Question

if tan A = 1, tan B = root 3 evaluate cos A cos B - sin A sin B

Answer 1

Answer:

The value of

$\cos A \cos B - \sin A \sin B = \frac { 183 } { 707 }$

Solution:

Given that the value of,

$\begin{array} { c } { \tan A = 1 } \\\\ { \tan A = \tan 45 ^ { \circ } } \end{array}$

We know that,

$\tan 45 ^ { \circ } = 1$

Thus the angle,

$A = 45 ^ { \circ }$

Similarly, it is given that,

$\begin{array} { c } { \tan B = \sqrt { 3 } } \\\\ { \tan B = \tan 60 ^ { \circ } } \end{array}$

We know that,

$\tan 60 ^ { \circ } = \sqrt { 3 }$

Thus the angle,

$\begin{array} { c } { B = 60 ^ { \circ } } \\\\ { \cos A \cos B - \sin A \sin B = \cos 45 ^ { \circ } \cos 60 ^ { \circ } - \sin 45 ^ { \circ } \sin 60 ^ { \circ } } \\\\ { \cos A \cos B - \sin A \sin B = \frac { 1 } { \sqrt { 2 } } \times \frac { 1 } { 2 } - \frac { 1 } { \sqrt { 2 } } \times \frac { \sqrt { 3 } } { 2 } } \end{array}$

$\cos A \cos B - \sin A \sin B = \frac { 1 } { 2 \sqrt { 2 } } - \frac { \sqrt { 3 } } { 2 \sqrt { 2 } }$

$\begin{array} { l } { \cos A \cos B - \sin A \sin B = \frac { 1 - \sqrt { 3 } } { 2 \sqrt { 2 } } } \\\\ { \cos A \cos B - \sin A \sin B = \frac { 1 - \sqrt { 3 } } { 2 \sqrt { 2 } } } \end{array}$

We know that the value of

$\begin{array} { l } { \sqrt { 3 } = 1.732 } \\\\ { \sqrt { 2 } = 1.414 } \end{array}$

Thus the term is simplified to,

$\begin{array} { c } { \cos A \cos B - \sin A \sin B = \frac { 1 - 1.732 } { 2 \times 1.414 } } \\\\ { = \frac { 0.732 } { 2 \times 1.414 } } \\\\ { = \frac { 0.366 } { 1.414 } } \end{array}$

$\cos A \cos B - \sin A \sin B = \frac { 366 } { 1414 } = \frac { 183 } { 707 }$

Thus, the value of $\cos A \cos B - \sin A \sin B$ is given by 183/707

Answer 2

Answer:

Value of cosAcosB-sinAsinB

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

Step-by-step explanation:

Given tanA = 1

=> tanA = tan45°

=> A = 45° ---(1)

tanB = √3

=> tanB = tan60°

=> B = 60° ----(2)

Now ,

Value of cosAcosB-sinAsinB

= cos45°cos60°-sin45°sin60°

/* from (1)& (2) */

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

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

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

Therefore,

Value of cosAcosB-sinAsinB

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

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