JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Medium

1 of 1

Application +4 | -1

Find the value of

k

for which the equation

4 \cos 36^{\circ}+\cot \left(\frac{15}{2}\right)^{\circ}=\sum\limits_{{r=1}}^k \sqrt{r}

is satisfied.

View answer and explanation

Correct option: A

Option A (6) is correct because by substituting the known values of

\cos 36^{\circ}

and

\cot (\frac{15}{2})^{\circ}

, the equation simplifies to

\sum_{r=1}^6 \sqrt{r}

, thus k=6.

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)=