3.Trigonometrical Ratios, Functions and Identities
easy

If ${\sin ^2}\theta = \frac{{{x^2} + {y^2} + 1}}{{2x}}$, then $x$ must be

A

$-3$

B

$-2$

C

$1$

D

None of these

Solution

(d) ${\sin ^2}\theta \le 1$ 

$\therefore$ $\frac{{{x^2} + {y^2} + 1}}{{2x}} \le 1$

$\Rightarrow$ ${x^2} + {y^2} – 2x + 1 \le 0$.

$\Rightarrow$ ${(x – 1)^2} + {y^2} \le 0$

It is possible, if $x = 1$ and $y = 0$, 

$i.e.$, It also depends on value of $y$.

Hence, option $(d)$ is correct.

Standard 11
Mathematics

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