Question

If $\sin \theta =\dfrac{3}{5}$ , then evaluate $\dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }$.

Answer 1

- Hint: In this question, we can use the basic relations that exist among the different trigonometric function which are mentioned as follows

& \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & \Rightarrow \tan x=\dfrac{\sin x}{\cos x} \\\ & \Rightarrow \cot x=\dfrac{1}{\tan x} \\\ \end{aligned}$$ Now, we can use the value of the sin function that is given above to evaluate the given expression. _Complete step-by-step solution_ - As mentioned in the question, we have to find the value of the expression that is given in the hint. Now, for finding the value of cos function, we can write as follows $$\begin{aligned} & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & \Rightarrow {{\left( \dfrac{3}{5} \right)}^{2}}+{{\cos }^{2}}\theta =1 \\\ & \Rightarrow \dfrac{9}{25}+{{\cos }^{2}}\theta =1 \\\ & \Rightarrow {{\cos }^{2}}\theta =1-\dfrac{9}{25} \\\ & \Rightarrow {{\cos }^{2}}\theta =\dfrac{25-9}{25} \\\ & \Rightarrow {{\cos }^{2}}\theta =\dfrac{16}{25} \\\ \end{aligned}$$ Now, on taking the square root of both the sides, we get the following $$\begin{aligned} & \Rightarrow \cos \theta =\sqrt{\dfrac{16}{25}} \\\ & \Rightarrow \cos \theta =\dfrac{4}{5} \\\ \end{aligned}$$ Now, as we know the value of cos as well as sin functions, hence, we can find the value of tan function as follows $$\begin{aligned} & \Rightarrow \tan x=\dfrac{\sin x}{\cos x} \\\ & \Rightarrow \tan \theta =\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} \\\ & \Rightarrow \tan \theta =\dfrac{3}{4} \\\ \end{aligned}$$ Now, as we have got the value of tan function, hence, we can find the value of cot function as well and it is as follows $$\begin{aligned} & \Rightarrow \cot x=\dfrac{1}{\tan x} \\\ & \Rightarrow \cot x=\dfrac{1}{\dfrac{3}{4}} \\\ & \Rightarrow \cot x=\dfrac{4}{3} \\\ \end{aligned}$$ Now, as we have the values of cos, tan and cot functions, we can thus evaluate the given expression as follows $$\begin{aligned} & \Rightarrow \dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }=\dfrac{\dfrac{4}{5}-\dfrac{1}{\dfrac{3}{4}}}{2\times \dfrac{4}{3}} \\\ & \Rightarrow \dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }=\dfrac{\dfrac{4}{5}-\dfrac{4}{3}}{\dfrac{8}{3}} \\\ & \Rightarrow \dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }=\dfrac{\dfrac{4}{5}-\dfrac{4}{3}}{\dfrac{8}{3}}=\dfrac{\dfrac{3\times 4-5\times 4}{5\times 3}}{\dfrac{8}{3}} \\\ & \Rightarrow \dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }=\dfrac{12-20}{5\times 8}=\dfrac{-8}{5\times 8}=\dfrac{-1}{5} \\\ \end{aligned}$$ Hence, the value of the given expression is $$\dfrac{-1}{5}$$ . Note:- Another method of doing this question is that - We can first convert the given expression in terms of sin function and then in the final step, we can simply just put the value of the sin function that is given in the question itself and hence, we can get to the final answer which will be the same as the one which we have obtained above. Also, the students can commit calculation errors, so, one should be extra careful while solving the question and while doing the calculations.