JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

Let

f(\theta)=\frac{\cot \theta}{1-\tan \theta}+\frac{\tan \theta}{1-\cot \theta}, \pi < \theta < \frac{3 \pi}{4}

, then

f\left(\frac{9 \pi}{8}\right)

is equal to:

View answer and explanation

Correct option: B

1+2 \sqrt{2}

Option B (

1+2 \sqrt{2}

) is correct because after simplifying the trigonometric expression and substituting the given value, we obtain

1+2 \sqrt{2}

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)=