If $A=130^{\circ}$ and $x=\sin A+\cos \mathrm{A}$, then

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

A=130^{\circ}

and

x=\sin A+\cos \mathrm{A}

, then

View answer and explanation

Correct option: A

x > 0

Option A (

x > 0

) is correct because

\sin A

is positive and

\cos A

is negative, but

\sin A

has a larger magnitude, resulting in a positive sum.

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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)=