Question

If tan A = 2 and tan B =3 . Evaluate ( cos A cosB - sinA sinB)^2

Answer 1

Answer:

$\dfrac{1}{2}$

Step-by-step explanation:

Given---->

$tan \alpha = 2 \: and \: tan \beta = 3$

To find---->

$value \: of {(cos \alpha \: cos \beta - \: sin \alpha \: sin \beta )}^{2}$

Concept used ---->

1)

$tan( \alpha + \beta ) = \dfrac{tan \alpha + tan \beta }{1 - tan \alpha \: tan \beta }$

2)

$cos( \alpha \: + \beta ) = cos \alpha \: cos \beta - sin \alpha \: sin \beta$

3)

${sec}^{2} \alpha = 1 + {tan}^{2} \alpha$

Solution---->

$\tan( \alpha + \beta ) = \dfrac{tan \alpha + tan \beta }{1 - tan \alpha \: tan \beta }$

$\tan( \alpha + \beta ) \: = \dfrac{2 + 3}{1 - (2) \: (3)}$

$\tan( \alpha + \beta ) \: = \dfrac{5}{1 - 6}$

$\tan( \alpha + \beta ) \: = \dfrac{5}{ - 5}$

$\tan( \alpha + \beta ) = - 1$

$now$

${(cos \alpha cos \beta - sin \alpha sin \beta )}^{2} = {(cos( \alpha + \beta ))}^{2}$

$= {cos}^{2}( \alpha + \beta )$

$= \dfrac{1}{ {sec}^{2}( \alpha + \beta ) }$

$= \dfrac{1}{1 + {tan}^{2}( \alpha + \beta ) }$

$= \dfrac{1}{1 + {( - 1)}^{2} }$

$= \dfrac{1}{1 + 1}$

$= \dfrac{1}{2}$