+0

# How did this happen?

0
142
2
+28

My math program says that you can transform....

$$1/sec\theta * tan\theta = sin\theta$$

Into...

$$tan^2\theta + 1 = sec^2\theta$$

Could someone explain how they made this transition? Thanks!

CurlyFry  Oct 20, 2017

#2
+18956
+2

My math program says that you can transform....
$$1/sec\theta * tan\theta = sin\theta$$
Into...
$$tan^2\theta + 1 = sec^2\theta$$
Could someone explain how they made this transition?

$$\begin{array}{|rcll|} \hline \frac{1}{\sec(\theta)} * \tan(\theta) &=& \sin(\theta) \quad & | \quad \cdot \sec(\theta) \\ \tan(\theta) &=& \sin(\theta) \cdot \sec(\theta) \quad & | \quad \text{square both sides} \\ \tan^2(\theta) &=& \sin^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sin^2(\theta) + \cos^2(\theta) = 1 \text{ or } \sin^2(\theta) = 1-\cos^2(\theta) \\ \tan^2(\theta) &=& [~1-\cos^2(\theta)~] \cdot \sec^2(\theta) \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \frac{\cos^2(\theta)}{\cos^2(\theta)} \quad & | \quad \frac{\cos^2(\theta)}{\cos^2(\theta)} = 1 \\ \tan^2(\theta) &=& \sec^2(\theta) - 1 \quad & | \quad +1 \\ \mathbf{\tan^2(\theta)+1} &\mathbf{=}& \mathbf{\sec^2(\theta)} \\ \hline \end{array}$$

heureka  Oct 20, 2017
Sort:

#1
+82811
+1

Here's a ( highly questionable ) proceedure.......

1/sec * tan  =  sin

cos * sin/cos  = sin

sin  = sin                multiply both sides by the sin

sin^2  = sin^2

sin^2  =  1 - cos^2       add cos^2  to both sides

sin^2 + cos^2  =  1       divide through by cos^2

sin^2 / cos^2    + 1  =  1 / cos^1

tan^2  +  1  =  sec^2

Note :  multiplying/dividing on both sides of an identity isn't usually allowed......!!!!

CPhill  Oct 20, 2017
edited by CPhill  Oct 20, 2017
edited by CPhill  Oct 20, 2017
#2
+18956
+2

My math program says that you can transform....
$$1/sec\theta * tan\theta = sin\theta$$
Into...
$$tan^2\theta + 1 = sec^2\theta$$
Could someone explain how they made this transition?

$$\begin{array}{|rcll|} \hline \frac{1}{\sec(\theta)} * \tan(\theta) &=& \sin(\theta) \quad & | \quad \cdot \sec(\theta) \\ \tan(\theta) &=& \sin(\theta) \cdot \sec(\theta) \quad & | \quad \text{square both sides} \\ \tan^2(\theta) &=& \sin^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sin^2(\theta) + \cos^2(\theta) = 1 \text{ or } \sin^2(\theta) = 1-\cos^2(\theta) \\ \tan^2(\theta) &=& [~1-\cos^2(\theta)~] \cdot \sec^2(\theta) \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \frac{\cos^2(\theta)}{\cos^2(\theta)} \quad & | \quad \frac{\cos^2(\theta)}{\cos^2(\theta)} = 1 \\ \tan^2(\theta) &=& \sec^2(\theta) - 1 \quad & | \quad +1 \\ \mathbf{\tan^2(\theta)+1} &\mathbf{=}& \mathbf{\sec^2(\theta)} \\ \hline \end{array}$$

heureka  Oct 20, 2017

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