TS EAMCET · Maths · Trigonometric Equations
If \(A+B+C=60^{\circ}\), then \(\cos \left(30^{\circ}-A\right)+\cos \left(30^{\circ}-B\right)+\cos \left(30^{\circ}-C\right)\) \(+\sin (A+B+C)=\)
- A \(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\)
- B \(4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\)
- C \(4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\)
- D \(4 \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}\)
Answer & Solution
Correct Answer
(A) \(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\)
Step-by-step Solution
Detailed explanation
For \(A+B+C=60^{\circ}\), then \(\left(\cos 30^{\circ}-A\right)+\cos \left(30^{\circ}-B\right)+\cos \left(30^{\circ}-C\right)\) \(+\sin (A+B+C)\) \(=2 \cos \left(30^{\circ}-\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\cos \left(30^{\circ}-C\right)\) \(+\cos 30^{\circ}\)…
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