| If \[tan\,(20{}^\circ -3\alpha )=cot\,\text{(}5\alpha -20{}^\circ )\], then find the value of \[\alpha \] and hence evaluate: |
| \[sin\text{ }\alpha \,.\text{ }sec\text{ }\alpha \,.\text{ }tan\text{ }\alpha -cosec\text{ }\alpha \text{ }.\text{ }cos\text{ }\alpha \text{ }.\text{ }cot\text{ }\alpha \]. |
Answer:
\[tan\,(20{}^\circ -3\alpha )=cot\,\text{(}5\alpha -20{}^\circ )\] \[\Rightarrow \tan \,(20{}^\circ -3\alpha )=\tan [90-(5\alpha -20{}^\circ )]\] \[[\because \,\,\cot \theta =\tan (90{}^\circ -\theta )]\] \[\Rightarrow 20{}^\circ -3\alpha =90{}^\circ -5\alpha +20{}^\circ \] \[\Rightarrow -3\alpha +5\alpha =90{}^\circ +20{}^\circ -20{}^\circ \] \[\Rightarrow 2\alpha =90{}^\circ \] \[\Rightarrow \alpha =45{}^\circ \] Now, \[sin\text{ }\alpha \,.\text{ }sec\text{ }\alpha \,.\text{ }tan\text{ }\alpha -cosec\text{ }\alpha \text{ }.\text{ }cos\text{ }\alpha \text{ }.\text{ }cot\text{ }\alpha \] \[=sin\text{ }45{}^\circ .\,sec\,45{}^\circ .\,tan\text{ }45{}^\circ -cosec\text{ }45{}^\circ .\cos \text{ }45{}^\circ .\,cot\text{ }45{}^\circ \] \[=\frac{1}{\sqrt{2}}\times \sqrt{2}\times 1-\sqrt{2}\times \frac{1}{\sqrt{2}}\times 1\] \[=1-1=0\]
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