JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

In a

\triangle A B C

, if

a, b

and

A

are given, then there are two triangles with third sides

c_1

and

c_2

, then

\left(c_1-c_2\right)

View answer and explanation

Correct option: C

2 \sqrt{a^2-b^2 \sin ^2 A}

Option C (

2 \sqrt{a^2-b^2 \sin ^2 A}

) is correct because it is derived from solving the quadratic equation for the third side

c

using the Law of Cosines and then finding the difference between the two possible solutions.

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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)=