Question

For any acute angle $\theta ,\cot (\frac{3\pi }{2}+\theta )=$, $\tan (\frac{3\pi }{2}-\theta )=$

• A. $-\tan \theta$
• B. $\cot \theta$
• C. $\tan \theta$
• D. $-\cot \theta$

Answer 1

Answer: (A), (B)

$\cot \theta$

Given an acute angle $\theta$ and we have to find the value of $\cot (\frac{3\pi }{2}+\theta )$ and
$\tan (\frac{3\pi }{2}-\theta ).$

Now, for $\cot (\frac{3\pi }{2}+\theta ),\frac{3\pi }{2}+\theta$ corresponds to an angle in the 4th quadrant, where both $\tan \&\cot$ of an angle is negative.
Also, $\tan$ and $\cot$ are co - functions of one another.
$\Rightarrow \cot (\frac{3\pi }{2}+\theta )=-\tan \theta$

Similarly, for $\tan (\frac{3\pi }{2}-\theta ),\frac{3\pi }{2}-\theta$ corresponds to an angle in the 3rd quadrant, where both $\tan \&\cot$ of an angle are positive.

$\Rightarrow \tan (\frac{3\pi }{2}-\theta )=\cot \theta$