Question

If $\tan (20-3\alpha )=\cot (5\alpha -20)$ then find the value of $\alpha $ and hence evaluate $\sin \alpha \,\sec \alpha \cdot \,\tan \alpha -\cos ec\alpha \cos \alpha \cdot \cot \alpha $

Answer 1

In general as we know tan and cot are positive in the first and third quadrant.
So we can write $\tan \left( \dfrac{\pi }{2}-\alpha \right)=\cot \alpha $ if $\alpha $is in first quadrant$\left( 0\le \alpha \le \dfrac{\pi }{2} \right)$.
 If $\alpha $ is in third quadrant$\left( \pi \le \alpha \le \dfrac{3\pi }{2} \right)$ then we can write $\tan \left( \dfrac{3\pi }{2}-\alpha \right)=\cot \alpha $.

Complete step-by-step answer:
Given equation is $\tan (20-3\alpha )=\cot (5\alpha -20)$
If $\alpha $ is in first quadrant$\left( 0\le \alpha \le \dfrac{\pi }{2} \right)$ , then we can write $\tan \left( \dfrac{\pi }{2}-\alpha \right)=\cot \alpha $
$\tan (20-3\alpha )=\tan \left( \dfrac{\pi }{2}-(5\alpha -20) \right)$
On comparing both side
$\Rightarrow (20-3\alpha )=\left( \dfrac{\pi }{2}-(5\alpha -20) \right)$
$\Rightarrow 20-3\alpha =\dfrac{\pi }{2}-5\alpha +20$
$\Rightarrow 20-3\alpha +5\alpha -20=\dfrac{\pi }{2}$
$\Rightarrow 2\alpha =\dfrac{\pi }{2}$
.$\Rightarrow \alpha =\dfrac{\pi }{4}$
So if $\alpha $ is in first quadrant $\left( 0\le \alpha \le \dfrac{\pi }{2} \right)$, then it’s value is $\dfrac{\pi }{4}$.
. If $\alpha $ is in third quadrant $\left( \pi \le \alpha \le \dfrac{3\pi }{2} \right)$ then we can write $\tan \left( \dfrac{3\pi }{2}-\alpha \right)=\cot \alpha $
$\tan (20-3\alpha )=\tan \left( \dfrac{3\pi }{2}-(5\alpha -20) \right)$
.On comparing both side
$\Rightarrow (20-3\alpha )=\left( \dfrac{3\pi }{2}-(5\alpha -20) \right)$
$\Rightarrow 20-3\alpha =\dfrac{3\pi }{2}-5\alpha +20$
$\Rightarrow 20-3\alpha +5\alpha -20=\dfrac{3\pi }{2}$
$\Rightarrow \alpha =\dfrac{3\pi }{4}$
So if $\alpha $ is in third quadrant $\left( \pi \le \alpha \le \dfrac{3\pi }{2} \right)$ then it’s value is $\dfrac{3\pi }{4}$.
.Given expression is
$\sin \alpha \,\sec \alpha \cdot \,\tan \alpha -\cos ec\alpha \cos \alpha \cdot \cot \alpha $
As we know $\sec \alpha =\dfrac{1}{\cos \alpha },\cos ec\alpha =\dfrac{1}{\sin \alpha }$.
So we can write given expression as
$\sin \alpha \,\times \dfrac{1}{\cos \alpha }\cdot \,\tan \alpha -\dfrac{1}{\sin \alpha }\times \cos \alpha \cdot \cot \alpha $
$\,{{\tan }^{2}}\alpha -{{\cot }^{2}}\alpha $
When $\alpha =\dfrac{\pi }{4}$
$\,\Rightarrow {{\tan }^{2}}\left( \dfrac{\pi }{4} \right)-{{\cot }^{2}}\left( \dfrac{\pi }{4} \right)$
$\,\Rightarrow {{\left( 1 \right)}^{2}}-{{\left( 1 \right)}^{2}}$
. $\,\Rightarrow 0$
When $\alpha =\dfrac{3\pi }{4}$
$\,\Rightarrow {{\tan }^{2}}\left( \dfrac{3\pi }{4} \right)-{{\cot }^{2}}\left( \dfrac{3\pi }{4} \right)$
$\,\Rightarrow {{\left( 1 \right)}^{2}}-{{\left( 1 \right)}^{2}}$
 $\,\Rightarrow 0$
Hence value is 0.

Note: In the given question we need to classify range of angle to get an exact answer. As we know $\tan (\theta )$ is a periodic function with period $\pi $ . That’s why it repeats it’s value after period of $\pi $
So we can write $\tan \left( \dfrac{3\pi }{4} \right)=\tan \left( \dfrac{\pi }{4} \right)=1$