JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

\theta

eliminant between the equations

x=a \sec ^3 \theta, y=b \tan ^3 \theta

View answer and explanation

Correct option: B

\left(\frac{x}{a}\right)^{2 / 3}-\left(\frac{y}{b}\right)^{2 / 3}=1

Option B (

\left(\frac{{x}}{{a}}\right)^{2 / 3}-\left(\frac{{y}}{{b}}\right)^{2 / 3}=1

) is correct because it is derived by expressing

\sec \theta

and

\tan \theta

in terms of x, y, a, and b, and then using the identity

\sec^2 \theta - \tan^2 \theta = 1

More Questions from TRIGONOMETRY RATIOS

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\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)=