Question

If $\sec \theta =\dfrac{5}{4}$, show that:
$\dfrac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }=\dfrac{12}{7}$

Answer 1

Hint: In this question, from the given values of secant function by using the trigonometric identities we can find the values of the sine, cosine and tangent functions. Then on substituting the respective values in the given expression of the question we can calculate the hand side and left hand side values.

Complete step-by-step answer:
Then on comparing the values obtained we get the result.

& \sec \theta =\dfrac{1}{\cos \theta } \\\ & {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \\\ & \sin \theta =\tan \theta \cdot \cos \theta \\\ & \cot \theta =\dfrac{1}{\tan \theta } \\\ \end{aligned}$$ Now, from the given question we have $$\sec \theta =\dfrac{5}{4}$$ Now, by using the trigonometric identity which gives the relation between secant function and cosine function we get, $$\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }$$ Now, this can also be written as $$\Rightarrow \cos \theta =\dfrac{1}{\sec \theta }$$ Let us now substitute the value of secant in this $$\Rightarrow \cos \theta =\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}$$ Now, this can be further written as $$\therefore \cos \theta =\dfrac{4}{5}$$ Now, using the relation between the secant function and tangent function from the trigonometric identities we have $$\Rightarrow {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$$ Now, this can also be written as $$\Rightarrow {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$$ Now, on substituting the value of secant we get, $$\Rightarrow {{\tan }^{2}}\theta ={{\left( \dfrac{5}{4} \right)}^{2}}-1$$ Now, on further simplification we have $$\Rightarrow {{\tan }^{2}}\theta =\dfrac{25-16}{16}$$ Let us now apply the square root on both the sides $$\Rightarrow \sqrt{{{\tan }^{2}}\theta }=\sqrt{\dfrac{9}{16}}$$ Now, on simplifying it we get, $$\therefore \tan \theta =\dfrac{3}{4}$$ Now, by using the relation between sine, cosine and tangent functions from the trigonometric identities we have, $$\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\tan \theta $$ Now, this can also be written as $$\Rightarrow \sin \theta =\tan \theta \times \cos \theta $$ Now, on substituting the respective values in the above equation we get, $$\Rightarrow \sin \theta =\dfrac{3}{4}\times \dfrac{4}{5}$$ Now, on simplifying further we get, $$\therefore \sin \theta =\dfrac{12}{20}=\dfrac{3}{5}$$ Now, from the given expression in the question we have $$\dfrac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }=\dfrac{12}{7}$$ Let us first consider the left hand side and calculate its value $$\begin{aligned} & L.H.S=\dfrac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta } \\\ & L.H.S=\dfrac{\dfrac{3}{5}-2\times \dfrac{4}{5}}{\tan \theta -\dfrac{1}{\tan \theta }} \\\ & L.H.S=\dfrac{\dfrac{3}{5}-\dfrac{8}{5}}{\dfrac{3}{4}-\dfrac{1}{\dfrac{3}{4}}} \\\ & L.H.S=\dfrac{\dfrac{-5}{5}}{\dfrac{3}{4}-\dfrac{4}{3}} \\\ & L.H.S=\dfrac{-1}{\dfrac{9-16}{12}} \\\ & L.H.S=\dfrac{-1}{\dfrac{-7}{12}} \\\ & L.H.S=\dfrac{12}{7} \\\ \end{aligned}$$ Thus, the value of right hand side is equal to left hand side Hence, it is verified that $$\dfrac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }=\dfrac{12}{7}$$ Note: Instead of calculating the values of right hand side and left hand side by substituting the respective values we can calculate either of them and then use proper trigonometric identities that both the expressions are equal. It is important to note that while calculating the values of respective functions we need to use the identities accordingly and solve them. Because neglecting any of the terms or writing it incorrectly changes the complete result.