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# Find the value of (x+y), if x = ${\text{tan1}}^\circ + \tan 2^\circ + .......... + \tan 45^\circ$ and y = $- \left( {{\text{cot46}}^\circ {\text{ + cot47}}^\circ {\text{ + }}........{\text{ + cot89}}^\circ } \right)$.${\text{A}}{\text{. 1}} \\ {\text{B}}{\text{. 0}} \\ {\text{C}}{\text{. - 1}} \\ {\text{D}}{\text{. }}\dfrac{{\sqrt 3 }}{2} \\$

Last updated date: 13th Jun 2024
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Hint: To find x + y, convert both x and y into the same trigonometric function/parameter using the trigonometric identities of either ${\text{tan}}\theta {\text{ or cot}}\theta$.

x = ${\text{tan1}}^\circ + \tan 2^\circ + .......... + \tan 45^\circ$ and y = $- \left( {{\text{cot46}}^\circ {\text{ + cot47}}^\circ {\text{ + }}........{\text{ + cot89}}^\circ } \right)$
⟹x + y = ${\text{tan1}}^\circ + \tan 2^\circ + .......... + \tan 45^\circ$+ ($- \left( {{\text{cot46}}^\circ {\text{ + cot47}}^\circ {\text{ + }}........{\text{ + cot89}}^\circ } \right)$)
⟹x + y = ${\text{tan1}}^\circ + \tan 2^\circ + .......... + \tan 45^\circ$- $\cot 46^\circ - \cot 47^\circ - .......... - \cot 89^\circ$
${\text{tan}}\left( {90^\circ - \theta } \right) = \cot \theta \\ \Rightarrow {\text{cot}}\left( {90^\circ - \theta } \right) = \tan \theta \\$ ---- (Put 90-θ in place of θ in the above to derive this)
⟹x + y= ${\text{tan1}}^\circ + \tan 2^\circ + .......... + \tan 45^\circ$-$\cot \left( {90 - 44} \right)^\circ - \cot \left( {90 - 43} \right)^\circ - .......... - \cot \left( {90 - 1} \right)^\circ$
⟹x + y=${\text{tan1}}^\circ + \tan 2^\circ + .......... + \tan 45^\circ$-${\text{tan}}\left( {44} \right)^\circ - \tan \left( {43} \right)^\circ - .......... - \tan \left( 1 \right)^\circ$
⟹x + y= 0 + 0 +………+ ${\text{tan45}}^\circ$
⟹x + y = 1 (from trigonometric table, ${\text{tan45}}^\circ$= 1)