Question

Find the values of other five trigonometric functions for the following: $sce x =\frac{13}{5}$, x lies in fourth quadrant.

Answer 1

given, secx = 13/5 , x lies in 4th quadrant.

we know, in 4th quadrant,
sin, cosec => negative
cos, sec => positive
tan , cot => negative.

secx = 13/5 , cosx = 1/secx = 5/13

we know, sec²x - tan²x = 1 [ from trigonometric identities]
(13/5)² - tan²x = 1
(13/5)² - 1 = tan²x
169/25 - 1 = tan²x
144/25 = tan²x
tanx = ±12/5 , but tanx ≠ 12/5 because x lies in 4th quadrant .
so, tanx = -12/5 and cotx = 1/tanx = -5/12

now, sin²x +cos²x = 1 [ from trigonometric identities]
sin²x + (5/13)² = 1
sin²x = 1 - (5/13)² = 144/169
sinx = ± 12/13 but sinx ≠ 12/13 because x lies in 4th quadrant.
so, sinx = -12/13 and cosecx = -13/12

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

$\tt{\rightarrow secx=\dfrac{13}{5}}$

As we know that :-

$\tt{\rightarrow cosx=\dfrac{1}{secx}}$

$\tt{\rightarrow cosx=\dfrac{1}{(13/5)}}$

$\tt{\rightarrow cosx=\dfrac{5}{13}}$

Also we know that :-

${\boxed{\sf\:{sin^2x+cos^2x=1}}}$

sin²x = 1 - cos²x

Hence,

${\boxed{\sf\:{Putting\;the\;values\;:-}}}$

$\tt{\rightarrow sin^2x=1-(\dfrac{5}{13})^2}$

$\tt{\rightarrow sin^2x=1-\dfrac{25}{169}}$

$\tt{\rightarrow sin^2x=\dfrac{144}{169}}$

$\tt{\rightarrow sinx=\pm\dfrac{12}{13}}$

Here we get,

x lies in 4th quadrant.

Hence,

Value of sinx will be negative.

Now,

$\tt{\rightarrow sinx=-\dfrac{12}{13}}$

Now,

$\tt{\rightarrow cosecx=\dfrac{1}{sinx}}$

$\tt{\rightarrow cosecx=\dfrac{1}{(-12/13)}}$

$\tt{\rightarrow cosecx=-\dfrac{13}{12}}$

Now,

$\tt{\rightarrow tanx=\dfrac{sinx}{cosx}}$

$\tt{\rightarrow tanx=\dfrac{(-12)/13}{(5/13)}}$

$\tt{\rightarrow tanx=-\dfrac{12}{5}}$

Now,

$\tt{\rightarrow cotx=\dfrac{1}{tanx}}$

$\tt{\rightarrow cotx=\dfrac{1}{-(12/5)}}$

$\tt{\rightarrow cotx=-\dfrac{5}{12}}$