Question

sec2theta-tan2theta =tan(π/4-theta)​

Answer 1

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

Consider LHS

$\rm :\longmapsto\:sec2\theta - tan2\theta$

$\rm \:  =  \: \dfrac{1}{cos2\theta} - \dfrac{sin2\theta}{cos2\theta}$

$\rm \:  =  \: \dfrac{1 - sin2\theta}{cos2\theta}$

We know,

$\boxed{\tt{ cos\bigg[\dfrac{\pi}{2} - x\bigg] = sinx}} \: \: and \: \: \boxed{\tt{ sin\bigg[\dfrac{\pi}{2} - x\bigg] = cosx}}$

So, using this, we get

$\rm \:  =  \: \dfrac{1 - cos\bigg[\dfrac{\pi}{2} -2\theta\bigg]}{sin\bigg[\dfrac{\pi}{2} - 2\theta\bigg]}$

Let assume that

$\red{\rm :\longmapsto\:\boxed{\tt{ \bigg[\dfrac{\pi}{2} - 2\theta\bigg] =2 y \: \: \rm \implies\:y = \bigg[\dfrac{\pi}{4} - \theta\bigg]}}}$

So, above can be rewritten as

$\rm \:  =  \: \dfrac{1 - cos2y}{sin2y}$

$\rm \:  =  \: \dfrac{ {2sin}^{2}y}{2 \: siny \: cosy}$

$\rm \:  =  \: \dfrac{siny}{cosy}$

$\rm \:  =  \: tany$

$\rm \:  =  \: tan\bigg[\dfrac{\pi}{4} - \theta \bigg]$

Hence,

$\red{\rm :\longmapsto\: \boxed{\tt{ \: sec2\theta - tan2\theta  =  \: tan\bigg[\dfrac{\pi}{4} - \theta \bigg] \: }}}$

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Additional Information

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

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

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

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

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

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

$\boxed{\tt{ tan3x = \frac{3tanx - {tan}^{3}x }{1 - {3tan}^{2}x }}}$