Question

If cos thetha = 3/5 , find the value of sin thetha-1/tan thetha / 2Tan thetha​

Answer 1

Step-by-step explanation:

Diagram:-

Let x be the opposite side of the triangle

$\sf cos \theta = \dfrac{3}{5} = \dfrac{Adjacent \: side}{Hypotenuse}$

For right angled triangle.

Hypotenuse² = (Adjacent side)² + (Opposite side)²

$\dashrightarrow \sf {AC}^{2} = {AB}^{2} + {BC}^{2}$

$\dashrightarrow \sf {AB}^{2} = {AC}^{2} - {BC}^{2}$

$\dashrightarrow \sf {AB}^{2} = {5}^{2} - {3}^{2}$

$\dashrightarrow \sf {AB}^{2} = 25 - 9$

$\dashrightarrow \sf {AB}^{2} = 16$

$\dashrightarrow \sf AB= \sqrt{16}$

$\dashrightarrow \sf AB = 4$

$\sf sin \theta = \dfrac{AB}{AC} = \dfrac{4}{5}$

$\sf tan\theta = \dfrac{sin\theta}{cos\theta} = \dfrac{4}{3}$

Now, the value of $\sf \dfrac{sin \theta - \dfrac{1}{tan \theta} }{2tan \theta}$

$\sf = \dfrac{ \dfrac{4}{5} - \dfrac{1}{ \frac{4}{3} } }{2 \times \dfrac{4}{3} }$

$\sf = \dfrac{ \dfrac{4}{5} - \dfrac{3}{4}}{ \dfrac{8}{3} }$

$\sf = \dfrac{ \dfrac{16 - 15}{20} }{ \dfrac{8}{3} }$

$\sf = \dfrac{ \dfrac{1}{20} }{ \dfrac{8}{3} } = \dfrac{1}{20} \times \dfrac{3}{8}$

$\sf = \dfrac{3}{160}$

$\:$

$\underline{\boxed{ \sf \dashrightarrow\dfrac{sin \theta - \dfrac{1}{tan \theta} }{2tan \theta} = \dfrac{3}{160}}}$