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Delicious77 [7]
2 years ago
15

Prove that 1/sec0-tan0= sec0+tan0

Mathematics
1 answer:
Leviafan [203]2 years ago
7 0

Answer:

Step-by-step explanation:

\text{LHS}=\frac{1}{\sec \theta-\tan \theta}\\=\frac{\sec \theta+\tan \theta}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}\\=\frac{\sec \theta+\tan \theta}{\sec^{2} \theta-\tan^{2} \theta}\\=\frac{\sec \theta+\tan \theta}{1}\\=\sec \theta+\tan \theta\\=\text{RHS}

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If you would like to write x^4y - 4x^2y - 5y in a completely factored form, you can do this using the following steps:

x^4y - 4x^2y - 5y = y * (x^4 - 4x^2 - 5) = y * (x^2 + 1) * (x^2 - 5)

The correct result would be <span>y * (x^2 + 1) * (x^2 - 5).</span>
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2 years ago
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3 years ago
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