Question Number 93952 by oustmuchiya@gmail.com last updated on 16/May/20
$${Express}\:{the}\:{following}\:{as}\:{functions} \\ $$$${of}\:\boldsymbol{{A}}: \\ $$$$\left(\mathrm{i}\right)\:{sec}\left({A}−\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:\:\left(\mathrm{ii}\right)\:{cosec}\left({A}−\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{iii}\right)\:{tan}\left({A}−\frac{\mathrm{3}\pi_{} }{\mathrm{2}}\right)\:\:\left(\mathrm{iv}\right)\:{cos}\left(\mathrm{720}°+{A}\right) \\ $$$$\left(\mathrm{v}\right)\:{tan}\:\left({A}+\pi\right) \\ $$
Answered by Kunal12588 last updated on 16/May/20
$${I}\:{think}\:{they}\:{already}\:{are}\:{functions}\:{of}\:{A} \\ $$$${anyway}\:{we}\:{can}\:{simplify} \\ $$$$\mathrm{1}\rangle\:{sec}\left(\mathrm{270}°−{A}\right)={cosec}\:{A} \\ $$$$\mathrm{2}\rangle\:−{cosec}\left(\mathrm{90}°−{A}\right)=−{sec}\:{A} \\ $$$$\mathrm{3}\rangle\:−{tan}\left(\mathrm{270}°−{A}\right)=−{cot}\:{A} \\ $$$$\mathrm{4}\rangle\:{cos}\left(\mathrm{360}°×\mathrm{2}+{A}\right)={cos}\:{A} \\ $$$$\mathrm{5}\rangle\:{tan}\left(\mathrm{180}°+{A}\right)=−{tan}\:{A} \\ $$
Commented by oustmuchiya@gmail.com last updated on 16/May/20
$${A}\:{million}\:{thanks}\:{to}\:{you}\:{indeed} \\ $$
Commented by Kunal12588 last updated on 16/May/20
well the last one is wrong tan(180°+A)=tan A anyway a piece of advice :- read a lot!