Question

Assertion :$ab+bc+ca=1$, then $\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}$ is equal to $\frac{2}{\sqrt{(1+a^2)(1+b^2)(1+c^2)}}$ Reason: In a triangle $ABC\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is false but Reason is true

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Reason is correct as
$\sin 2A+\sin 2B+\sin 2C=\sin 2A+\sin 2B+\sin (2\pi -2(A+B))=\sin 2A+\sin 2B-\sin \left(2(A+B)\right)=2\sin (A+B)\cos (A-B)-2\sin (A+B)\cos (A+B)=2\sin (A+B)(\cos (A-B)-\cos (A+B))=2\sin (\pi -C)\left(2\sin A\sin B\right)=4\sin A\sin B\sin C$
Let $\cot A=a.\cot B=b,\cot C=c$
$ab+bc+ca=\cot A\cot B+\cot B\cot C+\cot C\cot A=\frac{\cos A\cos B}{\sin A\sin B}+\frac{\cos B\cos C}{\sin B\sin C}+\frac{\cos C\cos A}{\sin C\sin A}=\frac{\cos A\cos B\sin C+\sin A\cos B\cos C+\cos A\sin B\cos C}{\sin A\sin B\sin C}=\frac{\sin A\sin B\sin C-\sin (A+B+C)}{\sin A\sin B\sin C}=1$
Now
$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{\cot A}{1+{\cot }^{2}A}+\frac{\cot B}{1+{\cot }^{2}B}+\frac{\cot C}{1+{\cot }^{2}C}=\frac{\cot A}{{\cos ec}^{2}A}+\frac{\cot B}{{\cos ec}^{2}B}+\frac{\cot C}{{\cos ec}^{2}C}=\frac{1}{2}(\sin 2A+\sin 2B+\sin 2C)=2\sin A\sin B\sin C=\frac{2}{\sqrt{{\cos ec}^{2}A{\cos ec}^{2}B{\cos ec}^{2}C}}=\frac{2}{\sqrt{(1+{\cot }^{2}A)(1+{\cot }^{2}B)(1+{\cot }^{2}C)}}=\frac{2}{\sqrt{(1+a^2)(1+b^2)(1+c^2)}}$
Hence Assertion is correct