Question

prove the following fast waiting for the solution​

Answer 1

Answer:

LHS = RHS .

Step-by-step explanation:

$lhs = \tan(45 - a) = \frac{ \tan(45) - \tan(a) }{1 + \tan(45) \tan(a) } \\ \\ = \frac{1 - \tan(a) }{1 + \tan(a) } \\ \\ = \frac{1 - \frac{ \sin(a) }{ \cos(a) } }{1 + \frac{ \sin(a) }{ \cos(a) } } \\ \\ = \frac{ \cos(a) - \sin(a) }{ \sin(a) + \cos(a) } \\ \\ \frac{ \cos(a) - \cos(a) }{ \cos(a) + \sin(a) } = rhs.$

Answer 2

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:tan(45 \degree \: - \: A) = \dfrac{cosA - sinA}{cosA + sinA}$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:tan(45\degree - A)$

We know,

$\boxed{ \sf{ \:tan(x - y) = \frac{tanx - tany}{1 + tanxtany}}}$

So using this identity, we get

$\rm \: = \: \:\dfrac{tan45\degree - tanA}{1 - tan45\degree \:tanA}$

We know,

$\boxed{ \sf{ \:tan45\degree = 1}}$

So, using this, we get

$\rm \: = \: \:\dfrac{1 - tanA}{1 + 1 \times tanA}$

$\rm \: = \: \:\dfrac{1 - tanA}{1 + tanA}$

We know,

$\boxed{ \sf{ \:tanx = \frac{sinx}{cosx}}}$

So, using this, we get

$\rm \: = \: \:\dfrac{1 - \dfrac{sinA}{cosA} }{1 + \dfrac{sinA}{cosA} }$

$\rm \: = \: \:\dfrac{\dfrac{cosA - sinA}{cosA} }{\dfrac{cosA + sinA}{cosA} }$

$\rm \: = \: \:\dfrac{cosA - sinA}{cosA + sinA}$

Hence, Proved

Alternative Method :-

Consider RHS

$\rm :\longmapsto\:\: \:\dfrac{cosA - sinA}{cosA + sinA}$

Divide numerator and denominator by cosA, we get

$\rm \: = \: \:\dfrac{\dfrac{cosA - sinA}{cosA} }{\dfrac{cosA + sinA}{cosA} }$

$\rm \: = \: \:\dfrac{1 - \dfrac{sinA}{cosA} }{1 + \dfrac{sinA}{cosA} }$

$\rm \: = \: \:\dfrac{1 - tanA}{1 + tanA}$

$\rm \: = \: \:\dfrac{1 - tanA}{1 + 1 \times tanA}$

$\rm \: = \: \:\dfrac{tan45\degree - tanA}{1 - tan45\degree \:tanA}$

$\rm \: = \: \:tan(45\degree - A)$

Hence, Proved

Additional Information :-

$\boxed{ \sf{ \:sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \sf{ \:sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \sf{ \:cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \sf{ \:cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{ \sf{ \:tan(x + y) = \frac{tanx + tany}{1 - tanxtany}}}$