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

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Medium

1 of 1

Application +4 | -1

\sin ^2 \theta=\frac{x^2+y^2+1}{2 x}

, then

x

must be

View answer and explanation

Correct option: C

Option C (1) is correct because the given equation implies that

x

must be 1 to satisfy the condition

0 \leq \sin^2 \theta \leq 1

Practice All Chapter MCQs →

\tan \alpha=\frac{p}{q}

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=