If sin^2 theta = frac{x^2 + y^2 + 1}{2x},then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that for any real $	heta$,$0 \le \sin^2 	heta \le 1$.Given $\sin^2 	heta = \frac{x^2 + y^2 + 1}{2x}$,we must have $\frac{x^2 + y^2 + 1}

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) We know that for any real $	heta$,$0 \le \sin^2 	heta \le 1$.Given $\sin^2 	heta = \frac{x^2 + y^2 + 1}{2x}$,we must have $\frac{x^2 + y^2 + 1}{2x} \le 1$.Assuming $x > 0$,we get $x^2 + y^2 + 1 \le 2x$,which simplifies to $(x - 1)^2 + y^2 \le 0$.Since squares of real numbers are non-negative,this inequality holds only if $(x - 1)^2 = 0$ and $y^2 = 0$,which implies $x = 1$ and $y = 0$.If $x Since the value of $x$ depends on $y$ (specifically $y$ must be $0$ for $x=1$),$x$ is not uniquely determined as a constant independent of $y$ in all cases,or rather,the condition requires $y=0$ and $x=1$. Given the options,$x=1$ is a specific case,but since $y$ is not defined,the most accurate choice is $D$.

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

Similar Questions

If $\left| a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta - \frac{1}{2}(a + c) \right| \le \frac{1}{2}k,$ then $k^2$ is equal to

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

In triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C =$

Find the maximum and minimum values of the function given by $f(x) = |\sin 4x + 3|$.

Assertion $(A)$: If $A=10^{\circ}, B=16^{\circ}, C=19^{\circ}$,then $\tan 2A \tan 2B + \tan 2B \tan 2C + \tan 2C \tan 2A = 1$.
Reason $(R)$: If $A+B+C=90^{\circ}$,then $\tan A \tan B + \tan B \tan C + \tan C \tan A = 1$.
Which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module