In ∆ABC, $\angle C=90^o$ and $tan A=\frac{1}{\sqrt3}$. Prove that $Sin ACos B+Cos A Sin B=1$.

Given:

In $\vartriangle ABC$, $\angle C=90^o$ and $tan A=\frac{1}{\sqrt3}$.
To do:

We have to prove that $Sin ACos B+Cos A Sin B=1$.

Solution:

$tan A=\frac{1}{\sqrt3}$

This implies,

$tan A=tan 30^o$

$\angle A=30^o$

Using angle sum property of a triangle,

$\angle A+\angle B+\angle C=180^o$

$30^0+\angle B+90^o=180^o$

$\angle B=180^o-120^o$

$\angle B=60^o$

Therefore,

$Sin ACos B+Cos A Sin B=sin30^oCos60^o+Cos30^oSin60^o$

$=\frac{1}{2}\times(\frac{1}{2})+\frac{\sqrt3}{2}\times(\frac{\sqrt3}{2})$

$=\frac{1}{4}+\frac{3}{4}$

$=\frac{1+3}{4}$

$=\frac{4}{4}$

$=1$

Hence proved.

Tutorialspoint

Updated on: 10-Oct-2022

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