Question

(sin theta-cos theat)/(sin theat+cos theta) + (sin theta+cos theta)/(sin theta-costheta) =2/(2sin^theta-1)​

Answer 1

Answer:

THIS IS YOUR REQUIRED SOLUTION OF REQUIRED QUESTION

Answer 2

We have,

$\frac{ \sinθ + \cosθ }{ \sinθ - \cos θ } + \frac{ \sinθ - \cos θ }{ \sinθ + \cosθ} \\ = \frac{ (\sinθ + \cosθ) ^{2} + ( \sinθ - \cosθ )^{2} }{( \sinθ - \cos θ)( \sinθ + \cosθ )} \\ = \frac{ { \sin }^{2}θ + { \cos}^{2}θ + 2 \sin θ \cos θ + { \sin }^{2}θ + \cos ^{2} θ - 2 \sin θ \cosθ }{ \sin^{2}θ - \cos^{2}θ } \\ = \frac{1 + 1}{ \sin^{2}θ - { \cos }^{2}θ } \\ = \frac{2}{ { \sin }^{2} θ - { \cos }^{2}θ} \\ = \frac{2}{ { \sin }^{2}θ - (1 - { \sin }^{2}θ) } \\ = \frac{2}{2 { \sin}^{2}θ - 1 }$

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