Question

sin theta -sin2theta/ 1-cos theta +cos2theta ​

Answer 1

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

$\rm :\longmapsto\:\dfrac{sin\theta - sin2\theta }{1 - cos\theta + cos2\theta }$

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

Given Trigonometric function is

$\rm :\longmapsto\:\dfrac{sin\theta - sin2\theta }{1 - cos\theta + cos2\theta }$

can be re-arranged as

$\rm \: = \: \: \dfrac{sin\theta - sin2\theta }{(1 + cos2\theta ) - cos\theta }$

We know,

$\boxed{ \rm{ sin2x = 2sinxcosx}}$

and

$\boxed{ \rm{ 1 + cos2x = {2cos}^{2}x}}$

Using these two Identities, we get

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

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

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

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

$\rm \: = \: \: - \: tan\theta$

Hence,

$\bf :\longmapsto\:\dfrac{sin\theta - sin2\theta }{1 - cos\theta + cos2\theta } = - \: tan\theta$

Additional Information :-

$\boxed{ \rm{ sin2x = \frac{2tanx}{1 + {tan}^{2} x}}}$

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

$\boxed{ \rm{ cos2x = {cos}^{2} x - {sin}^{2} x}}$

$\boxed{ \rm{ cos2x = 1 - {2sin}^{2} x}}$

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

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

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

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