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If X= 1^{0}+ 2^{0}+........+ 45^{0} and y= -( 46^{0}+ 47^{0}

Question

If $X = tan 1^{0} + tan 2^{0} + ........ + tan 4 5^{0}$ and $y = - (cot 4 6^{0} + cot 4 7^{0} + ....... + cot 8 9^{0})$ then find the value of $(x + y)$ .

1

0

- 1

\frac{3}{2}

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Oliver Discussed

X = tan 1^{0} + tan 2^{0} + ........ + tan 4 5^{0}

and

y = - (cot 4 6^{0} + cot 4 7^{0} + ....... + cot 8 9^{0})

then find the value of

(x + y)

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Since

tan (9 0^{o} - x) = cot 9 0^{0}

we have

x + y = tan 1^{o} + tan 2^{o} + ....... + tan 4 5^{0} - (cot 4 6^{0} + cot 4 7^{0} + ... + cot 8 9^{0})

=> x + y = tan 1^{o} + tan 2^{o} + .... + tan 4 5^{0} - (cot (9 0^{0} - 4 4^{0}) + cot (9 0^{0} - 4 3^{0}) + .. + cot (9 0^{0} - 1^{0})

=> x + y = tan 1^{o} + tan 2^{o} + .... + tan 4 5^{0} - (tan 4 4^{0} + tan 4 3^{0} + .. + tan 1^{0})

=> x + y = 0 + 0 + .. + tan 4 5^{0} = 1

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Question Text	If $X = tan 1^{0} + tan 2^{0} + ........ + tan 4 5^{0}$ and $y = - (cot 4 6^{0} + cot 4 7^{0} + ....... + cot 8 9^{0})$ then find the value of $(x + y)$ .
Answer Type	Text solution:1
Upvotes	150