Question

If \[\cos 9\alpha = \sin \alpha \] and $9\alpha < 90^\circ $, then the value of $\tan 5\alpha $ is?
A) $\dfrac{1}{{\sqrt 3 }}$
B) $\sqrt 3 $
C) $1$
D) $0$

Answer 1

Hint: We have given any equation involving sine and cosine functions. We can convert the $\sin \alpha $ by replacing it with $\cos (90^\circ - \alpha )$. Then we get an equation involving cosine only. Simplifying this we get the value of $\alpha $. Knowing the value of $\alpha $, we can find $5\alpha $ and thus $\tan 5\alpha $.

Formula used :
For any $\alpha $ we have, $\cos (90^\circ - \alpha ) = \sin \alpha $
$\tan 45^\circ = 1$

Complete step-by-step answer:
It is given that \[\cos 9\alpha = \sin \alpha \].
We know that $\cos (90^\circ - \alpha ) = \sin \alpha $
This gives,
\[\cos 9\alpha = \cos (90^\circ - \alpha )\]
Comparing both sides we get,
$9\alpha = 90^\circ - \alpha $, since we have $9\alpha < 90^\circ $.
Adding $\alpha $ on both sides we get,
$9\alpha + \alpha = 90^\circ - \alpha + \alpha $
Simplifying we get,
$10\alpha = 90^\circ $
Dividing both sides by $10$ we get,
$\alpha = \dfrac{{90^\circ }}{{10}} = 9^\circ $
We need to find $\tan 5\alpha $.
We have, $\alpha = 9^\circ $
Multiplying both sides by $5$ we get,
$5\alpha = 5 \times 9^\circ = 45^\circ $
This gives,
$\tan 5\alpha = \tan 45^\circ $
We know that $\tan 45^\circ = 1$.
Substituting this, we have,
$\tan 5\alpha = 1$
$\therefore $ The answer is option C.

Note: We can do the problem in another way as well. Instead of converting the sine term to cosine term, we can convert cosine term to sine.
For that, we have $\sin (90^\circ - x) = \cos x$
Letting $x = 9\alpha $ we get,
$\sin (90^\circ - 9\alpha ) = \cos 9\alpha $
Given that $\cos 9\alpha = \sin \alpha $.
Substituting for $\cos 9\alpha $ using above result, we have
$\sin (90^\circ - 9\alpha ) = \sin \alpha $
So comparing both sides we get,
$90^\circ - 9\alpha = \alpha $
Adding $9\alpha $ on both sides we get,
$90^\circ - 9\alpha + 9\alpha = \alpha + 9\alpha $
Simplifying we get,
$90^\circ = 10\alpha $
Dividing both sides by $10$ we have,
$\alpha = 9^\circ $
Then we can proceed as above to get the answer.
Sine and cosine are periodic functions with period $360^\circ $. So, the same value can be taken for different angles. But here we have the condition $9\alpha < 90^\circ $. So we can restrict this to get a single value. Otherwise this is not possible.