Question

In a triangle ABC right angled at C, if tan A = 1/√3 and tan B =√3, then
find the value of sin A cos B-cos A sin B​

Answer 1

Solution:-

$\rm \tan(A) = \frac{1}{ \sqrt{3} } = \frac{p}{b}$

Now

$\rm \: p = 1 \: \: b = \sqrt{3} \: \: and \: \: h = x$

Using phythogoeros theorem

$\rm \: {h}^{2} = {p}^{2} + {b}^{2}$

We get

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

$\rm \: {h}^{2} = 1 + 3$

$\rm \: {h}^{2} = 4$

$\rm \: h = 2$

So we get ,

$\rm \: p = 1 \: \: b = \sqrt{3} \: \: and \: \: h = 2$

Now

$\rm \: \sin(A) = \frac{1}{2} = \frac{p}{h}$

$\rm \cos(A) = \frac{ \sqrt{3} }{2} = \frac{b}{h}$

Now take

$\rm \tan(B) = \frac{ \sqrt{3} }{1} = \frac{p}{b}$

$\rm \: p = \sqrt{3} \: \: \: \: b \: = 1 \: \: and \: h = x$

Using phythogoeros theorem

$\rm \: {h}^{2} = {b}^{2} + {p}^{2}$

$\rm \: {h}^{2} = {1}^{2} + ( \sqrt{3} ) {}^{2}$

$\rm \: {h}^{2} = 1 + 3$

$\rm \: {h}^{2} = 4$

$\rm \: h = 2$

We get

$\rm \: p = \sqrt{3} \: \: \: \: b \: = 1 \: \: and \: h = 2$

then

$\rm \: \sin(B) = \frac{ \sqrt{3} }{2} = \frac{p}{h}$

$\rm \: \cos(B) = \frac{1}{2} = \frac{b}{h}$

We have to find

$\rm \sin(A) \cos(B) - \sin(B) \cos(A)$

Put the value

$\rm \: \frac{1}{2} \times \frac{1}{2} - \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2}$

$\rm \: \frac{1}{4} - \frac{3}{4}$

$\rm \: \frac{ - 2}{4}$

$\to \: \frac{ - 1}{2}$

Answer:-

$\implies \frac{ - 1}{2}$