Question

Prove that tan 225 cot 405 + cot 675 tan 765 = 0

Answer 1

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Answer 2

Answer:  

Hence proved that,  

$\tan 225 ^ { \circ } \cot 405 ^ { \circ } + \cot 675 ^ { \circ } \tan 765 ^ { \circ } = 0$

To find:

Prove that  $\tan 225 ^ { \circ } \cot 405 ^ { \circ } + \cot 675 ^ { \circ } \tan 765 ^ { \circ } = 0$

Solution:

$\begin{array} { c } { \tan 225 ^ { \circ } \cot 405 ^ { \circ } + \cot 675 ^ { \circ } \tan 765 ^ { \circ } } \\\\ { = \tan \left( 180 ^ { \circ } + 45 ^ { \circ } \right) \cot \left( 2 \times 180 ^ { \circ } + 45 ^ { \circ } \right) } \\ { + \cot \left( 4 \times 180 ^ { \circ } - 45 ^ { \circ } \right) \tan \left( 4 \times 180 ^ { \circ } + 45 ^ { \circ } \right) } \\\\ { = \tan 45 ^ { \circ } \cot 45 ^ { \circ } + \cot \left( - 45 ^ { \circ } \right) \tan 45 ^ { \circ } } \end{array}$

$= 1 - \cot 45 ^ { \circ } \tan 45 ^ { \circ }$

We know that the value of $\tan 45^{\circ}=1$ and also $\cot 45^{\circ}=1$. Substituting this, we get the value as,

= 1 - 1

= 0.

Hence proved that the value of ,  

$\tan 225 ^ { \circ } \cot 405 ^ { \circ } + \cot 675 ^ { \circ } \tan 765 ^ { \circ } = 0$