Question

Question 1 Find the values of other five trigonometric functions if cos x = -1/2, x lies in third quadrant.

Class X1 - Maths -Trigonometric Functions Page 63

Answer 1

Cosx = -1/2 , where x lies in third quadrant.
We know, in 3rd quadrant tang∅, and cot∅ are positive.

Cosx = -1/2 = b/h
So, p = √(h² - b²) = √(2²-1²)
= ±√3
Now, sinx = P/h = ±√3/2
But sinx is negative in 3rd quadrant so, sinx = -√3/2

cosecx = 1/sinx = 1/(-√3/2) = -2/√3

Tanx = P/b = ±√3/1
we know, tanx is positive so,
Tanx = √3
Cotx = 1/tanx = 1/√3

Secx = 1/cosx = 1/(-1/2) =-2

Answer 2

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

$\tt{\rightarrow cosx=-\dfrac{1}{2}}$

As we know that :-

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

$\tt{\rightarrow secx=\dfrac{1}{-(1/2)}}$

secx = -2

Also we know that :-

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

Hence,

sin²x = 1 - cos²x

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

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

$\tt{\rightarrow sin^2x=1-\dfrac{1}{4}}$

$\tt{\rightarrow sin^2x=\dfrac{3}{4}}$

Hence,

$\tt{\rightarrow sinx=\pm\dfrac{\sqrt{3}}{2}}$

Here we get that x lies in third quadrant.

Hence, value of sinx will be negative.

Now,

$\tt{\rightarrow sinx=-\dfrac{\sqrt{3}}{2}}$

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

$\tt{\rightarrow cosecx=\dfrac{1}{(-\sqrt{3}/2)}}$

$\tt{\rightarrow cosecx=-\dfrac{2}{\sqrt{3}}}$

Now,

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

$\tt{\rightarrow tanx=\dfrac{-(\sqrt{3}/2)}{-(1/2)}}$

tanx = √3

Now,

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

$\tt{\rightarrow cotx=\dfrac{1}{\sqrt{3}}}$