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If A+B=45°than show that (1+tanA)(1+tanB)=2 explain in telugu ​

Answer 1

Answer:

$\huge\frak\red{Answer}$

Given that,

$</p><p>\begin{gathered}\sf{x + y = 45} \\ \\ \longrightarrow \: \sf{tan(x + y) = tan 45}\end{gathered}x+y=45⟶tan(x+y)=tan45</p><p>$

To Prove

$\sf (1 + tan \: x)(1 + tan \: y) = 2(1+tanx)(1+tany)=2$

Now,

$\begin{gathered} \longrightarrow \: \sf{ \dfrac{tan \: x + tan \: y}{1 - tan \: x.tan \: y} = 1 } \\ \\ \longrightarrow \: \sf{tan \: x + tan \: y = 1 - tan \: x.tan \: y} \\ \\ \longrightarrow \: \underline{\boxed{\sf{tan \: x + \: tan \: y + tan \: x.tan \: y = 1}}}\end{gathered}⟶1−tanx.tanytanx+tany=1⟶tanx+tany=1−tanx.tany⟶tanx+tany+tanx.tany=1$

Adding one on both sides,we get :

$\begin{gathered} \longrightarrow \: \sf{tan \: x + tan \: y + tan \: x.tan \: y + 1 = 1 + 1} \\ \\ \longrightarrow \: \sf{1(1 + tan \: x) + tan \: y(1 + tan \: x) = 2} \\ \\ \longrightarrow \: \tt{(1 + tan \: x)(1 + tan \: y) = 2}\end{gathered}⟶tanx+tany+tanx.tany+1=1+1⟶1(1+tanx)+tany(1+tanx)=2⟶(1+tanx)(1+tany)=2$

Henceforth,Proved✅