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Question

$$\tan{{1}^{\circ}} \tan{{2}^{\circ}} \tan{{3}^{\circ}} ...........\tan{{89}^{\circ}}=$$


A
0
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B
1
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C
1
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D
2
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Solution

The correct option is B $$1$$
$$\tan{{1}^{\circ}}\tan{{2}^{\circ}}\tan{{3}^{\circ}}...\tan{{89}^{\circ}}$$

$$=\tan{{1}^{\circ}}\tan{{2}^{\circ}}\tan{{3}^{\circ}}...\tan{{87}^{\circ}}\tan{{88}^{\circ}}\tan{{89}^{\circ}}$$

$$=\tan{{1}^{\circ}}\tan{{2}^{\circ}}\tan{{3}^{\circ}}...\tan{\left({90}^{\circ}-{3}^{\circ}\right)}\tan{\left({90}^{\circ}-{2}^{\circ}\right)}\tan{\left({90}^{\circ}-{1}^{\circ}\right)}$$

$$=\tan{{1}^{\circ}}\tan{{2}^{\circ}}\tan{{3}^{\circ}}...\cot{{3}^{\circ}}\cot{{2}^{\circ}}\cot{{1}^{\circ}}$$

$$=\left(\tan{{1}^{\circ}}\cot{{1}^{\circ}}\right)\left(\tan{{2}^{\circ}}\cot{{2}^{\circ}}\right)\left(\tan{{2}^{\circ}}\cot{{2}^{\circ}}\right)...\left(\tan{{45}^{\circ}}\cot{{45}^{\circ}}\right)$$ since $$\tan{\theta}\cot{\theta}=1$$

$$=1\times 1\times 1\times ...\times 1$$

$$=1$$

Mathematics

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