JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard

1 of 1

Application +4 | -1

The minimum value of the function

f(x)=3 \tan x+4 \cot x

where

x \in\left(0, \frac{\pi}{2}\right)

View answer and explanation

Correct option: B

2 \sqrt{12}

Option B (2 sqrt(12)) is correct because the minimum value of the function can be found using AM >= GM inequality, which gives 3tan(x) + 4cot(x) >= 2*sqrt(3tan(x)*4cot(x)) = 2*sqrt(12).

More Questions from TRIGONOMETRY RATIOS

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

, where

\alpha=6 \beta

and

\alpha

being an acute angle, then

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

is equal to The value of

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

is If the

\triangle A B C

is acute angled at

C

, then The value of the expression

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

equals

\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)=