Question

If $\cos P = \dfrac{1}{7}$ and $\cos Q = \dfrac{{13}}{{14}}$ ,where P and Q both are acute angles. Then the value of P-Q is,
A. ${30^o}$
B. ${60^o}$
C ${45^o}$
D. ${75^o}$

Answer 1

As we know that by using the trigonometric formulas as $\cos (P - Q) = \cos P\cos Q + \sin P\sin Q$ . So, here the given values are $\cos P$ and $\cos Q$ . You only need to calculate the value of $\sin P,\sin Q$ using the formula of ${\sin ^2}x + {\cos ^2}x = 1$ .And then substitute it in the given above formula and then take its inverse so the value of angle P-Q is calculated.

Complete step-by-step answer:
As the given values are $\cos P = \dfrac{1}{7}$ and $\cos Q = \dfrac{{13}}{{14}}$ ,
Now, using the formula of ${\sin ^2}x + {\cos ^2}x = 1$ .
Calculate the value of $\sin P,\sin Q$ .
Hence, ${\sin ^2}Q + {\cos ^2}Q = 1$
The value of ${\cos ^2}Q$ is calculated as,
$\Rightarrow$ ${\cos ^2}Q = {\left( {\dfrac{{13}}{{14}}} \right)^2}$
On simplifying, we get,
$\Rightarrow$ ${\cos ^2}Q = \dfrac{{169}}{{196}}$
Putting the value in ${\sin ^2}Q + {\cos ^2}Q = 1$ , we get,
$\Rightarrow$ ${\sin ^2}Q + \dfrac{{169}}{{196}} = 1$
On simplifying, we get,
$\Rightarrow$ ${\sin ^2}Q = 1 - \dfrac{{169}}{{196}}$
On taking LCM and solving we get,
$\Rightarrow$ ${\sin ^2}Q = \dfrac{{196 - 169}}{{196}} = \dfrac{{27}}{{196}}$
Now, taking the square root of above equation, we get,
$\Rightarrow$ $\sin Q = \sqrt {\dfrac{{27}}{{196}}} = \dfrac{{3\sqrt 3 }}{{14}}$
Now, calculating same for $\sin P$
The value of ${\cos ^2}P$ is calculated as,
${\cos ^2}P = {\left( {\dfrac{1}{7}} \right)^2}$
On simplifying, we get,
$\Rightarrow$ ${\cos ^2}P = \dfrac{1}{{49}}$
Putting the value in ${\sin ^2}P + {\cos ^2}P = 1$ , we get,
$\Rightarrow$ ${\sin ^2}P + \dfrac{1}{{49}} = 1$
On simplifying, we get,
$\Rightarrow$ ${\sin ^2}P = 1 - \dfrac{1}{{49}}$
On taking LCM and solving we get,
${\sin ^2}P = \dfrac{{49 - 1}}{{49}} = \dfrac{{48}}{{49}}$
Now, taking the square root of above equation, we get,
$\Rightarrow$ $\sin P = \sqrt {\dfrac{{48}}{{49}}} = \dfrac{{4\sqrt 3 }}{7}$
Hence, now putting all the values in the above formula of $\cos (P - Q) = \cos P\cos Q + \sin P\sin Q$ , we get,
$\Rightarrow$ $\cos (P - Q) = \left( {\dfrac{1}{7}} \right)\left( {\dfrac{{13}}{{14}}} \right) + \left( {\dfrac{{3\sqrt 3 }}{{14}}} \right)\left( {\dfrac{{4\sqrt 3 }}{7}} \right)$
On calculating the above value,
$\Rightarrow$ $\cos (P - Q) = \left( {\dfrac{{13}}{{98}}} \right) + \left( {\dfrac{{12 \times 3}}{{98}}} \right)$
On simplifying, we get,
$= \left( {\dfrac{{13}}{{98}}} \right) + \left( {\dfrac{{36}}{{98}}} \right)$
Hence, $\cos (P - Q) = \left( {\dfrac{{49}}{{98}}} \right)$
Which can also be given as, $\cos (P - Q) = \dfrac{1}{2}$ .
As, $\cos {60^o} = \dfrac{1}{2}$
So we have $\theta = {60^o}$
Hence, $P - Q = {60^o}$
So, option (B) is the correct answer.

Note: Other trigonometric formulas similar to the one’s which we used to solve this problem are:
$\cos (P + Q) = \cos P\cos Q - \sin P\sin Q$
$\sin (P + Q) = \sin P\cos Q + \cos P\sin Q$
$\sin (P - Q) = \sin P\cos Q - \cos P\sin Q$
${\sec ^2}x - {\tan ^2}x = 1$
$co{\sec ^2}x - \operatorname{co} {\operatorname{t} ^2}x = 1$
Remember the trigonometric formula such as $\cos (P - Q) = \cos P\cos Q + \sin P\sin Q$ . Also remember the correct method of using various values such as ${\sin ^2}x + {\cos ^2}x = 1$ . Substitute and use the correct calculation and hence the required answer will be obtained. Remember both the above concepts and apply them and place the value correctly in the above formed equations so that the correct answer can be obtained.