Question

if cot theta=cos (x+y) and cot phy =cos (x-y) then prove it tan(theta-phy)=2sinxsiny/cos square x+cos square y​

Answer 1

Answer:

$tan(\theta-\phi)=\frac{2sinx\:siny}{cos^2x+cos^2y}$

Step-by-step explanation:

Formula used:

$tan(A-B)=\frac{tanA-tanB}{1+tanA\:tanB}$

$cos(A-B)=cosA\:cosB+sinA\:sinB$

$cos(A+B)=cosA\:cosB-sinA\:sinB$

$cos(A+B).cos(A-B)=cos^2A-sin^2B$

Given:

$cot\theta=cos(x+y)$

Taking reciprocals, we get

$tan\theta=\frac{1}{cos(x+y)}$

and

$cot\phi=cos(x-y)$

Taking reciprocals, we get

$tan\phi=\frac{1}{cos(x-y)}$

Now,

$tan(\theta-\phi)$

$=\frac{tan\theta-tan\phi}{1+tan\theta.tan\phi}$

$=\frac{\frac{1}{cos(x+y)}-\frac{1}{cos(x-y)}}{1+\frac{1}{cos(x+y)}.\frac{1}{cos(x-y)}}$

$=\frac{\frac{cos(x-y)-cos(x+y)}{cos(x+y)\:cos(x-y)}}{1+\frac{1}{cos(x+y)\:cos(x-y)}}$

$=\frac{cos(x-y)-cos(x+y)}{cos(x+y)\:cos(x-y)+1}$

$=\frac{(cosx\:cosy+sinx\:siny)-(cosx\:cosy-sinx\:siny)}{cos^2x-sin^2y+1}$

$=\frac{cosx\:cosy+sinx\:siny-cosx\:cosy+sinx\:siny}{cos^2x+(1-sin^2y)}$

$=\frac{2sinx\:siny}{cos^2x+(1-sin^2y)}$

$=\frac{2sinx\:siny}{cos^2x+cos^2y}$

Answer 2

Answer=(c da attachment)

So, lhs = rhs

Hence proved.