# Cot 16. Cot 44+cot 44.cot76 - cot 76.cot16 =Trigonometry problem

Arun
25757 Points
4 years ago
Dear student

E = cot(76)cot(44) + cot(44)cot(16) - cot(76)cot(16)

First, use the formula, cot(A) = tan(90 - A). Then we have :

E = tan(14)tan(46) + tan(46)tan(74) - tan(14)tan(74)

Rearrange this so the larger numbers are first :
E = tan(46)tan(14) + tan(74)tan(46) - tan(74)tan(14)

Now we need these two formulae :

tan(A + B) = [tan(A) + tan(B)]/[1 - tan(A)tan(B)]
which rearranges to : tan(A)tan(B) = 1 - [tan(A) + tan(B)]/tan(A + B) .......... (1)
and
tan(A - B) = [tan(A) - tan(B)] / [1 + tan(A)tan(B)]
which rearranges to : tan(A)tan(B) = [tan(A) - tan(B)]/tan(A - B) - 1 ............ (2)

For the 1st expression of E, use (1) : tan(46)tan(14) = 1 - [tan(46) + tan(14)]/tan(60)
But tan(60) = √3, so,
tan(46)tan(14) = 1 - [tan(46 + tan(14)]/√3 = 1 - tan(46)/√3 - tan(14)/√3 ........... (3)

For the 2st expression of E, use (1) : tan(74)tan(46) = 1 - [tan(74) + tan(46)]/tan(120)
But tan(120) = -√3, so,
tan(74)tan(46) = 1 - [tan(74) + tan(46)]/(-√3) = 1 + tan(74)/√3 + tan(46)/√3 ..... (4)

For the 3st expression of E, use (2) : -tan(74)tan(14) = -{[tan(74) - tan(14)]/tan(60) - 1}
But tan(60) = √3, so,
-tan(74)tan(14) = -{[tan(74) - tan(14)]/√3 - 1} = -tan(74)/√3 + tan(14)/√3 + 1 .... (5)

Now add, (3) + (4) + (5), which gives the answer, E = 3.

Regards