Question

How do you find the exact value of \[\sec \left( {v - u} \right)\] given that \[\sin u = \dfrac{5}{{13}}\] and \[\cos v = \dfrac{{ - 3}}{5}\] ?

Answer 1

Hint: Here we are given with a tricky question. We are familiar with the trigonometric sum and difference formula. Here we will use the formula for \[\cos \left( {v - u} \right)\] since \[\sec \left( {v - u} \right) = \dfrac{1}{{\cos \left( {v - u} \right)}}\]. We are given the value of sin and cos function.

Complete step-by-step solution:
Given that \[\sec \left( {v - u} \right)\]
We know that \[\sec \left( {v - u} \right) = \dfrac{1}{{\cos \left( {v - u} \right)}}\]
But the difference identity is \[\cos \left( {v - u} \right) = \cos v.\cos u + \sin v.\sin u\]
Given that \[\sin u = \dfrac{5}{{13}}\] and \[\cos v = \dfrac{{ - 3}}{5}\]
Just putting the values we get,
\[\Rightarrow \cos \left( {v - u} \right) = \dfrac{{ - 3}}{5}.\cos u + \sin v.\dfrac{5}{{13}}\]
But we don’t know the remaining two values. We will find them also by using standard identity function \[\Rightarrow {\sin ^2}x + {\cos ^2}x = 1\]
On modifying it we get,
\[\Rightarrow \sin x = \sqrt {1 - {{\cos }^2}x} \] and \[\cos x = \sqrt {1 - {{\sin }^2}x} \]
Thus to find the values of remaining functions ,

To find the value of \[\cos u\]
\[\Rightarrow \cos u = \sqrt {1 - {{\sin }^2}u} \]
Putting the value of sin function we get,
\[\Rightarrow \cos u = \sqrt {1 - {{\left( {\dfrac{5}{{13}}} \right)}^2}} \]
Taking the square
\[\Rightarrow \cos u = \sqrt {1 - \dfrac{{25}}{{169}}} \]
Taking the LCM we get,
\[\Rightarrow \cos u = \sqrt {\dfrac{{169 - 25}}{{169}}} \]
On subtracting we get,
\[\Rightarrow \cos u = \sqrt {\dfrac{{144}}{{169}}} \]
Taking the square root we get,
\[\Rightarrow \cos u = \dfrac{{12}}{{13}}\]

To find the value of \[\sin v\]
\[\Rightarrow \sin v = \sqrt {1 - {{\cos }^2}v} \]
Putting the value of cos function we get,
\[\Rightarrow \sin v = \sqrt {1 - {{\left( {\dfrac{{ - 3}}{5}} \right)}^2}} \]
Taking the square
\[\Rightarrow \sin v = \sqrt {1 - \dfrac{9}{{25}}} \]
Taking the LCM we get,
\[\Rightarrow \sin v = \sqrt {\dfrac{{25 - 9}}{{25}}} \]
On subtracting we get,
\[\Rightarrow \sin v = \sqrt {\dfrac{{16}}{{25}}} \]
Taking the square root we get,
\[\Rightarrow \sin v = \dfrac{4}{5}\]

Putting these two values we get,
\[\Rightarrow \cos \left( {v - u} \right) = \dfrac{{ - 3}}{5}.\dfrac{{12}}{{13}} + \dfrac{4}{5}.\dfrac{5}{{13}}\]
Now on simplifying we get,
\[\Rightarrow \cos \left( {v - u} \right) = \dfrac{{ - 3 \times 12}}{{65}} + \dfrac{{4 \times 20}}{{65}}\]
On multiplying the numerators we get,
\[\Rightarrow \cos \left( {v - u} \right) = \dfrac{{ - 36}}{{65}} + \dfrac{{80}}{{65}}\]
Since the denominators are same we can add them directly,
\[\Rightarrow \cos \left( {v - u} \right) = \dfrac{{ - 36 + 80}}{{65}}\]
\[\Rightarrow \cos \left( {v - u} \right) = \dfrac{{54}}{{65}}\]
This is the value of cos function. We will substitute in the reciprocal function.
\[\Rightarrow \sec \left( {v - u} \right) = \dfrac{1}{{\dfrac{{54}}{{65}}}}\]
On rearranging we get,
\[\Rightarrow \sec \left( {v - u} \right) = \dfrac{{65}}{{54}}\]
This is the correct answer.

Thus the correct answer is \[\sec \left( {v - u} \right) = \dfrac{{65}}{{54}}\]

Note: Note that this function is not available directly. We can use other trigonometric functions as we have used it above. Also note that in multiple choices we can tick the value of cos function so obtained by mistake but that is not the correct answer. Also in order to find the values of remaining functions in the trigonometric difference formula we can use Pythagorean triplet if known instead of doing the calculations. Such that \[\left( {5,12,13} \right)\] and \[\left( {3,4,5} \right)\] are the Pythagorean triplets that satisfy the Pythagoras theorem. This is just to save our time.