Question

# If $$\cot^{-1} (\displaystyle \frac{1-x^{2}}{2x})+\cos^{-1} (\displaystyle \frac{1-x^{2}}{1+x^{2}})=\frac{2\pi}{3}, x>0$$ , $$x\neq 1$$ then $$x=$$

A
12
B
±12
C
±13
D
13

Solution

## The correct option is D $$\displaystyle \frac{1}{\sqrt{3}}$$$$\displaystyle cot^{-1}(\frac{1-x^{2}}{2x})=\frac{2\pi }{3}-cos ^{-1}(\frac{1-x^{2}}{1+x^{2}})$$$$\displaystyle cos ^{-1}(\frac{1-x^{2}}{1+x^{2}})=\frac{2\pi }{3}-cot^{-1}(\frac{1-x^{2}}{2x})$$take cos on both sidesSimplify, $$\displaystyle \frac{1-x^{2}}{1+x^{2}}=-\frac{1}{2}(\frac{1-x^{2}}{1+x^{2}})+\frac{\sqrt{3}}{2} \frac{2x}{1+x^{2}} \quad [cos(A-B)=cos A cos B+sin A sin B]$$$$\displaystyle \frac{2-2x^{2}-2\sqrt{3}x}{2(1+x^{2})}=-\frac{1}{2}\frac{(1-x^{2})}{(1+x^{2})}$$$$2-2x^{2}-2\sqrt{3}x=x^{2}-1$$$$3x^{2}+2\sqrt{3}x-3=0$$$$\therefore x=\displaystyle \frac{-2\sqrt{3}\pm \sqrt{12+36}}{6}$$$$\displaystyle x=\frac{-2\sqrt{3}\pm 4\sqrt{3}}{6}$$$$\displaystyle x=-\sqrt{3} , \frac{\sqrt{3}}{3}$$$$\displaystyle x=-\sqrt{3} , \frac{1}{\sqrt{3}}$$Mathematics

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