Question

The value of ${(cot\frac{x}{2}-tan\frac{x}{2})}^2$ (1-2 tan x cot 2x) is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D

4

We have,

${(cot\frac{x}{2}-tan\frac{x}{2})}^2(1-2tanxcot2x)(co{t}^2\frac{x}{2}-2cot\frac{x}{2}tan\frac{x}{2}+ta{n}^2\frac{x}{2})\{1-2tanx\left(\frac{co{t}^{2}x-1}{2cotx}\right)\}$

$(co{t}^2\frac{x}{2}-2+ta{n}^2\frac{x}{2})\{1-tanx\left(\frac{co{t}^{2}x-1}{cotx}\right)\}(co{t}^2\frac{x}{2}+ta{n}^2\frac{x}{2}-2)(1-\frac{cotx-tanx}{cotx})(co{t}^2\frac{x}{2}+ta{n}^2\frac{x}{2}-1)\left(ta{n}^{2}x\right)(co{t}^2\frac{x}{2}+ta{n}^2\frac{x}{2}-2){\left(\frac{2tan\frac{x}{2}}{1-ta{n}^2\frac{x}{9}}\right)}^2$

$=\frac{1}{1-ta{n}^2{\left(\frac{x}{2}\right)}^2}(4+4ta{n}^4\frac{x}{2}-8ta{n}^2\frac{x}{2})=\frac{1}{{(1-ta{n}^2\frac{x}{2})}^2}(4-8ta{n}^2\frac{x}{2}+4ta{n}^2\frac{x}{2})=\frac{4}{{(1-ta{n}^2\frac{x}{2})}^2}\{{\left(ta{n}^2\frac{x}{2}\right)}^2-2(ta{n}^2\frac{x}{2}+1)\}=\frac{4{(ta{n}^2\frac{x}{2}-1)}^2}{{(1-ta{n}^2\frac{x}{2})}^2}$
= 4