Question Number 211575 by BaliramKumar last updated on 13/Sep/24
Answered by som(math1967) last updated on 13/Sep/24
$$\:\left(\boldsymbol{{a}}\right)\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{m}}+\boldsymbol{{n}}+\frac{\mathrm{1}}{\boldsymbol{{m}}}+\frac{\mathrm{1}}{\boldsymbol{{n}}}\right) \\ $$
Answered by A5T last updated on 13/Sep/24
$${sec}^{\mathrm{2}} \theta−{tan}^{\mathrm{2}} \theta=\left({sec}\theta−{tan}\theta\right)\left({sec}\theta+{tan}\theta\right)=\mathrm{1} \\ $$$$\Rightarrow{sec}\theta+{tan}\theta=\frac{\mathrm{1}}{{n}} \\ $$$${cosec}^{\mathrm{2}} \theta−{cot}^{\mathrm{2}} \theta=\left({cosec}\theta−{cot}\theta\right)\left({cosec}\theta+{cot}\theta\right)=\mathrm{1} \\ $$$$\Rightarrow{cosec}\theta+{cot}\theta=\frac{\mathrm{1}}{{m}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left({m}+{n}+\frac{\mathrm{1}}{{m}}+\frac{\mathrm{1}}{{n}}\right)={cosec}\theta+{sec}\theta\Rightarrow\left({a}\right) \\ $$