Question

(1/sin theta +cos theta ) + (1/sin theta - cos theta ) = 2 sin theta /1-2cos square theta

Answer 1

$\bigg( \bf{ \dfrac{1}{sin( \theta) + cos( \theta)} } \bigg) + \bigg( \bf{ \dfrac{1}{sin( \theta) - cos( \theta)} } \bigg) = \bf{ \dfrac{2sin( \theta)}{1 - {2cos}^{2} ( \theta)} }$

Taking LHS__

$\bigg( \bf{ \dfrac{1}{sin( \theta) + cos( \theta)} } \bigg) + \bigg( \bf{ \dfrac{1}{sin( \theta) - cos( \theta)} } \bigg) \\ \\ \\ \longrightarrow \: \dfrac{ \text{sin} (\theta) - \text{cos} (\theta) + \text{sin} (\theta) + \text{cos} (\theta) }{(\text{sin} (\theta) + \text{cos} (\theta))(\text{sin} (\theta) - \text{cos} (\theta))}$

Comparing the denominator with__

• a² – b² = ( a + b )( a - b )

$\longrightarrow \: \dfrac{ \text{2 sin} (\theta)}{ { \text{sin}}^{2} ( \theta) -{ \text{cos}}^{2} ( \theta) }$

From 1st trigonometric identity__

$\: \: \: \: \: \: \: \: \sf{ \small{ {sin}^{2} ( \theta) + {cos}^{2} ( \theta) = 1}} \\ \\ \: \: \: \: \: \: \: \: \rightarrow \: \sf{ \small{ {sin}^{2} ( \theta) = 1 - {cos}^{2}( \theta) }}$

$\longrightarrow \: \dfrac{ \text{2 sin} (\theta)}{ { 1 - \text{cos}}^{2} ( \theta) -{ \text{cos}}^{2} ( \theta) } \\ \\ \\ \longrightarrow \: \dfrac{ \text{2 sin} (\theta)}{ 1 -{ \text{2cos}}^{2} ( \theta) }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf{ \dfrac{2 \: sin \: ( \theta) }{1 - 2 {cos}^{2} ( \theta)} = RHS}}$