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If cos $x$ = m and 0 < $x$ < 90 $°$ then find tan $x$

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a

\frac{\sqrt{1 - m^{2}}}{m}

b

\frac{m}{\sqrt{1 - m^{2}}}

c

\frac{\sqrt{1 + m^{2}}}{m}

d

\frac{1 - m^{2}}{m}

answer is A.

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Detailed Solution

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

< 9

0 °

.
And let us consider a right-angled triangle ABC where the acute angle is

x

so that the relation is changed as cos

x

=

\frac{Base}{Hypotenuse}

As per the above condition of 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,
⇒ 12 = m2 + Perpendicular2
Now we are going to find perpendicular by solving the above equation, we get,
⇒ Perpendicular2 = 12 − m2
⇒ Perpendicular =

\sqrt{1 - m^{2}}

Now, substitute the value of perpendicular and base in the tan

x

formula, we get,
⇒ tan

x

=

\frac{Perpendicular}{Base}

⇒ tan

x

=

\frac{\sqrt{1 - m^{2}}}{m}

Hence, option (1) is the correct option.

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