নিচের অপশন গুলা দেখুন
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1 উত্তর
সঠিক উত্তর হচ্ছে:
ব্যাখ্যা:
ব্যাখ্যা:
$${\text{ tan 1}}{5^ \circ }{\text{cot 7}}{5^ \circ } + {\text{tan 7}}{5^ \circ }{\text{cot 1}}{5^ \circ }$$
$$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$ $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$ $${\text{cot1}}{5^ \circ }$$
$$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$
$$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$ $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$ $${\text{cot1}}{5^ \circ }$$
$$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$