Question

If $cosx-sin\alpha cot\beta sinx=cos\alpha$, then the value of $tan\left(x/2\right)$ is

• A. $-tan\left(\alpha /2\right)cot\left(\beta /2\right)$
• B. $tan\left(\alpha /2\right)tan\left(\beta /2\right)$
• C. $-cot\left(\alpha /2\right)tam\left(\beta /2\right)$
• D. $cot\left(\alpha /2\right)cot\left(\beta /2\right)$

Answer 1

Answer: (A), (B)

The correct options are
A $-tan\left(\alpha /2\right)cot\left(\beta /2\right)$
B $tan\left(\alpha /2\right)tan\left(\beta /2\right)$

$cosx-sin\alpha cot\beta sinx=cos\alpha \Rightarrow \frac{1-ta{n}^2\left(\frac{x}{2}\right)}{1+ta{n}^2\left(\frac{x}{2}\right)}-sin\alpha cot\beta \frac{2tan\left(\frac{x}{2}\right)}{1+ta{n}^2\left(\frac{x}{2}\right)}=cos\alpha \Rightarrow ta{n}^2\left(\frac{x}{2}\right)(1+cos\alpha )+sin\alpha cot\beta \cdot 2tan\left(\frac{x}{2}\right)-(1-cos\alpha )=0\Rightarrow ta{n}^2\left(\frac{x}{2}\right)+\frac{2sin\alpha cot\beta }{1+cos\alpha }tan\left(\frac{x}{2}\right)-\frac{1-cos\alpha }{1+cos\alpha }=0\Rightarrow ta{n}^2\left(\frac{x}{2}\right)+2tan\left(\frac{\alpha }{2}\right)cot\beta tan\left(\frac{x}{2}\right)-ta{n}^2\left(\frac{\alpha }{2}\right)=0\Rightarrow ta{n}^2\left(\frac{x}{2}\right)+2tan\left(\frac{\alpha }{2}\right)\cdot \frac{1}{2}(cot\left(\frac{\beta }{2}\right)-tan\left(\frac{\beta }{2}\right))tan\left(\frac{x}{2}\right)-ta{n}^2\left(\frac{\alpha }{2}\right)=0\Rightarrow (tan\left(\frac{x}{2}\right)+cot\left(\frac{\beta }{2}\right)tan\left(\frac{\alpha }{2}\right))(tan\left(\frac{x}{2}\right)-tan\left(\frac{\beta }{2}\right)tan\left(\frac{\alpha }{2}\right))=0$
$\Rightarrow tan\left(\frac{x}{2}\right)=-tan\left(\frac{\alpha }{2}\right)cot\left(\frac{x}{2}\right)$ or $tan\left(\frac{x}{2}\right)=tan\left(\frac{\alpha }{2}\right)tan\left(\frac{x}{2}\right)$
Hence, options 'A' and 'B' are correct.