Chapter 7: Q. 103 (page 475)

Show $\sqrt{16 + 16 \tan^{2} θ} = 4 s e c θ i f \frac{- π}{2} < θ < \frac{π}{2}$

Short Answer

Expert verified

It is proved that $\sqrt{16 + 16 \tan^{2} θ} = 4 s e c θ$

Step by step solution

Step 1. Given Information

The given equation and interval is $\sqrt{16 + 16 \tan^{2} θ} = 4 s e c θ, \frac{- π}{2} < θ < \frac{π}{2}$

Step 2. Calculation

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

$\sqrt{16 + 16 \tan^{2} θ} = \sqrt{16 (1 + \tan^{2} θ)} = 4 \sqrt{1 + \tan^{2} θ} = 4 \sqrt{1 + \frac{\sin^{2} θ}{\cos^{2} θ}} = 4 \sqrt{\frac{\cos^{2} θ + \sin^{2} θ}{\cos^{2} θ}} = 4 \sqrt{\frac{1}{\cos^{2} θ}} = 4 \sqrt{s e c^{2} θ} = 4 s e c θ$

Hence, proved

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Show $\sqrt{16 + 16 \tan^{2} θ} = 4 s e c θ i f \frac{- π}{2} < θ < \frac{π}{2}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

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Show16+16tan2θ=4secθif-π2<θ<π2

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

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Show $\sqrt{16 + 16 \tan^{2} θ} = 4 s e c θ i f \frac{- π}{2} < θ < \frac{π}{2}$