Question

if tan alpha=1 and tan beta=√3 evaluate cos alpha cos beta - sin alpha sin bheta​

Answer 1

Answer:$\frac{\sqrt{2}-\sqrt{6} }{4}$

Step-by-step explanation:

tanA = 1

=> A=45 deg  (tan45 = 1)

tanB = $\sqrt{3}$

=> B=60 deg  (tan60 =$\sqrt{3}$)

putting values of A and B in cosAcosB-sinAsinB,

= cos45.cos60 - sin45.sin60

= $\frac{1}{\sqrt{2} }$ x $\frac{1}{2}$ -  $\frac{1}{\sqrt{2} }$x$\frac{\sqrt{3} }{2}$

=$\frac{1}{\sqrt{2} }$ [ $\frac{1}{2}$ - $\frac{\sqrt{3} }{2}$]

=$\frac{1-\sqrt{3} }{2\sqrt{2} }$= $\frac{\sqrt{2}-\sqrt{6} }{4}$