Question

if cot x = -3/4 x is in 2 quadrant then find sinx/2 cosx/2 tanx/2​

Answer 1

$\textbf{Given:}$

$cotx=\dfrac{-3}{4}$

$\text{x lies in 2 quadrant}$

$\textbf{To find:}$

$sin\frac{x}{2}\,cos\frac{x}{2}\,tan\frac{x}{2}$

$\textbf{Solution:}$

$tanx=\dfrac{-4}{3}$

$\text{Consider,}$

$sec^2x=1+tan^2x$

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

$sec^2x=\dfrac{25}{16}$

$sec^2x=\dfrac{25}{16}$

$secx=\pm\dfrac{5}{4}$

$\text{Taking reciprocals we get}$

$cosx=\pm\dfrac{4}{5}$

$\text{But x lies in 2 nd quadrant}$

$\implies\,cosx=\dfrac{-4}{5}$

$\text{Using the identity}$

$\boxed{cosA=1-2\,sin^2\frac{A}{2}}$

$1-2\,sin^2\frac{x}{2}=\dfrac{-4}{5}$

$2\,sin^2\frac{x}{2}=1+\dfrac{4}{5}$

$\implies\,sin^2\frac{x}{2}=\dfrac{9}{10}</p><p>$

$\text{Now}$

$sin\frac{x}{2}\,cos\frac{x}{2}\,tan\frac{x}{2}$

$=sin\frac{x}{2}\,cos\frac{x}{2}\,\dfrac{sin\frac{x}{2}}{cos\frac{x}{2}}$

$=sin\frac{x}{2}{\times}sin\frac{x}{2}$

$=sin^2\frac{x}{2}$

$=\dfrac{9}{10}$

$\textbf{Answer:}$

$\bf\,sin\frac{x}{2}\,cos\frac{x}{2}\,tan\frac{x}{2}=\dfrac{9}{10}$