Q55 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
With usual notations, in $\triangle ABC$, $(b - c)^2 \cos^2 \dfrac{A}{2} + (b + c)^2 \sin^2 \dfrac{A}{2} =$
- Correct answerA$a^2$
- B$b^2 - c^2$
- C$a^2 + b^2 + c^2$
- D$0$
Solution
Expanding the given expression:
$(b^2 + c^2 - 2bc) \cos^2 \dfrac{A}{2} + (b^2 + c^2 + 2bc) \sin^2 \dfrac{A}{2}$
Grouping the terms with $(b^2 + c^2)$ and $2bc$:
$(b^2 + c^2)(\cos^2 \dfrac{A}{2} + \sin^2 \dfrac{A}{2}) - 2bc(\cos^2 \dfrac{A}{2} - \sin^2 \dfrac{A}{2})$
Using the trigonometric identities $\cos^2 \theta + \sin^2 \theta = 1$ and $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$:
$b^2 + c^2 - 2bc \cos A$
By the cosine rule in $\triangle ABC$, $a^2 = b^2 + c^2 - 2bc \cos A$.
Thus, the given expression is equal to $a^2$.
Answer: $a^2$
$(b^2 + c^2 - 2bc) \cos^2 \dfrac{A}{2} + (b^2 + c^2 + 2bc) \sin^2 \dfrac{A}{2}$
Grouping the terms with $(b^2 + c^2)$ and $2bc$:
$(b^2 + c^2)(\cos^2 \dfrac{A}{2} + \sin^2 \dfrac{A}{2}) - 2bc(\cos^2 \dfrac{A}{2} - \sin^2 \dfrac{A}{2})$
Using the trigonometric identities $\cos^2 \theta + \sin^2 \theta = 1$ and $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$:
$b^2 + c^2 - 2bc \cos A$
By the cosine rule in $\triangle ABC$, $a^2 = b^2 + c^2 - 2bc \cos A$.
Thus, the given expression is equal to $a^2$.
Answer: $a^2$
Hence, the correct answer is option: 'A'.