Question

sin2thetha=2cot thetha/cot square thetha+1​

Answer 1

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

Consider,

$\rm :\longmapsto\:sin2\theta$

can be rewritten as

$\rm \: = \:sin(\theta + \theta )$

$\rm \: = \:sin\theta cos\theta + cos\theta sin\theta$

$\rm \: = \:2sin\theta cos\theta$

can be rewritten as

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

We know,

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

So, using this, we get

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

can be further rewritten as

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

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

We know,

$\boxed{ \tt{ \: \frac{cosx}{sinx} = cotx \: \: }}$

So, using this, we get

$\rm \: = \:\dfrac{2 \: cot\theta }{1 + {cot}^{2} \theta }$

Hence,

$\bf\implies \:\rm \boxed{ \tt{ \: \:sin2\theta \: = \:\dfrac{2 \: cot\theta }{1 + {cot}^{2} \theta } \: \: }}$

Hence, Proved

Additional Information :-

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

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

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

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

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

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

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