoras theorem in ∆ABC,
sinA = $\frac{1}{2}$
cos A = $\frac{\sqrt 3}{2}$
sinB = $\frac{\sqrt 3}{2}$
cosB = $\frac{1}{2}$
sinAcosB +cosAsinB
$\frac{1}{2}$ x $\frac{1}{2}$ + $\frac{\sqrt 3}{2}$ x $\frac{\sqrt 3}{2}$
=1

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