Question

proove that:-

a) sin (A – B) / cos A cos B = tan A – tan B



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Answer 1

Question :-

Prove that

$\rm \: \dfrac{sin(A - B)}{cosAcosB} = tanA - tanB$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \dfrac{sin(A - B)}{cosAcosB}$

We know,

$\boxed{\sf{ \:sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{sinA \: cosB \: - \: sinB \: cosA}{cosA \: cosB}$

$\rm \: = \: \dfrac{sinA \: cosB}{cosA \: cosB} \: - \: \dfrac{sinB \: cosA}{cosA \: cosB}$

$\rm \: = \: \dfrac{sinA}{cosA} \: - \: \dfrac{sinB}{cosB}$

$\rm \: = \: tanA - tanB$

Hence,

$\rm\implies \:\boxed{\rm{ \: \frac{sin(A - B)}{cosA \: cosB} \rm \: = \: tanA - tanB \: }}$

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Additional Information :-

$\boxed{\sf{ \:sin(x + y) = sinx \: cosy \: + \: siny \: cosx \: }}$

$\boxed{\sf{ \:cos(x + y) = cosx \: cosy - siny \: sinx \: }}$

$\boxed{\sf{ \:cos(x - y) = cosx \: cosy + siny \: sinx \: }}$

$\boxed{\sf{ \:tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: }}$

$\boxed{\sf{ \:tan(x - y) = \frac{tanx - tany}{1 + tanx \: tany} \: }}$

$\boxed{\sf{ \: {sin}^{2}x - {sin}^{2}y = sin(x + y) \: sin(x - y) \: }}$

$\boxed{\sf{ \: {cos}^{2}x - {sin}^{2}y = cos(x + y) \: cos(x - y) \: }}$

Answer 2

Step-by-step explanation:

$\boxed{\displaystyle{ \sin{{\left({x}-{y}\right)}}}={\sin{{x}}}{\cos{{y}}}-{\cos{{x}}}{\sin{{y}}})} \\ \\ \frac{\sin(a-b)}{\cos a\cos b} =\frac{\sin a\cos b-\cos a\sin b}{\cos a\cos b}\\ \\\bf \:Separate \: as:\\ (\frac{\sin(a-b)}{\cos a\cos b} ) =\left( \: \frac{\sin a\cos b}{\cos a\cos b}\right)-\left(\frac{\cos a\sin b}{\cos a\cos b}\right)\\ \\ \bf \: Cancel \: common \: factors: \\ \frac{\sin(a-b)}{\cos a\cos b} =\left(\frac{\sin a}{cos a}\right)-\left(\frac{\sin b}{\cos b}\right)\\ \\ \bf \: Use \: the \: quotient \: identity: \\ (\displaystyle{\tan{{x}}}=(\frac{{\sin{{x}}}}{{\cos{{x}}}}) \\ \boxed{ \red{\frac{ \: \sin(a-b)}{ \cos a\cos b}= \tan a-\tan b}}$