Find the value of $$\cot\frac{\pi}{32}-\tan\frac{\pi}{32}-2\cot\frac{\pi}{16}$$

Expression : $$\cot\frac{\pi}{32}-\tan\frac{\pi}{32}-2\cot\frac{\pi}{16}$$

= $$(\frac{cos\frac{\pi}{32}}{sin\frac{\pi}{32}}-\frac{sin\frac{\pi}{32}}{cos\frac{\pi}{32}})-2cot\frac{\pi}{16}$$

= $$(\frac{cos^2\frac{\pi}{32}-sin^2\frac{\pi}{32}}{sin\frac{\pi}{32}\times cos\frac{\pi}{32}})-2cot\frac{\pi}{16}$$

$$\because cos^2\theta-sin^2\theta=cos2\theta$$ and multiplying and divide by 2,

= $$(\frac{2cos\frac{\pi}{16}}{2\sin\frac{\pi}{32}.cos\frac{\pi}{32}})-2cot\frac{\pi}{16}$$

Also, $$2sin\theta.cos\theta=sin2\theta$$

= $$(\frac{2cos\frac{\pi}{16}}{\sin\frac{\pi}{16}})-2cot\frac{\pi}{16}$$

= $$2cot\frac{\pi}{16}-2cot\frac{\pi}{16}=0$$

=> Ans - (B)