Question

Question 3 Find the values of other five trigonometric functions if cot x = 3/4 , x lies in third quadrant.

Class X1 - Maths -Trigonometric Functions Page 63

Answer 1

Cotx = 3/4 where x lies in 3rd quadrant.

In 3rd quadrant :- tanx, cotx are positive . all rest are negative.

cotx = 3/4 = b/P
h = √(b²+P²) = √(3²+4²)=±5
tanx = 1/cotx = 4/3
Cosx = b/h = - 3/5
Secx = 1/cosx = -5/3
sinx = P/h = -4/5
Cosecx = -5/4

Answer 2

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

$\tt{\rightarrow cotx=\dfrac{3}{4}}$

As we know that :-

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

$\tt{\rightarrow tanx=\dfrac{1}{(3/4)}}$

$\tt{\rightarrow tanx=\dfrac{4}{3}}$

Also we know that :-

${\boxed{\sf\:{1+tan^2x=sec^2x}}}$

Hence,

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

$\tt{\rightarrow 1+(\dfrac{4}{3})^2=sec^2x}$

$\tt{\rightarrow 1+\dfrac{16}{9}=sec^2x}$

$\tt{\rightarrow\dfrac{25}{9}=sec^2x}$

$\tt{\rightarrow secx=\pm\dfrac{5}{3}}$

Here we get,

x lies in third quadrant.

Hence,

Value of secx will be negative.

Now,

$\tt{\rightarrow secx=-\dfrac{5}{3}}$

Now,

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

$\tt{\rightarrow cosx=\dfrac{1}{(-5/3)}}$

$\tt{\rightarrow cosx=-\dfrac{3}{5}}$

Now,

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

$\tt{\rightarrow tanx=\dfrac{(3/5)}{-(4/5)}}$

$\tt{\rightarrow\dfrac{4}{3}=\dfrac{-3}{5}}$

$\tt{\rightarrow sinx=\dfrac{4}{3}\times\dfrac{-3}{5}}$

$\tt{\rightarrow sinx=-\dfrac{4}{5}}$

Now,

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

$\tt{\rightarrow cosecx=-\dfrac{5}{4}}$