Question

when sec$\theta$ = 13/12,then find all trigonometric ratios. ​

Answer 1

SOLUTION:-

______________________

=> In ΔABC,

=> <ABC= 90° , <ACB = θ

=> sec θ = AC/BC

=> .°. sec θ= 13/12

=> .°. AC/BC = 13/12

=> .°. AC = 13k and BC = 12k

=> in Right Δ ABC,

=> AC² = AB²+BC² -----> ( According to Pythagoras theorem)

=> (13k) ² = (AB)² + (12k)²

=> 169k² = AB² + 144k²

=> AB² = 169k² - 144k²

=> AB²= 25k²

=> √(AB)² = √(25k²)------> ( take square root On both side)

=> AB = 25k.

_______________________

=> Now, find all trigonometric ratios ,

=> (1) sinθ = AB/AC = 5k/13k = 5/13.

=> (2) cosθ = BC/AC = 12k/13k = 12/13.

=> (3) tanθ = AB/BC = 5k/12k = 5/12.

=> (4) cosecθ = AC/AB = 13k/5k = 13/5.

=> (5) secθ = AC/BC = 13k/12k = 13/12.

=> (6) cotθ = BC/AB = 12k/5k = 12/5.

________________________

Answer:-

=> sinθ = 5/13 , cosθ = 12/13 , tanθ = 5/12 ,cosecθ= 13/5 , secθ = 13/12 , cotθ = 12/5.

________________________

Answer 2

Solution :

$\sf \sec \theta = \frac{13}{12} = \frac{ac}{bc} \\ \\ \sf ac = 13 \\ \sf bc = 12$

Using Pythagoras theorem :

$\sf {ac}^{2} = {ab}^{2} + {bc}^{2} \sf \\ \\ \sf {13}^{2} = {ab}^{2} + {12}^{2} \\ \\ \sf 169 = {ab}^{2} + 144 \\ \\ \sf ab = \sqrt{25} \\ \\ \sf ab = 5$

___________________

$\sf \sin \theta = \frac{ab}{ac} \\ \\ \boxed{ \sf \sin \theta = \frac{5}{13}}$

___________________

$\sf \cos \theta = \frac{bc}{ac} \\ \\ \boxed{\sf \cos \theta = \frac{12}{13} }$

___________________

$\sf \tan \theta = \frac{ab}{bc} \\ \\ \boxed{\sf \tan \theta = \frac{5}{12}}$

___________________

$\sf \cosec \theta = \frac{ac}{ab} \\ \\ \boxed{ \sf \cosec \theta = \frac{13}{5} }$

___________________

$\sf \cot \theta = \frac{bc}{ab} \\ \\ \boxed{\sf \cot \theta = \frac{12}{5}}$