JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

The value of

\cos 15^{\circ}-\sin 15^{\circ}

is equal to

View answer and explanation

Correct option: A

\frac{1}{\sqrt{2}}

Option A (

\frac{{1}}{{\sqrt{{2}}}}

) is correct because

\cos 15^{\circ} - \sin 15^{\circ}

can be simplified to

\sin 75^{\circ} - \sin 15^{\circ}

, which then simplifies to

\frac{{1}}{{\sqrt{{2}}}}

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

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\triangle A B C

C

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\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)=