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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

Prove that ta...

Question

Prove that $\tan 75 ° + c o t 75 ° = 4$ .

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Solution

To prove $\tan 75 ° + c o t 75 ° = 4$
Consider L.H.S :
$\begin{array}{rcl} \tan 75 ° + \cot 75 ° & = & \tan 75 ° + \frac{1}{\tan 75 °} \\ = \tan (30 ° + 45 °) + \frac{1}{\tan (30 ° + 45 °)} [∵ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}] \\ = & \frac{\tan 30 ° + \tan 45 °}{1 - \tan 30 ° \tan 45 °} + \frac{1}{\frac{\tan 30 ° + \tan 45 °}{1 - \tan 30 ° \tan 45 °}} \\ = & \frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}} 1} + \frac{1}{\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}} 1}} [∵ \tan 30 ° = \frac{1}{\sqrt{3}} a n d \tan 45 ° = 1] \\ = & \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} + \frac{1}{\frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}} \\ = & \frac{1 + \sqrt{3}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{1}{\frac{1 + \sqrt{3}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3} - 1}} \\ = & \frac{1 + \sqrt{3}}{\sqrt{3} - 1} + \frac{1}{\frac{1 + \sqrt{3}}{\sqrt{3} - 1}} \\ = & \frac{1 + \sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3} - 1}{1 + \sqrt{3}} \\ = & \frac{{(1 + \sqrt{3})}^{2} + {(\sqrt{3} - 1)}^{2}}{{(\sqrt{3})}^{2} - 1^{2}} [∵ {(a + b)}^{2} = a^{2} + b^{2} + 2 a b and {(a - b)}^{2} = a^{2} + b^{2} - 2 ab and (a^{2} - b^{2}) = (a + b) (a - b)] \\ = & \frac{1 + 3 + 2 \sqrt{3} + 1 + 3 - 2 \sqrt{3}}{3 - 1} \\ = & \frac{8}{2} \\ = & 4 \end{array}$
Hence, it is proved that $\tan 75 ° + c o t 75 ° = 4$

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