Question

Given sec θ = 13/12 Calculate all other trigonometric ratios

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Answer 1

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

Let us assume a right angled triangle ABC, right angled at B

sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC2=AB2 + BC2

Substitute the value of AB and AC

(13k)2= (12k)2 + BC2

169k2= 144k2 + BC2

169k2= 144k2 + BC2

BC2 = 169k2 – 144k2

BC2= 25k2

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5

Answer 2

sec theta=1/cos =hypo/adj

We know that, sec theta=13/12

Let AC be=13k and AB=12k

BC2 = AC2 -AB2

BC2=169k2 - 144k2

BC=√25k=5k.

sin theta=side opposite /hypo

=5k/13k=5/13

cos theta=adj/hypo

=12k/13k=12/13

tan theta=sin/cos=side opposite/adj

=5k/12k=5/12

cosec theta=1/sin=hypo/sin

=13k/5k=13/5

cot theta=1/tan=cos/sin

=12k/5k=12/5