Question

If tan A : cot B=1 : 2, then cos(A+B) : cos(A-B)​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\dfrac{tan(A)}{cot(B)}=\dfrac{1}{2}}$

$\sf{tan(A)\,tan(B)=\dfrac{1}{2}\,\,\,\,\,\,\,...(1)}$

Now, we have,

$\tt{\dfrac{cos(A+B)}{cos(A-B)}}$

$\sf{=\dfrac{cos(A)\,cos(B)-sin(A)\,sin(B)}{cos(A)\,cos(B)+sin(A)\,sin(B)}}$

Divide num. and den. by cos(A).cos(B)

$\sf{=\dfrac{\dfrac{cos(A)\,cos(B)-sin(A)\,sin(B)}{cos(A)\,cos(B)}}{\dfrac{cos(A)\,cos(B)+sin(A)\,sin(B)}{cos(A)\,cos(B)}}}$

$\sf{=\dfrac{1-tan(A)\,tan(B)}{1+tan(A)\,tan(B)}}$

From (1), we get,

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

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

$\sf{=\dfrac{\dfrac{1}{2} }{\dfrac{3}{2}}}$

$\sf{=\dfrac{1}{3}}$