Question

$\tan {69}^o+\tan {66}^o-\tan {69}^o\tan {66}^o=2k$, then $k=?$

• A. $-1$
• B. $1/2$
• C. $-1/2$
• D. None of these

Answer 1

Answer: (C)

$-1/2$

Let If put $k=-\frac{1}{2}$ given function.

Then,

$\tan {69}^0+\tan {66}^0-\tan {69}^0\tan {66}^0=2k$

$\Rightarrow \tan {69}^0+\tan {66}^0-\tan {69}^0\tan {66}^0=2\times (-\frac{1}{2})$

$\Rightarrow \tan {69}^0+\tan {66}^0-\tan {69}^0\tan {66}^0=-1$

$\Rightarrow \tan {69}^0+\tan {66}^0=-1+\tan {69}^0\tan {66}^0$

$\Rightarrow \frac{\tan {69}^0+\tan {66}^0}{1-\tan {69}^0\tan {66}^0}=-1$

$\Rightarrow \tan ({69}^0+{66}^0)=-1$

$\Rightarrow \tan {135}^0=-1$

$\Rightarrow \tan ({180}^0-{45}^0)=-1$

$\Rightarrow -\tan {45}^0=-1$

$\Rightarrow -1=-1$

Hence proved.

Then $k=-\frac{1}{2}$ is the answer.

Option $C$ is correct.