Question : If $P = 7 + 4 \sqrt{3}$ and $P Q = 1$ , then what is the value of $\frac{1}{P^{2}} + \frac{1}{Q^{2}}$ ?
Option 1: 196
Option 2: 194
Option 3: 206
Option 4: 182

Question

Question : If $\sec θ - \tan θ = \frac{1}{\sqrt{3}}$ , then the value of $\sec θ . \tan θ$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{2}{\sqrt{3}}$
Option 3: $\frac{4}{\sqrt{3}}$
Option 4: $\frac{1}{\sqrt{3}}$

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Answer 1

Correct Answer: 194

Solution : Given: $P Q = 1, P = 7 + 4 \sqrt{3}$
⇒ $Q = \frac{1}{P}$
⇒ $P^{2} = (7 + 4 \sqrt{3})^{2} = 97 + 56 \sqrt{3}$
Taking the inverse on both sides and rationalising the denominators,
⇒ $\frac{1}{P^{2}} = 97 - 56 \sqrt{3}$
⇒ $\frac{1}{P^{2}} + \frac{1}{Q^{2}} = \frac{1}{P^{2}} + P^{2}$
⇒ $\frac{1}{P^{2}} + \frac{1}{Q^{2}} = (97 - 56 \sqrt{3} + 97 + 56 \sqrt{3})$
⇒ $\frac{1}{P^{2}} + \frac{1}{Q^{2}} = 194$
Hence, the correct answer is 194.

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Question : If P=7+43 and PQ=1, then what is the value of 1P2+1Q2?Option 1: 196Option 2: 194Option 3: 206Option 4: 182

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Question : If $P = 7 + 4 \sqrt{3}$ and $P Q = 1$ , then what is the value of $\frac{1}{P^{2}} + \frac{1}{Q^{2}}$ ?
Option 1: 196
Option 2: 194
Option 3: 206
Option 4: 182