Question

For any acute angle $\theta ,\tan (\theta -\frac{7\pi }{2}).\sec (\theta -\frac{7\pi }{2})=$

• A. $\text{cosec}\theta {\sec }^2\theta$
• B. $\sin \theta {\sec }^2\theta$
• C. $\sin \theta {\cos }^2\theta$
• D. $\text{cosec}\theta {\cos }^2\theta$

Answer 1

The correct option is B $\sin \theta {\sec }^2\theta$

Given: $\tan (\theta -\frac{7\pi }{2}).\sec (\theta -\frac{7\pi }{2})$
We have given an acute angle $\theta$ and we have to find the value of $\tan (\theta -\frac{7\pi }{2}).$
Now, we know for any angle $\varphi ,$
$\tan (-\varphi )=-\tan \varphi \cdots \left(a\right)$
Also, $\tan (\theta -\frac{7\pi }{2})=\tan \{-(\frac{7\pi }{2}-\theta )\}$
Thus, using the condition $\left(a\right):$
$\tan \{-(\frac{7\pi }{2}-\theta )\}=-\tan (\frac{7\pi }{2}-\theta )$
Since, $\theta$ is an acute angle, $\frac{7\pi }{2}-\theta$ will correpsond to an angle in the 3rd quadrant.
$\tan (\frac{7\pi }{2}-\theta )=-\tan \theta ...\left(i\right)$

Similarly, for $\sec (\theta -\frac{7\pi }{2})$
From the relation: $\sec (-\varphi )=\sec \varphi \cdots \left(b\right)$

$\sec (\theta -\frac{7\pi }{2})=\sec \{-(\frac{7\pi }{2}-\theta )\}$
Thus, using the condition $\left(b\right):$

$\sec (\theta -\frac{7\pi }{2})=\sec (\frac{7\pi }{2}-\theta )$
And, since $\frac{7\pi }{2}-\theta$ corresponds to an angle in the second quadrant, where $\sin$ of any angle is positive.

$\Rightarrow \sec (\frac{7\pi }{2}-\theta )=-\sec \theta ...\left(ii\right)$

Now, use the value of $\left(i\right),\left(ii\right)$ in given
$\tan (\theta -\frac{7\pi }{2}).\sec (\theta -\frac{7\pi }{2})=(-\tan \theta ).(-\sec \theta )$
$\Rightarrow \tan (\theta -\frac{7\pi }{2}).\sec (\theta -\frac{7\pi }{2})=\sin \theta {\sec }^2\theta$