Question

If 13 cos 0 = 5, find the value of tan square theata- sec square theata​

Answer 1

Answer:

-5

Step-by-step explanation:

Cos = Base/ Hypotenuse = 5/13

By using pythagors theorem,

${H}^{2} = {P }^{2} + { B}^{2}$

${13}^{2} = {5}^{2} + {P}^{2}$

$169 - 25 = {P}^{2}$

$\sqrt{144 } = 12 \: cm$

We know that,

Tan = Perpendicular/Base = 12/5

Sec = Hypotenuse/Base = 13/5

${ \tan}^{2} - { \sec}^{2}$

$( { \frac{12}{5} })^{2} - ( { \frac{13}{5} )}^{2}$

$\frac{144}{25} - \frac{169}{25}$

$\frac{ - 25}{5} = - 5$