Question

If cos A = 1/2 and sin B = 1/√2,find value of
$\frac{ \tan(a) - \tan(b) }{1 + \tan(a) \tan(b) }$
Here angles A and B are from different right triangles.​

Answer 1

Answer:

$2-\sqrt{3}$

Step-by-step explanation:

given,

$cosA=\frac{1}{2} , sinB=\frac{1}{\sqrt{2} }$

now,

⇒ $cosA=\frac{1}{2} \\But, cos60^o=\frac{1}{2} \\\\A=60^o$

similarly,

⇒ $sinB=\frac{1}{\sqrt{2} } \\but, sin45^o=\frac{1}{\sqrt[]{2} } \\B=45^o$

now, $\frac{tanA-tanB}{1+tanA.tanB} = \frac{tan60^o-tan45^o}{1+tan60^o.tan45^o}$

 = $\frac{\sqrt[]{3}-1 }{1+\sqrt{3}.1 }$           $[tan60^o=\sqrt{3}, tan45^o=1]$

=$= \frac{\sqrt{3} -1}{\sqrt{3}+1 } \\=\frac{\sqrt{3} -1}{\sqrt{3}+1 } .\frac{\sqrt{3} -1}{\sqrt{3}-1 }--- [rationalisation]\\=\frac{(\sqrt{3} -1)^{2} }{(\sqrt{3})^2-(1)^2 } \\=\frac{3+1-2\sqrt{3} }{3-1} \\=\frac{4+2\sqrt{3} }{2} \\=\frac{4-2\sqrt{3} }{2} \\=2-\sqrt{3}$