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Chapter 1: Problem 13

If \(\sin ^{2} \theta=\frac{x^{2}+y^{2}+1}{2 x}\), then \(x\) must be (a) \(-3\) (b) \(-2\) (c) 1 (d) None of these

Short Answer

Expert verified
The correct answer is (c) 1.

Step by step solution

01

Introduce the Trigonometric Identity

Recall the trigonometric identity \(\sin^2{\theta} + \cos^2{\theta} = 1\). We can use it to first express \(\cos^2{\theta}\) in terms of \(\sin^2{\theta}\), and then substitute the given expression for \(\sin^2{\theta}\).
02

Express Cosine Squared in terms of Sine Squared

From the trigonometric identity, we can find \(\cos^2{\theta}\) as follows: \(\cos^2{\theta} = 1 - \sin^2{\theta}\) Now, substitute the given expression for \(\sin^2{\theta}\): \(\cos^2{\theta} = 1 - \frac{x^2 + y^2 + 1}{2x}\)
03

Simplify the Equation

To simplify the equation, multiply both sides by \(2x\): \(2x\cos^2{\theta} = 2x - (x^2 + y^2 + 1)\) Now, distribute and rearrange the terms: \(2x\cos^2{\theta} + x^2 + y^2 + 1 = 2x\)
04

Apply Trigonometric Identity

Substitute \(\sin^2{\theta}\) in terms of \(\cos^2{\theta}\) and rearrange the equation: \(2x\sin^2{\theta} - 2x\cos^2{\theta} = x^2 + y^2 + 1 - 2x\) Now, use the given equation for \(\sin^2{\theta}\): \(\frac{x^2 + y^2 + 1}{x} - 2x\cos^2{\theta} = x^2 + y^2 + 1 - 2x\)
05

Eliminate the Variables

Observe that the equation above contains both \(y^2\) and \(\cos^2{\theta}\). We can eliminate these two terms by substituting \(\cos^2{\theta} = 1 - \sin^2{\theta}\) and then using the given equation for \(\sin^2{\theta}\): \(\frac{x^2 + y^2 + 1}{x} - 2x(1 - \frac{x^2 + y^2 + 1}{2x}) = x^2 + y^2 + 1 - 2x\)
06

Solve for x

To solve for \(x\), first expand the terms and then simplify the equation: \(x^2 + y^2 + 1 - (x^2 + y^2 + 1) = 2x^2 - 2x\) After canceling the terms, and simplifying further: \(2x^2 - 2x = 0\) Factor out \(2x\): \(2x(x - 1) = 0\) This gives us two possible solutions for \(x\): 1. \(x = 0\) 2. \(x = 1\) However, since the denominator of the given equation cannot be zero, \(x\) must be: \(x = 1\) Therefore, the correct answer is (c) 1.

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Most popular questions from this chapter

If \(A B C\) is a triangle such that angle \(A\) is obtuse, then (a) \(\tan B \tan C>1\) (b) \(\tan B \tan C<1\) (c) \(\tan B \tan C=1\) (d) None of theseSolution: (b) Let us take \(\tan A=-\tan (B+C)\) \(\Rightarrow \tan A=\frac{\tan B+\tan C}{\tan B \tan C-1}\) Since \(A\) is obtuse therefore \(\tan B \tan C-1<0\) \(\Rightarrow \tan B \tan C<1\)

If \(A=130^{\circ}\) and \(x=\sin A+\cos A\), then (a) \(x>0\) (b) \(x<0\) (c) \(x=0\) (d) \(x \geq 0\)

\(\sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}\) equals (a) \(\frac{1}{64}\) (b) \(-\frac{1}{64}\) (c) \(\frac{1}{8}\) (d) \(\frac{-1}{8}\)

If \(a_{n+1}=\sqrt{\frac{1}{2}\left(1+a_{n}\right)}\) then \(\cos \left(\frac{\sqrt{1-a_{0}^{2}}}{a_{1} a_{2} a_{3} \ldots \ldots . \text { to } \infty}\right)\) is equal to (a) 1 (b) \(-1\) (c) \(a_{0}\) (d) \(1 / a_{0}\)

Which of the following statement(s) is/are correct? (a) \(\sum_{r=1}^{7} \tan ^{2} \frac{r \pi}{16}=\sum_{t=1}^{7} \cot ^{2} \frac{r \pi}{16}=35\) (b) \(\sum_{r=1}^{10} \cos ^{3} \frac{r \pi}{3}=-\frac{9}{8}\) (c) \(\sum_{r=1}^{3} \tan ^{2}\left(\frac{2 r-1}{7}\right)=\sum_{r=1}^{3} \cot ^{2}\left(\frac{2 r-1}{7}\right)\) (d) \(\sum_{r=1}^{3} \tan ^{2}\left(\frac{2 r-1}{7}\right) \cdot \sum_{r=1}^{3} \cot ^{2}\left(\frac{2 r-1}{7}\right)=105\)
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