Question

If \[\sec \alpha =\dfrac{13}{5}\] where \[\alpha <{{90}^{\circ }}\], then the value of \[\dfrac{2-3\cot \alpha }{4-9\tan \alpha }\] is?

Answer 1

Hint: In this problem, we are given \[\sec \alpha =\dfrac{13}{5}\],where \[\alpha <{{90}^{\circ }}\], we have to find the value of the given expression. We are already given the value of secant which is of the form \[\dfrac{Hypotenuse}{Adjacent}\] from which we can find the value of the tangent and the cotangent using Pythagoras theorem to find one of the sides and substitute them in the given expression to get the answer for the given expression.

Complete step by step answer:
Here we have to find the value of,
\[\dfrac{2-3\cot \alpha }{4-9\tan \alpha }\] ……. (1)
where \[\sec \alpha =\dfrac{13}{5}\].
We know that,
\[\cos \alpha =\dfrac{Adjacent}{Hypotenuse}\], where \[\sec \alpha =\dfrac{1}{\cos \alpha }\].
We can now write secant as,
 \[\Rightarrow \sec \alpha =\dfrac{Hypotenuse}{Adjacent}=\dfrac{13}{5}\]
We can now plot it in the triangle.

We can see that we have hypotenuse and adjacent sides, we can use the Pythagoras theorem to find the other side.
\[\Rightarrow Opposite=\sqrt{{{\left( 13 \right)}^{2}}-{{5}^{2}}}=\sqrt{169-25}=12\]
We know that,
\[\Rightarrow \tan \alpha =\dfrac{opposite}{adjacent}=\dfrac{12}{5}\] …….. (2)
We also know that cotangent is the inverse of tangent, we can write it as
\[\Rightarrow \cot \alpha =\dfrac{5}{12}\] …….. (3)
 We can now substitute (2) and (3) in (1), we get
\[\Rightarrow \dfrac{2-3\left( \dfrac{5}{12} \right)}{4-9\left( \dfrac{12}{5} \right)}\]
We can now simplify the above step, we get
\[\Rightarrow \dfrac{2-\dfrac{5}{4}}{4-\dfrac{108}{5}}=\dfrac{\dfrac{3}{4}}{-\dfrac{88}{5}}=\dfrac{3}{4}\times -\dfrac{5}{88}=-\dfrac{15}{352}\]
Therefore, the value of \[\dfrac{2-3\cot \alpha }{4-9\tan \alpha }=-\dfrac{15}{352}\] .

Note: We should always remember some of the trigonometric formulas such as \[\cos \alpha =\dfrac{Adjacent}{Hypotenuse}\], \[\sec \alpha =\dfrac{1}{\cos \alpha }\]. We should also know that cotangent is the inverse of tangent, \[\tan \alpha =\dfrac{1}{\cot \alpha }\]. We should also know that we can use Pythagoras theorem to find one of the sides of a right-triangle if we have any two sides.