Answer
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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 $.
Complete step-by-step answer:
Given data,
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 $
We know, from the trigonometric table of tangent function,
$
{\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)
Hence, Option A is the correct answer.
Note: In order to solve these types of questions, look out for all the given trigonometric functions in the question and find out the relation between them. Using that convert one function into the other and then solve for the answer. A good knowledge in trigonometric table and its identities helps arrive at the answer faster.
Complete step-by-step answer:
Given data,
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 $
We know, from the trigonometric table of tangent function,
$
{\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)
Hence, Option A is the correct answer.
Note: In order to solve these types of questions, look out for all the given trigonometric functions in the question and find out the relation between them. Using that convert one function into the other and then solve for the answer. A good knowledge in trigonometric table and its identities helps arrive at the answer faster.
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