Question

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

Answer 1

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

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side so proof is complete

Explanation:

Answer 2

`(sin\theta + cos\theta)/(sin\theta-cos\theta)+(sin\theta-cos\theta)/(sin\theta+cos\theta)=(2sec^2\theta)/(tan^2\theta-1)`

From Left side:

`(sin\theta + cos\theta)/(sin\theta-cos\theta)\bigg((sin\theta+cos\theta)/(sin\theta+cos\theta)\bigg)+(sin\theta-cos\theta)/(sin\theta+cos\theta)\bigg((sin\theta-cos\theta)/(sin\theta-cos\theta)\bigg)`

`(sin^2\theta+2cos\thetasin\theta+cos^2\theta)/(sin^2\theta-cos^2\theta)+(sin^2\theta-2cos\thetasin\theta+cos^2\theta)/(sin^2\theta-cos^2\theta)`

NOTE: sin²θ + cos²θ = 1

`(1 + 2cos\theta sin\theta)/(sin^2\theta-cos^2\theta)+(1-2cos\theta sin\theta)/(sin^2\theta-cos^2\theta)`

`(1 + 2cos\theta sin\theta+1-2cos\theta sin\theta)/(sin^2\theta-cos^2\theta)`

`(2)/(sin^2\theta-cos^2\theta)`

`(2)/(\bigg(sin^2\theta-cos^2\theta\bigg)\bigg((cos^2\theta)/(cos^2\theta)\bigg))`

`(2sec^2\theta)/((sin^2\theta)/(cos^2\theta)-(cos^2\theta)/(cos^2\theta))`

`(2sec^2\theta)/(tan^2\theta-1)`

Left side = Right side so proof is complete