Question

If $\dfrac{{\sin A}}{{\sin B}} = p{\text{ and }}\dfrac{{\cos A}}{{\cos B}} = q,{\text{then }}\tan A{\text{ and}}\tan B.$

Answer 1

Hint: In the given question we had given the value of \[2\] functions \[p\] and \[q\] which are trigonometric functions or they are in the form of trigonometric ratios. And we have to find out the value of other trigonometric ratios in terms of given ratios. Using identities of various trigonometric ratios we will find the value of asked trigonometric ratios.

Complete step-by-step answer:
 In the question , we are given the values of P and q which are in the terms of trigonometric ratios but they are having different angles. As in the question, there are two angles used Given the value of $p$ and \[q\] is
$p = \dfrac{{\operatorname{sin} A}}{{\operatorname{sin} B}}.............{\text{(1)}}$
and $P = \dfrac{{\cos A}}{{\cos B}}..............{\text{(2)}}$
From there, we have to find out the value of tan A and tan B.
It is clear that two parts are asked of us . Use will solve our problems after another.
${\text{From (1) p = }}\dfrac{{\operatorname{Sin} A}}{{\operatorname{Sin} B}}$
On cross multiplying it, we get
\[Sin{\text{ }}A{\text{ }} = \;P{\text{ }}sin{\text{ }}B..............(3)\]
${\text{From (2) q = }}\dfrac{{\cos A}}{{\cos B}}$
and on cross multiply it an get
\[cos{\text{ }}A = q\cos B..........{\text{(4)}}\]
Also we know the identity
\[ \Rightarrow {\operatorname{sin} ^2}A + {\operatorname{cos} ^2}A = 1\]
On substituting the value of \[\operatorname{sin} A\] and \[\operatorname{cos} B\] from
${\text{(3) and (4)}}$ her, up will get
$ \Rightarrow {\left( {p\operatorname{sin} B} \right)^2} + {\left( {q\cos B} \right)^2} = 1$
On squaring we get
$ \Rightarrow {P^2}{\operatorname{sin} ^2}B + {q^2}{\cos ^2}B = 1$
Dividing each and every term by ${\operatorname{Cos} ^2}B$, use get
\[ \Rightarrow {P^2}\dfrac{{{{\operatorname{sin} }^2}B}}{{{{\operatorname{cos} }^2}B}} + {q^2}\dfrac{{{{\operatorname{cos} }^2}B}}{{{{\operatorname{cos} }^2}B}} = \dfrac{1}{{{{\operatorname{cos} }^2}B}}\]
\[\;{P^2}{\tan ^2}B + {q^2}(1) = \sec B\]
This is because we know \[\;{\tan ^2}B = \dfrac{{{{\operatorname{sin} }^2}B}}{{{{\operatorname{cos} }^2}B}}\]
and ${\operatorname{sec} ^2}B = \dfrac{1}{{{{\operatorname{cos} }^2}B}}$
Therefore we get
\[\;{P^2}{\tan ^2}B + {q^2} = {\sec ^2}B\]
Also use know ${\operatorname{Sec} ^2}B = 1 + {\tan ^2}B.$ So we get
\[\; \Rightarrow {P^2}{\tan ^2}B + {q^2} = 1 + {\tan ^2}B\]
Taking all the terms to left hand side, We get
 $ \Rightarrow $\[\;{P^2}{\tan ^2}B + {q^2} - 1 - {\tan ^2}B = 0\]
Taking \[{\tan ^2}B\]common
\[ \Rightarrow {\tan ^2}B({P^2} - 1) + {q^2} - 1 = 0\]
taking second & third term to right hand side
$ \Rightarrow ({p^2} - 1){\tan ^2}B = 1 - {q^2}$
Also taking $({p^2} - 1)$ to Right hand side
$\tan B = \dfrac{{1 - q}}{{{p^2} - 1}}$
Taking square root on both sides we get
$\tan B = \pm \sqrt {\dfrac{{1 - {q^2}}}{{{p^2} - 1}}} $
Hence, we get the value of \[tan{\text{ }}B\]
Now we will find the value of \[tan{\text{ }}A\]
Dividing ${\text{(3) and (4)}}$ , we get
$ \Rightarrow \dfrac{{\operatorname{sin} A}}{{\operatorname{cos} A}} = \dfrac{{p\operatorname{sin} B}}{{q\operatorname{cos} B}}$
Which implies
$\tan A = \dfrac{p}{q}\tan B{\text{ }}..........{\text{(6)}}$
because $\dfrac{{\sin A}}{{\cos A}} = \tan A$
So, on substituting the value of $\tan B$ from $..........{\text{(5)}}$
$\tan A = \pm \dfrac{p}{q}\sqrt {\dfrac{{1 - {q^2}}}{{{p^2} - 1}}} $
Which is the value of $\tan A$

Note: In the above question, we had used various trigonometric identities, which we applied to find out the values of $\tan A$and $\tan B$. Where angels were different. That was $A$ and $B$. Trigonometric identities used were ${\sin ^2}A + {\operatorname{Cos} ^2}A = 1$ .
$\tan A = \dfrac{{\sin A}}{{\cos A}},{\tan ^2}A + 1 = {\sec ^2}A$ and after square root, we used the signs of positive and negative both.