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Question: If \(\tan (20-3\alpha )=\cot (5\alpha -20)\) then find the value of \(\alpha \) and hence evaluate \...

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

Explanation

Solution

In general as we know tan and cot are positive in the first and third quadrant.
So we can write tan⁡(π2−α)=cot⁡α\tan \left( \dfrac{\pi }{2}-\alpha \right)=\cot \alpha if α\alpha is in first quadrant(0≤α≤π2)\left( 0\le \alpha \le \dfrac{\pi }{2} \right).
If α\alpha is in third quadrant(π≤α≤3π2)\left( \pi \le \alpha \le \dfrac{3\pi }{2} \right) then we can write tan⁡(3π2−α)=cot⁡α\tan \left( \dfrac{3\pi }{2}-\alpha \right)=\cot \alpha .

Complete step-by-step answer:
Given equation is tan⁡(20−3α)=cot⁡(5α−20)\tan (20-3\alpha )=\cot (5\alpha -20)
If α\alpha is in first quadrant(0≤α≤π2)\left( 0\le \alpha \le \dfrac{\pi }{2} \right) , then we can write tan⁡(π2−α)=cot⁡α\tan \left( \dfrac{\pi }{2}-\alpha \right)=\cot \alpha
tan⁡(20−3α)=tan⁡(π2−(5α−20))\tan (20-3\alpha )=\tan \left( \dfrac{\pi }{2}-(5\alpha -20) \right)
On comparing both side
⇒(20−3α)=(π2−(5α−20))\Rightarrow (20-3\alpha )=\left( \dfrac{\pi }{2}-(5\alpha -20) \right)
⇒20−3α=π2−5α+20\Rightarrow 20-3\alpha =\dfrac{\pi }{2}-5\alpha +20
⇒20−3α+5α−20=π2\Rightarrow 20-3\alpha +5\alpha -20=\dfrac{\pi }{2}
⇒2α=π2\Rightarrow 2\alpha =\dfrac{\pi }{2}
.⇒α=π4\Rightarrow \alpha =\dfrac{\pi }{4}
So if α\alpha is in first quadrant (0≤α≤π2)\left( 0\le \alpha \le \dfrac{\pi }{2} \right), then it’s value is π4\dfrac{\pi }{4}.
. If α\alpha is in third quadrant (π≤α≤3π2)\left( \pi \le \alpha \le \dfrac{3\pi }{2} \right) then we can write tan⁡(3π2−α)=cot⁡α\tan \left( \dfrac{3\pi }{2}-\alpha \right)=\cot \alpha
tan⁡(20−3α)=tan⁡(3π2−(5α−20))\tan (20-3\alpha )=\tan \left( \dfrac{3\pi }{2}-(5\alpha -20) \right)
.On comparing both side
⇒(20−3α)=(3π2−(5α−20))\Rightarrow (20-3\alpha )=\left( \dfrac{3\pi }{2}-(5\alpha -20) \right)
⇒20−3α=3π2−5α+20\Rightarrow 20-3\alpha =\dfrac{3\pi }{2}-5\alpha +20
⇒20−3α+5α−20=3π2\Rightarrow 20-3\alpha +5\alpha -20=\dfrac{3\pi }{2}
⇒α=3π4\Rightarrow \alpha =\dfrac{3\pi }{4}
So if α\alpha is in third quadrant (π≤α≤3π2)\left( \pi \le \alpha \le \dfrac{3\pi }{2} \right) then it’s value is 3π4\dfrac{3\pi }{4}.
.Given expression is
sin⁡α sec⁡α⋅ tan⁡α−cos⁡ecαcos⁡α⋅cot⁡α\sin \alpha \,\sec \alpha \cdot \,\tan \alpha -\cos ec\alpha \cos \alpha \cdot \cot \alpha
As we know sec⁡α=1cos⁡α,cos⁡ecα=1sin⁡α\sec \alpha =\dfrac{1}{\cos \alpha },\cos ec\alpha =\dfrac{1}{\sin \alpha }.
So we can write given expression as
sin⁡α ×1cos⁡α⋅ tan⁡α−1sin⁡α×cos⁡α⋅cot⁡α\sin \alpha \,\times \dfrac{1}{\cos \alpha }\cdot \,\tan \alpha -\dfrac{1}{\sin \alpha }\times \cos \alpha \cdot \cot \alpha
 tan⁡2α−cot⁡2α\,{{\tan }^{2}}\alpha -{{\cot }^{2}}\alpha
When α=π4\alpha =\dfrac{\pi }{4}
 ⇒tan⁡2(π4)−cot⁡2(π4)\,\Rightarrow {{\tan }^{2}}\left( \dfrac{\pi }{4} \right)-{{\cot }^{2}}\left( \dfrac{\pi }{4} \right)
 ⇒(1)2−(1)2\,\Rightarrow {{\left( 1 \right)}^{2}}-{{\left( 1 \right)}^{2}}
.  ⇒0\,\Rightarrow 0
When α=3π4\alpha =\dfrac{3\pi }{4}
 ⇒tan⁡2(3π4)−cot⁡2(3π4)\,\Rightarrow {{\tan }^{2}}\left( \dfrac{3\pi }{4} \right)-{{\cot }^{2}}\left( \dfrac{3\pi }{4} \right)
 ⇒(1)2−(1)2\,\Rightarrow {{\left( 1 \right)}^{2}}-{{\left( 1 \right)}^{2}}
 ⇒0\,\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⁡(θ)\tan (\theta ) is a periodic function with period π\pi . That’s why it repeats it’s value after period of π\pi
So we can write tan⁡(3π4)=tan⁡(π4)=1\tan \left( \dfrac{3\pi }{4} \right)=\tan \left( \dfrac{\pi }{4} \right)=1