Question

If \(\cos A = \frac{3}{5}\), then the value of \(\tan A\) is :

• A. \(\frac{4}{5}\)
• B. \(\frac{5}{4}\)
• C. \(\frac{3}{4}\)
• D. \(\frac{4}{3}\)

Hint:

Trigonometric ratios are frequently built on the classic \((3, 4, 5)\) right-angled Pythagorean triple.
Since \(\cos A = \frac{3}{5}\), the base is 3, the hypotenuse is 5, and the missing perpendicular side must be 4.
Using \(\tan A = \frac{\text{Perpendicular}}{\text{Base}}\), you can immediately write \(\frac{4}{3}\) without drawing a triangle or writing down identities!

Answer 1

The Correct Option is D

Step 1: Understand what is being asked.
We are given $\cos A=\frac{3}{5}$ and need $\tan A$. Instead of building a right triangle, let us use the identity that connects $\sec A$ and $\tan A$ directly.

Step 2: Find $\sec A$ first.
Since $\sec A$ is the reciprocal of $\cos A$:
\[ \sec A = \frac{1}{\cos A} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]
Step 3: Use the identity $1+\tan^2 A=\sec^2 A$.
This is one of the three standard Pythagorean trigonometric identities. Rearranged for $\tan^2 A$:
\[ \tan^2 A = \sec^2 A - 1 \]
Substitute $\sec A=\frac{5}{3}$:
\[ \tan^2 A = \left(\frac{5}{3}\right)^2 - 1 = \frac{25}{9} - 1 = \frac{25-9}{9} = \frac{16}{9} \]
Step 4: Take the square root.
Since $A$ is an angle in a right triangle (so $0^\circ<A<90^\circ$), $\tan A$ is positive:
\[ \tan A = \sqrt{\frac{16}{9}} = \frac{4}{3} \]
Final Answer:
The value of $\tan A$ is $\frac{4}{3}$, matching option (D).
\[ \boxed{\frac{4}{3}} \]