Question

if tan A equal to root 2 then the value of Cos A minus sin a by Cos A + sin a is​

Answer 1

Answer:

Tan A=√2

sinA/cosA=√2

sinA=√2×cosA

Answer 2

$\dfrac{\cos A-\sin A}{\cos A+\sin A}$ = $\dfrac{1-\sqrt{2}}{1+\sqrt{2}}$ or, - 3 + 2$\sqrt{2}$

Step-by-step explanation:

We have,

$\tan A$ = $\sqrt{2}$

To find, the value of $\dfrac{\cos A-\sin A}{\cos A+\sin A}$ = ?

∴ $\dfrac{\cos A-\sin A}{\cos A+\sin A}$

Dividing numerator and denominator by $\cos A$, we get

= $\dfrac{\dfrac{\cos A-\sin A}{\cos A} }{\dfrac{\cos A+\sin A}{\cos A}}$

= $\dfrac{1-\tan A}{1+\tan A}$

Put $\tan A$ = $\sqrt{2}$, we get

= $\dfrac{1-\sqrt{2}}{1+\sqrt{2}}$

Rationalizing numerator and denominator, we get

= $\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\times \dfrac{1-\sqrt{2}}{1-\sqrt{2}}$

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

Using the algebraic identity,

$a^{2} -b^{2}$ = (a + b) (a - b) and

$(a-b)^{2}$ = $a^{2} +b^{2} -2ab$

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

= $\dfrac{1+2-2\sqrt{2}}{1-2}$

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

= - 3 + 2$\sqrt{2}$

∴ $\dfrac{\cos A-\sin A}{\cos A+\sin A}$ = $\dfrac{1-\sqrt{2}}{1+\sqrt{2}}$ or, - 3 + 2$\sqrt{2}$