Question

sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

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

Step-by-step explanation:

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

$=\frac{(sin^\theta+cos\theta)^{2}+(sin\theta-cos\theta)^{2}}{(sin\theta-cos\theta)(sin\theta+cos\theta)}$

$=\frac{2(sin^{2}\theta+cos^{2}\theta)}{(sin^{2}\theta-cos^{2}\theta)}$

$=\frac{2\times 1}{(sin^{2}\theta-cos^{2}\theta)}$

$\boxed { Since , \: sin^{2}\theta+cos^{2}\theta = 1}$

$=\frac{2}{(sin^{2}\theta-cos^{2}\theta)}$

$Divide \: numerator \: and \: denominator \\by \: cos^{2}\theta,we\:get$

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

$= \frac{2sec^{2}\theta}{tan^{2}\theta-1}\\=RHS$

Therefore.,

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

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