Question

If \(\alpha, \beta, \gamma\) are angles of inclinations of a line with \(x, y\) and \(z\)-axis respectively, then which of the following are correct?
A. \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\)
B. \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1\)
C. \(\sin^2\alpha + \sin^2\beta + \sin^2\gamma = -2\)
D. \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 2\)
Choose the correct answer from the options given below:

• A. A and B only
• B. A, B and C only
• C. B and D only
• D. A and D only

Hint:

Use \(l^2+m^2+n^2=1\) for direction cosines, then \(\cos 2\theta = 2\cos^2\theta-1\).

Answer 1

The Correct Option is A

Step 1: Take a concrete line.
Pick the line along the x-axis. Then $\alpha=0$ and $\beta=\gamma=90^\circ$. The direction cosine identity holds for every line, so a special case is a quick test of each statement.
Here $\cos\alpha=1$, $\cos\beta=0$, $\cos\gamma=0$.

Step 2: Test A and D.
The sum of squares of the cosines is $1+0+0=1$. So A gives $1=1$, which holds. D says the sum is $2$, which fails.

Step 3: Test B.
$\cos 2\alpha=\cos 0=1$. $\cos 2\beta=\cos 180^\circ=-1$. $\cos 2\gamma=-1$. The sum is $1-1-1=-1$. So B holds.

Step 4: Test C.
$\sin^2\alpha=0$, $\sin^2\beta=1$, $\sin^2\gamma=1$. The sum is $2$, not $-2$. So C fails. A sum of squares is never negative anyway.

Step 5: Why the special case is enough.
The identity $l^2+m^2+n^2=1$ is true for every line. So A is always true and D is always false. B and C follow from A by simple algebra, so their truth does not depend on which line we pick.

Step 6: Choose the option.
A and B are the correct statements. That is option 1.

Final Answer:
A and B are correct. \[ \boxed{\text{Option 1: A and B only}} \]