Chapter 7: Q. 104 (page 475)

Show $\sqrt{9 s e c^{2} θ - 9} = 3 \tan θ i f π \leq θ < \frac{3 π}{2}$

Short Answer

Expert verified

It is proved that $\sqrt{9 s e c^{2} θ - 9} = 3 \tan θ$

Step by step solution

Step 1. Given Information

The given equation and interval is $\sqrt{9 s e c^{2} θ - 9} = 3 \tan θ, π \leq θ < \frac{3 π}{2}$

Step 2. Calculation

Consider the left hand side of the equation and prove it to the right side of the given equation.

$\sqrt{9 s e c^{2} θ - 9} = \sqrt{9 (s e c^{2} θ - 1)} = 3 \sqrt{s e c^{2} θ - 1} = 3 \sqrt{\frac{1}{\cos^{2} θ} - 1} = 3 \sqrt{\frac{1 - \cos^{2} θ}{\cos^{2} θ}} = 3 \sqrt{\frac{\sin^{2} θ}{\cos^{2} θ}} = 3 \sqrt{\tan^{2} θ} = 3 \tan θ$

Hence, Proved

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Show $\sqrt{9 s e c^{2} θ - 9} = 3 \tan θ i f π \leq θ < \frac{3 π}{2}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

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Show9sec2θ-9=3tanθifπ≤θ<3π2

Short Answer

Step by step solution

Step 1. Given Information

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Show $\sqrt{9 s e c^{2} θ - 9} = 3 \tan θ i f π \leq θ < \frac{3 π}{2}$