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Question

Mathematics Question on Trigonometric Functions

In a triangle ABC,a[b cos C−c cos B]ABC, a[b \,cos \,C - c\, cos\, B] =

A

a2a^2

B

b2b^2

C

00

D

b2−c2b^{2}-c^{2}

Answer

b2−c2b^{2}-c^{2}

Explanation

Solution

a[bcos⁡C−ccos⁡B]a[b \cos C-c \cos B]
=(bcos⁡C+ccos⁡B)(bcos⁡C−ccos⁡B)=(b \cos C+c \cos B)(b \cos C-c \cos B)
=b2cos⁡2C−c2cos⁡2B=b^{2} \cos ^{2} C-c^{2} \cos ^{2} B
=b2(1−sin⁡2C)−c2(1−sin⁡2B)=b^{2}\left(1-\sin ^{2} C\right)-c^{2}\left(1-\sin ^{2} B\right)
=b2(1−c24R2)−c2(1−b24R2)=b^{2}\left(1-\frac{c^{2}}{4 R^{2}}\right)-c^{2}\left(1-\frac{b^{2}}{4 R^{2}}\right)
(∵sin⁡C=C2R)\left(\because \sin C=\frac{C}{2 R}\right)
=b2−b2c24R2−c2+c2b24R2=b^{2}-\frac{b^{2} c^{2}}{4 R^{2}}-c^{2}+\frac{c^{2} b^{2}}{4 R^{2}}
=b2−c2=b^{2}-c^{2}