A line makes an angle $\alpha ,\beta ,\gamma $ with the coordinate axes, if $\alpha + \beta = {90^ \circ }$ then $\gamma = $
A. $0$
B. ${90^ \circ }$
C. ${180^ \circ }$
D. ${45^ \circ }$

Hint: Given, A line makes an angle $\alpha ,\beta ,\gamma $ with the coordinate axes, if $\alpha + \beta = {90^ \circ }$. We have to find the value $\gamma $. First, we will use the concept ${\cos ^2}a + {\cos ^2}b + {\cos ^2}c = 1$. Then, we will use $\alpha + \beta = {90^ \circ }$ to find the relation between $\cos \alpha $ and $\cos \beta $ then, will eliminate $\cos \alpha $ and $\cos \beta $ so that we left with only $\cos \gamma $. At last, we will find the value of $\gamma $.

Formula Used: \[{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1\]

Complete step by step solution: Given, A line makes an angle $\alpha ,\beta ,\gamma $ with the coordinate axes
We know if a line make an angle $a,b,c$
Then, ${\cos ^2}a + {\cos ^2}b + {\cos ^2}c = 1$
Using above concept
\[{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1\] (1)
It is given that $\alpha + \beta = {90^ \circ }$
Shifting $\beta $ to other side
$\alpha = {90^ \circ } - \beta $
Taking cos on both sides
\[\cos \alpha = \cos ({90^ \circ } - \beta )\]
We know that \[\cos ({90^ \circ } - \theta ) = \sin \theta \]
Using the above identity
\[\cos \alpha = \sin \beta \]
Squaring both sides
\[{\cos ^2}\alpha = {\sin ^2}\beta \]
We know that ${\sin ^2}x = 1 - {\cos ^2}x$
Using above property
\[{\cos ^2}\alpha = 1 - {\cos ^2}\beta \]
Putting in the equation (1)
\[1 - {\cos ^2}\beta + {\cos ^2}\beta + {\cos ^2}\gamma = 1\]
After simplifying the above equation
\[{\cos ^2}\gamma = 0\]
Taking square root on both sides
\[\cos \gamma = 0\]
We know $\cos {90^ \circ } = 0$
$\cos \gamma = \cos {90^ \circ }$
$ \Rightarrow \gamma = {90^ \circ }$

Hence option B is correct.

Note:Students should use concepts correctly so that they do not get stuck anywhere while calculating the answer. First, they should clearly think about the identities they can apply in order to get smooth error-free calculations and solve questions step-wise. Students should use \[\cos \alpha = \cos ({90^ \circ } - \beta )\] this identity correctly. This is one of the main parts of the solution.