Question

prove that cot16cot44 + cot44cot76 - cot76cot16 = 3

(all are degree)

Answer 1

cot16°cot44°+cot44°cot76°-cot76°cot16°
=cot(90°-74°)cot(90°-46°)+cot(90°-46°)cot(90°-14°)-cot(90°-14°)cot(90°-74°)
=tan74°tan46°+tan46°tan14°-tan14°tan74°
Now, tan(A+B)=(tanA+tanB)/(1-tanAtanB)
or, 1-tanAtanB=tanA+tanB/tan(A+B)
or, -tanAtanB=tanA+tanB/tan(A+B)-1
or, tanAtanB=1-[(tanA+tanB)/tan(A+B)]
∴, tan74°tan46°
=1-(tan74°+tan46°)/tan(74°+46°)
=1-(tan74°+tan46°)/tan120°
=1-(tan74°+tan46°)/tan{(90°×1)+30°}
=1-(tan74°+tan46°)/-cot30°
=1-tan74°/(-√3)-tan46°/(-√3)
=1+tan74°/√3+tan46°/√3 -------------(1)
Similarly, tan46°tan14°
=1-(tan46°+tan14°)/tan60°
=1-tan46°/√3-tan14°/√3 ---------------(2)
Again, tan(A-B)=(tanA-tanB)/(1+tanAtanB)
or, 1+tanAtanB=(tanA-tanB)/tan(A-B)
or, tanAtanB=[(tanA-tanB)/tan(A-B)]-1
∴, tan14°tan74°
=[(tan14°-tan74°)/tan(-60°)]-1
=-[(tan14°-tan74°)/√3]-1
=-tan14°/√3+tan74°/√3-1
∴,

Answer 2

Answer:

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)    eq(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 .    eq(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.

 MARK BRAINLIEST