Given that cot A = 1/t and that A is acute, find in terms of t,?

Question

(i) cos (-A)

Please show workings so that I can understand.

Answer 1

cot A=1/t
therefore, 1/tan A =1/t
tan A= t
Drawing a right angled triangle with an angle A,adjacent side to angle 1 and opposite side to angle t will give the hypotenuse through pythagoras' theorem
threfore, Using pythagoras' theorem
hypotenuse=square root(t*2 +1)

Note: t*2 is t to the power of two

cos A=adjacent/hypotenuse
= 1/[square root(t*2 +1)]

cos (-A) = cos(360-A)
=cos A
=1/[square root(t*2 +1)]

Answer 2

Note below square (t) is written as t^ 2

Method 1.
Cot A = 1/t implies Cot A =1/tan A= 1/t {by definition Cot A = Cos A/Sin A)
Hence tan A = t.
(Here A is acute or 0 0 as all ratios are positive in 1st quadrant)
Now tan A = t or SinA/Cos A = t as Sin A /Cos A = tan A by definition
or Sin A = t Cos A
Sin^2 A + Cos^2 A = 1 can be used to find Cos A
Substituting for Sin A
(t Cos A) ^2 + (Cos A)^2 = 1
(t^2 + 1) Cos^2A = 1
Cos A = 1/( t^ 2+ 1)
Considering A is acute Cos A is positive
(- A) is nothing but angle A clockwise from positive X axis (zero deg) (-A) is in quadrant IV and Cos is positive in this quadrant IV
Cos A = Cos (-A) because In a circle drawn with unit value of radius and origin as center of circle the x cordinate at the point where angle meets the circle is the cosine or cos value of the angle.

Method 2.
Draw a Right angle traingle ABC right angled at B
By definition Cot A = [length adjacent side / length opposite side]
In our case this will be side(AB)/side(BC)
Consider side AB = 1 and side BC =t
(To be mathematically correct it should be a the two sides should be multiple of t say x and t. x but above is accepted)
Using Phytoghoras theorem we can find AC or hypotenuse as [1+ t^2]
Now Cos A is by definition Adjacent side / hypotenuse
which gives Cos A = [Side (AB) / Side AC] = 1/ t^2 + 1
as mentioned above the value of cos (- A) = Cos (A)=1/ (t^2 + 1)
Subhash

Answer 3

cos (-A) = cos A for any real A as the cosine is symmetrical about A=0
then the real question is to find cos A.
If cot A = cos A / sin A = 1/t
imagine a triangle with angle A and the side opposite A with length at ,for convenience take a=1 since this is the case to which all others reduce to, and adjacent with length a
then the hypotenuse has length at a=1 of V(1+t^2)
hence cos A = 1/[V(1+t^2)]

Answer 4

If cot A = 1/t, in a right angle triangle, cot A = Adjacent side/Opposite side.
Hence length of adjacent side is 1 and opposite side is t.

We know cos(-A) = -COS A

Then cos A = adjacent side/Hypotenus

= 1/root of 1+ t^2

Then -cosA = cos (-A) = -1/root of 1+ t^2