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f(x)=(tan(x)^2-sin(x)^2)/(1-cos(x)^2)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

tan2x−sin2x1−cos2x\frac{tan^2x-sin^2x}{1-cos^2x}

Step-by-step Solution

1

Apply the trigonometric identity: 1−cos⁡(θ)21-\cos\left(\theta \right)^2=sin⁡(θ)2=\sin\left(\theta \right)^2

tan⁡(x)2−sin⁡(x)2sin⁡(x)2\frac{\tan\left(x\right)^2-\sin\left(x\right)^2}{\sin\left(x\right)^2}
Why is 1 - cos(x)^2 = sin(x)^2 ?

Final answer to the exercise

tan⁡(x)2−sin⁡(x)2sin⁡(x)2\frac{\tan\left(x\right)^2-\sin\left(x\right)^2}{\sin\left(x\right)^2}

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tan2x−sin2x1−cos2x 
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