JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Analyzing +4 | -1

x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}{{\tan }^{2n}}\theta }

and

y = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta }

for

0 < \theta < \frac{\pi }{4}

then

View answer and explanation

Correct option: B

y(1 - x) = 1

Option B (

y(1 - x) = 1

) is correct because by simplifying the given infinite sums for x and y, we can derive the relationship

y(1-x) = 1

More Questions from TRIGONOMETRY RATIOS

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)=