Question

sin theta-cos theta/sin theta + cos theta + sin theta + cos theta/sin theta-cos theta is equal to 2 / 2 sin squared theta minus one ​

Answer 1

ANSWER

to prove

$\bold{\frac{ \sin \theta - \cos \theta}{ \sin \theta + \cos \theta} + \frac{ \sin \theta + \cos \theta}{ \sin \theta - \cos \theta} = \frac{2}{2 \sin^{2} \theta - 1} }$

we have,

$\bold{LHS = \frac{ \sin \theta - \cos \theta}{ \sin \theta + \cos \theta} + \frac{ \sin \theta + \cos \theta}{ \sin \theta - \cos \theta} }$

and

$\bold{RHS = \frac{2}{2 \sin^{2} \theta - 1 } }$

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now solve LHS

$\bold{LHS = \frac{ \sin \theta - \cos \theta}{ \sin \theta + \cos \theta} + \frac{ \sin \theta + \cos \theta}{ \sin \theta - \cos \theta} }$

now, by taking LCM

$\bold{LHS = \frac{(sin \theta - \cos \theta)( \sin \theta - \cos \theta) + ( \sin \theta + \cos \theta)( \sin \theta + \cos \theta)}{( \sin \theta + \cos \theta)( \sin \theta - \cos \theta)} }$

$\bold{: \implies LHS = \frac{( { \sin}^{2} \theta + { \cos}^{2} \theta - 2 \sin \theta \cos \theta) + ( { \sin}^{2} \theta + { \cos}^{2} \theta + 2 \sin \theta \cos \theta) }{ { \sin}^{2} \theta - { \cos}^{2} \theta } }$

$\bold{ : \implies LHS = \frac{ { \sin}^{2} \theta + { \cos}^{2} \theta - \cancel{ 2 \sin \theta \cos \theta }+ { \sin}^{2} \theta + { \cos}^{2} \theta + \cancel {2 \sin \theta \cos \theta} }{ { \sin}^{2} \theta - { \cos}^{2} \theta } }$

$\bold{: \implies LHS = \frac{ { (\sin}^{2} \theta + { \cos}^{2} \theta) + ( { \sin}^{2} \theta + { \cos}^{2 } \theta)}{ { \sin}^{2} \theta - { \cos }^{2} \theta } }$

and

$\boxed{ \bold{{ \sin}^{2} \theta + { \cos}^{2} \theta = 1}}$

$\bold{: \implies LHS = \frac{1 + 1}{ { \sin}^{2} \theta - { \cos}^{2 } \theta } }$

and

$\boxed{ \bold {{ \cos}^{2} \theta = 1 - { \sin}^{2} \theta}}$

$\bold{ : \implies LHS = \frac{1 + 1}{ { \sin}^{2} \theta - (1 - { \sin}^{2} \theta) } }$

$\bold{ : \implies LHS = \frac{2}{ { \sin}^{2} \theta - 1 + { \sin}^{2} \theta } }$

$\bold{ : \implies LHS = \frac{2}{2 { \sin}^{2} \theta - 1 } =RHS }$

so, LHS = RHS

hence, proved!

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☯︎ some identities :

• sin² θ + cos² θ = 1

• cos² θ = 1 - sin² θ

• sin² θ = 1 - cos² θ

• 1 + tan² θ = sec² θ

• sec² θ - tan² θ = 1

• sec² θ - 1 = tan² θ

• 1 + cot² θ = cosec² θ

• cosec² θ - cot² θ = 1

• cosec² θ - 1 = cot² θ

• sec θ + tan θ = 1/sec θ - tan θ

• cosec θ + cot θ = 1/cosec θ - cot θ