JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\sin (x-y)=\cos (x+y)=1 / 2

then

x

and

y

lying between

0^{\circ}

and

90^{\circ}

are given by -

View answer and explanation

Correct option: D

x=45^{\circ}, y=15^{\circ}

Option D (

x=45^{\circ}, y=15^{\circ}

) is correct because solving the system of equations

\sin (x-y) = 1/2

and

\cos (x+y) = 1/2

yields these values for x and y within the given range.

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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)=