Question

If $\theta$ and $\phi$ are angles in the first quadrant such that $\tan \theta =\dfrac{1}{7}$ and $\sin \phi =\dfrac{1}{\sqrt{10}}$, then
A. $\theta +2\phi ={{90}^{\circ }}$
B. $\theta +2\phi ={{30}^{\circ }}$
C. $\theta +2\phi ={{75}^{\circ }}$
D. $\theta +2\phi ={{45}^{\circ }}$

Answer 1

we need to find the value of $\theta +2\phi$. Therefore, using the value $\sin \phi =\dfrac{1}{\sqrt{10}}$ we find the value of $\cos \phi$, $\tan \phi$ and $\tan 2\phi$. Then we use the identity of $\tan \left( \theta +2\phi \right)=\dfrac{\tan \theta +\tan \left( 2\phi \right)}{1-\tan \theta \tan \left( 2\phi \right)}$ to find the value of $\theta +2\phi$.

Complete step-by-step solution:
It is given that $\tan \theta =\dfrac{1}{7}$ and $\sin \phi =\dfrac{1}{\sqrt{10}}$. From the value of $\sin \phi =\dfrac{1}{\sqrt{10}}$, we try to find the value of $\cos \phi$ and $\tan \phi$.
We have the identity where ${{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1$. This gives $\cos \phi =\sqrt{1-{{\sin }^{2}}\phi }$.
We take only the positive values as $\theta$ and $\phi$ are angles in the first quadrant which means any ratio of those angles will give positive value.
Putting the value, we get $\cos \phi =\sqrt{1-{{\left( \dfrac{1}{\sqrt{10}} \right)}^{2}}}=\sqrt{1-\dfrac{1}{10}}=\dfrac{3}{\sqrt{10}}$.
Now we try to find the value of $\tan \phi =\dfrac{\sin \phi }{\cos \phi }=\dfrac{\dfrac{1}{\sqrt{10}}}{\dfrac{3}{\sqrt{10}}}=\dfrac{1}{3}$.
From the value of $\tan \phi$, we try to find the value of $\tan 2\phi$ where $\tan 2\phi =\dfrac{2\tan \phi }{1-{{\tan }^{2}}\phi }$.
Putting the values, we get $\tan 2\phi =\dfrac{2\left( \dfrac{1}{3} \right)}{1-{{\left( \dfrac{1}{3} \right)}^{2}}}=\dfrac{3}{4}$.
From the given equations in options, we can evaluate that we need the value of $\theta +2\phi$.
We have the associate angle formula of $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.
We put the values of $A=\theta ,B=2\phi$ and get
$\tan \left( \theta +2\phi \right)=\dfrac{\tan \theta +\tan \left( 2\phi \right)}{1-\tan \theta \tan \left( 2\phi \right)}$.
Now we put the values of the required ratios as $\tan \theta =\dfrac{1}{7}$ and $\tan \left( 2\phi \right)=\dfrac{3}{4}$.
$\tan \left( \theta +2\phi \right)=\dfrac{\tan \theta +\tan \left( 2\phi \right)}{1-\tan \theta \tan \left( 2\phi \right)}=\dfrac{\dfrac{1}{7}+\dfrac{3}{4}}{1-\dfrac{1}{7}\times \dfrac{3}{4}}$.
Now we simplify the equation to get $\tan \left( \theta +2\phi \right)=\dfrac{4+21}{28-3}=\dfrac{25}{25}=1$.
We now equate the value for our known values of ratio tan where $\tan \left( \theta +2\phi \right)=1=\tan \left( {{45}^{\circ }} \right)$
This gives $\left( \theta +2\phi \right)={{45}^{\circ }}$. The correct option is D.

Note: We need to be careful about the sign of the ratios. The quadrant dictates the signs and therefore if anything is not mentioned then we have to consider all possible choices.