If sin-x-y-cos-x-y-1-m-1-m prove that tan- Pi-4-x-tan- Pi-4-y-m

Sinx+ycosx y=1 m1+m #8658;cosx ysinx+y=1+m1 mApplying #160;componendo #160;and #160;dividendo #8658;cosx y+sinx+ycosx y sinx+y=1+m+1 m1+m 1+m #8658; ...

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Byju's Answer

Standard XII

Mathematics

Differentiation of a Function

if sin(x+y)/c...

Question

if sin(x+y)/cos(x-y)=1-m/1+m
prove that tan(Π/4-x).tan(Π/4-y)=m

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Solution

$\frac{\sin (x + y)}{\cos (x - y)} = \frac{1 - m}{1 + m} \Rightarrow \frac{\cos (x - y)}{\sin (x + y)} = \frac{1 + m}{1 - m} Applying componendo and dividendo \Rightarrow \frac{\cos (x - y) + \sin (x + y)}{\cos (x - y) - \sin (x + y)} = \frac{1 + m + 1 - m}{1 + m - 1 + m} \Rightarrow \frac{\cos (x - y) + \sin (x + y)}{\cos (x - y) - \sin (x + y)} = \frac{1}{m} \Rightarrow \frac{\cos (x - y) - \sin (x + y)}{\cos (x - y) + \sin (x + y)} = m \Rightarrow \frac{\cos x \cos y + \sin x \sin y - \sin x \cos y - \cos x \sin y}{\cos x \cos y + \sin x \sin y + \sin x \cos y + \cos x \sin y} = m \Rightarrow \frac{\cos y (\cos x - \sin x) - \sin y (\cos x - \sin x)}{\cos y (\cos x + \sin x) - \sin y (\cos x + \sin x)} = m \Rightarrow \frac{(\cos x - \sin x) (\cos y - \sin y)}{(\cos x + \sin x) (\cos y + \sin y)} = m \Rightarrow (\frac{1 - \tan x}{1 + \tan x}) (\frac{1 - \tan y}{1 + \tan y}) = m \Rightarrow (\frac{\tan \frac{π}{4} - \tan x}{1 + \tan \frac{π}{4} \tan x}) (\frac{\tan \frac{π}{4} - \tan y}{1 + \tan \frac{π}{4} \tan y}) = m \Rightarrow \tan (\frac{π}{4} - x) \tan (\frac{π}{4} - y) = m (Since \tan (a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}) Hence Proved .$

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if sin(x+y)/cos(x-y)=1-m/1+m prove that tan(Π/4-x).tan(Π/4-y)=m

if sin(x+y)/cos(x-y)=1-m/1+m
prove that tan(Π/4-x).tan(Π/4-y)=m