Question

1+cos2theta/sin2theta=1/2(cot theta/2-tan theta/2

Answer 1

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

$\rm \: \dfrac{1 + cos2\theta}{sin2\theta } = \dfrac{1}{2}\bigg(cot\dfrac{\theta }{2} - tan\dfrac{\theta }{2} \bigg)$

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

Consider, LHS

$\rm \: \dfrac{1 + cos2\theta }{sin2\theta }$

We know,

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

and

$\boxed{\sf{ \:sin2x = 2 \: sinx \: cosx \: }}$

So, on substituting the values, in above expression, we get

$\rm \: = \: \dfrac{1 + {2cos}^{2}\theta - 1 }{2 \: sin\theta \: cos\theta }$

$\rm \: = \: \dfrac{ {2cos}^{2} \theta }{2 \: sin\theta cos\theta }$

$\rm \: = \: \dfrac{cos\theta }{sin\theta }$

$\rm \: = \: cot\theta$

$\rm \: = \: \dfrac{1}{tan\theta }$

We know,

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

So, using this, we get

$\rm \: = \: \dfrac{1}{\dfrac{2tan\dfrac{\theta }{2}}{1 - {tan}^{2}\dfrac{\theta }{2} } }$

$\rm \: = \: \dfrac{1}{2}\bigg(\dfrac{1 - {tan}^{2} \dfrac{\theta }{2}}{tan\dfrac{\theta }{2}} \bigg)$

$\rm \: = \: \dfrac{1}{2}\bigg(\dfrac{1}{tan\dfrac{\theta }{2}} - tan\dfrac{\theta }{2} \bigg)$

$\rm \: = \: \dfrac{1}{2}\bigg(cot\dfrac{\theta }{2} - tan\dfrac{\theta }{2} \bigg)$

Hence,

$\rm\implies \: \: \boxed{\sf{ \:\: \rm \: \dfrac{1 + cos2\theta}{sin2\theta } = \dfrac{1}{2}\bigg(cot\dfrac{\theta }{2} - tan\dfrac{\theta }{2} \bigg) \: \: }}$

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

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

$\boxed{\sf{ \:cos2x = \frac{1 + {tan}^{2}x }{1 + {tan}^{2} x} \: }}$

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

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

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

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