अगर $\tan \mathrm{A}+\cot \mathrm{A}=2$, तो $\tan ^{10} \mathrm{~A}+$

$\cot ^{10} \mathrm{~A}$ मूल्य है

tanA + cotA = 2 $\Rightarrow \tan \mathrm{A}+\frac{1}{\tan \mathrm{A}}=2$ $\Rightarrow \frac{\tan ^{2} \mathrm{~A}+1}{\tan \mathrm{A}}=2$ $\Rightarrow \tan ^{2} \mathrm{~A}+1=2 \tan \mathrm{A}$ $\Rightarrow \tan ^{2} \mathrm{~A}-2 \tan \mathrm{A}+1=0$ $\Rightarrow(\tan \mathrm{A}-1)^{2}=0$ $\Rightarrow \tan \mathrm{A}-1=0 \Rightarrow \tan \mathrm{A}=1$ $\Rightarrow \cot \mathrm{A}=1$ $\therefore \tan ^{10} \mathrm{~A}+\cot ^{10} \mathrm{~A}=1+1=2$ विकल्प: tan A + cot A = 2 मान लीजिए A = 45° तब, tan 45° + cot 45° = 2

अभी,$\tan ^{10} \mathrm{~A}+\cot ^{10} \mathrm{~A}$

$=\tan ^{10} 45^{\circ}+\cot ^{10} 45^{\circ}$

$=2$