Question

If \[\cos x = m\]and \[0 < x < {90^ \circ }\] then find \[\tan x\]
A) \[\dfrac{{\sqrt {1 - {m^2}} }}{m}\]
B) \[\dfrac{m}{{\sqrt {1 - {m^2}} }}\]
C) \[\dfrac{{\sqrt {1 + {m^2}} }}{m}\]
D) \[\dfrac{m}{{\sqrt {1 - {m^2}} }}\]
E) \[\dfrac{{1 - {m^2}}}{m}\]

Answer 1

Hint: Let us consider a triangle \[ABC\]where \[\angle B\]is right angle. With respect to the acute angle\[\angle C\], the relation between the side and angle of the triangle is given as,
\[\tan \angle C = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}\] and \[\cos \angle C = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}\]

Pythagoras theorem: Let us consider a triangle \[ABC\]whose \[\angle B\]is right angle. Then, the theorem states that,
\[{\text{Hypotenus}}{{\text{e}}^{\text{2}}}{\text{ = Bas}}{{\text{e}}^{\text{2}}}{\text{ + Perpendicula}}{{\text{r}}^{\text{2}}}\].

Complete step by step answer:
To find the value of \[\tan x\], initially, we are in need to find the value of perpendicular.
It is given that \[\cos x = m\]and \[0 < x < {90^ \circ }\].
And let us consider a right angled triangle \[ABC\]where the acute angle is \[x\] so that the relation is changed as \[\cos x = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}\].
As per the above condition of \[{\text{cos }}x\], we can conclude that,
Base \[ = m\] and hypotenuse\[ = 1\].
In Pythagoras theorem we are going to substitute the value of base \[ = m\] and hypotenuse\[ = 1\] we get,
\[\Rightarrow {1^2} = {m^2} + {\text{Perpendicula}}{{\text{r}}^{\text{2}}}\]
Now we are going to find perpendicular by solving the above equation, we get,
\[\Rightarrow {\text{Perpendicula}}{{\text{r}}^2} = {1^2} - {m^2}\]
\[\Rightarrow {\text{Perpendicular}} = \sqrt {1 - {m^2}} \]
Now, substitute the value of perpendicular and base in the \[{\text{tan }}x\] formula, we get,
\[\Rightarrow \tan x = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}\]
\[\Rightarrow \tan x = \dfrac{{\sqrt {1 - {m^2}} }}{m}\]

Hence, the correct option is (A) \[\dfrac{{\sqrt {1 - {m^2}} }}{m}\].

Note:
We can solve this problem in a different manner, which is nothing but the trigonometric form.
We know the following trigonometric identity, \[{\sin ^2}x + {\cos ^2}x = 1\]
It is given that \[\cos x = m\]
So, substitute the value of \[\cos x\]in the trigonometric identity, we get,
\[\Rightarrow {\sin ^2}x + {m^2} = 1\]
Let us solve the above equation to find \[\sin x\] we get,
\[\Rightarrow {\sin ^2}x = 1 - {m^2}\]
\[\Rightarrow \sin \theta = \sqrt {1 - {m^2}} \]
We know that\[\tan x = \dfrac{{\sin x}}{{\cos x}}\],
Substitute the values of \[\sin x\] and \[\cos x\] in the above identity,
\[\Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}} = \dfrac{{\sqrt {1 - {m^2}} }}{m}\]
Here is the answer for \[\tan {\text{ }}x\] where \[{\text{cos }}x{\text{ }} = m\] is given.