Question

1- sin2a/cos2a = tan ( 45 - a)

Answer 1

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

$\boxed{ \rm{ \: \frac{1 - sin2a}{cos2a} = tan(45 \degree \: - \: a)}}$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\underbrace{\boxed{ \tt{ sin2x = 2sinxcosx\: }}}$

$\underbrace{\boxed{ \tt{1 - cos2x \: = \: {2sin}^{2}x}}}$

$\underbrace{\boxed{ \tt{sin(90 \degree - x) = cosx \: }}}$

$\underbrace{\boxed{ \tt{cos(90 \degree - x) = sinx \: }}}$

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

Consider LHS

$\rm :\longmapsto\:\dfrac{1 - sin2a}{cos2a}$

can be rewritten as

$\rm \:  =  \:  \: \dfrac{1 - cos(90 \degree - 2a)}{sin(90 \degree - 2a)}$

Let assume that 90° - 2a = 2x so that 45° - a = x

$\rm \:  =  \:  \: \dfrac{1 - cos2x}{sin2x}$

$\rm \:  =  \:  \: \dfrac{ {2sin}^{2} x}{2sinxcosx}$

$\rm \:  =  \:  \: \dfrac{ \cancel{2} \:\cancel{ sinx} \: sinx}{\cancel{2} \: \cancel{sinx} \: cosx}$

$\rm \:  =  \:  \: \dfrac{sinx}{cosx}$

$\rm \:  =  \:  \: tanx$

On substituting the value of x = 45° - a, we get

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

Hence,

$\: \: \: \: \underbrace{\boxed{ \bf{ \qquad\: \frac{1 - sin2a}{cos2a} = tan(45 \degree \: - \: a) \qquad}}}$

Additional Information :-

$\underbrace{\boxed{ \tt{cos2x = 1 - {2sin}^{2}x \: }}}$

$\underbrace{\boxed{ \tt{cos2x ={2cos}^{2}x - 1 \: }}}$

$\underbrace{\boxed{ \tt{cos2x = \frac{1 - {tan}^{2}x}{1 + {tan}^{2}x} \: }}}$

$\underbrace{\boxed{ \tt{sin2x = \frac{2{tan}x}{1 + {tan}^{2}x} \: }}}$

$\underbrace{\boxed{ \tt{tan2x = \frac{2{tan}x}{1 - {tan}^{2}x} \: }}}$

$\underbrace{\boxed{ \tt{sin3x\: = \: 3sinx - {4sin}^{3}x}}}$

$\underbrace{\boxed{ \tt{cos3x = {4cos}^{3}x - 3cosx}}}$