Question

If $ta{n}^{-1}\frac{a}{x}+ta{n}^{-1}\frac{b}{x}+ta{n}^{-1}\frac{c}{x}+ta{n}^{-1}\frac{d}{x}=\frac{\pi }{2}$ then $x^4-x^2\sum ab+abcd$ is equal to

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

0

$ta{n}^{-1}\frac{a}{x}+ta{n}^{-1}\frac{b}{x}+ta{n}^{-1}\frac{c}{x}+ta{n}^{-1}\frac{d}{x}=\frac{\pi }{2}\Rightarrow ta{n}^{-1}\left(\frac{(a+b)x}{x^2-ab}\right)+ta{n}^{-1}\left(\frac{(c+d)x}{x^2-cd}\right)=\frac{\pi }{2}\Rightarrow ta{n}^{-1}\left(\frac{(a+b)x}{x^2-ab}\right)=\frac{\pi }{2}-ta{n}^{-1}\left(\frac{(c+d)x}{x^2-cd}\right)\Rightarrow ta{n}^{-1}\left(\frac{(a+b)x}{x^2-ab}\right)=co{t}^{-1}\left(\frac{(c+d)x}{x^2-cd}\right)\Rightarrow ta{n}^{-1}\left(\frac{(a+b)x}{x^2-ab}\right)=ta{n}^{-1}\left(\frac{x^2-cd}{(c+d)x}\right)\Rightarrow \frac{(a+b)x}{x^2-ab}=\frac{x^2-cd}{(c+d)x}\Rightarrow x^4-x^2\sum ab+abcd=0$