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JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard

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Application +4 | -1

If

x=\sin \theta|\sin \theta|, y=\cos \theta|\cos \theta|

, where

\frac{99 \pi}{2} \leq \theta \leq 50 \pi

, then

View answer and explanation

Correct option: D

y-x=1

Option D (

y-x=1

) is correct because within the specified range of

\theta

, x becomes

-\sin^2 \theta

and y becomes

\cos^2 \theta

, leading to

y-x = \cos^2 \theta + \sin^2 \theta = 1

.

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

\alpha=6 \beta

\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}

\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

\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)=