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4cos 36^@ +cot (7 1/2)^@=...

`4cos 36^@ +cot (7 1/2)^@=`

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4 cos 36^(@) + cot 7 (1^(@))/(2)=

Prove that 4 cos 36^@ +cot 7 1^@/2 = sqrt1+sqrt2 +sqrt3+ sqrt4+ sqrt5+ sqrt6 .

4cos36^(@)+cot(7(1)/(2))^(@)=

If 4 cos 36^(@) + cot(7(1^(@))/(2)) = sqrt(n_(1)) + sqrt(n_(2)) + sqrt(n_(3)) + sqrt(n_(4)) + sqrt(n_(5)) + sqrt(n_(6)) then the product of the digits in sum_(i = 1)^(6) n_(i)^(2) =

4cos36^(@)+cot7(1)/(2^(@))=

4cos36 ^ (@) + cos (7 (1) / (2)) ^ (0) =

4 cos 36^(circ)+cot (7 (1)/(2))^(circ)=

If 4cos36^(@)+cot(7(1^(@))/(2))=sqrt(n_(1))+sqrt(n_(2))+sqrt(n_(3))+sqrt(n_(4))+sqrt(n_(5))+sqrt(n_(6)) , then the value of ((Sigma_(i=1)^(6)n_(i)^(2))/(10)) is equal to