Question

If m= sinx/siny and n= cosx/cosy then find (m2-n2)sin2y

Answer 1

Answer:

$1 - {n}^{2}$

Step-by-step explanation:

plz rerer to the pic

Answer 2

Given: $m= \dfrac{\sin x}{\sin y} \text { and } m= \dfrac{\cos x}{\cos y}$

To find: $(m^2-n^2) \sin^2 y$

Step-by-step explanation:

Therefore substitution the value of m and n in $(m^2-n^2) \sin^2 y$ we get

$((\dfrac{\sin x}{\sin y} )^2-(\dfrac{\cos x}{\cos y} )^2)\sin^2 y \\\\\Rightarrow (\dfrac{\sin^2 x}{\sin^2 y} -\dfrac{\cos^2 x}{\cos^2 y} )\sin^2 y \\\\\Rightarrow (\dfrac{\cos^2y \sin^2 x-\cos^2x \sin^2y}{\sin^2 y \cos^2 y} )\sin^2 y \\\\\Rightarrow (\dfrac{\cos^2y (1-\cos^2 x)-\cos^2x (1-\cos^2y)}{\sin^2 y \cos^2 y} )\sin^2 y$

$\because \sin^2 A +\cos^2 A= 1$

Now we have

$\\\\\Rightarrow \dfrac{\cos^2 y -\cos^2x \cos^2y -\cos^2 x+\cos^2x \cos^2 y}{\cos^2 y}\\\\\Rightarrow \dfrac{\cos^2 y -\cos^2 x }{\cos^2 y}\\\\\Rightarrow 1- \dfrac{\cos^2x}{\cos ^2 y} \\\\\Rightarrow 1-n^2$

Hence the value of  $(m^2-n^2) \sin^2 y$  is $1-n^2$