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**Hint**: We know that the cosine function $ \cos A $ can also be written in terms of sine function as $ \sin \left( {{{90}^ \circ } - A} \right) $ . As the given trigonometric function has a sine function and cosine function, convert the cosine function into sine using the given relation. And then find the values of $ \theta $ .

**:**

__Complete step-by-step answer__We are given a trigonometric equation $ \sin a\theta + \cos b\theta = 0 $ .

We have to find the values of $ \theta $ are in AP or in GP.

$ \sin a\theta + \cos b\theta = 0 \to eq\left( 1 \right) $

Here write $ \cos b\theta $ in terms of sine using the relation $ \cos A = \sin \left( {{{90}^ \circ } - A} \right) $ , here the value of A is $ b\theta $

So therefore,

$

\cos b\theta = \sin \left( {{{90}^ \circ } - b\theta } \right) \\

{90^ \circ } = \dfrac{\pi }{2}radians \\

\Rightarrow \cos b\theta = \sin \left( {\dfrac{\pi }{2} - b\theta } \right) \\

$

On substituting the above value in equation 1, we get

$

\sin a\theta + \sin \left( {\dfrac{\pi }{2} - b\theta } \right) = 0 \\

\Rightarrow \sin a\theta = - \sin \left( {\dfrac{\pi }{2} - b\theta } \right) \\

$

Sending the minus inside in the above right hand side term, which is $ - \sin \left( {\dfrac{\pi }{2} - b\theta } \right) $

$ \to \sin a\theta = \sin \left( {b\theta - \dfrac{\pi }{2}} \right) $

Both the sides, the functions are sine, so equate their angle measures.

$

a\theta = b\theta - \dfrac{\pi }{2} \\

\Rightarrow \dfrac{\pi }{2} = b\theta - a\theta \\

\Rightarrow \theta \left( {a - b} \right) = \dfrac{\pi }{2} \\

\Rightarrow \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} \\

\Rightarrow \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 2n\pi \\

$

When n is equal to 0, $ \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 0 = \dfrac{\pi }{{2\left( {a - b} \right)}} $

When n is equal to 1, $ \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 2\pi $

When n is equal to 2, $ \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 4\pi $

When n is equal to 3, $ \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 6\pi $

As we can see, for two every two consecutive values of $ \theta $ , there is a difference of $ 2\pi $ .

So this difference can also be called a common difference.

Therefore, we can say that the possible values of $ \theta $ are in an $AP$.

**So, the correct answer is “Option A”.**

**Note**: Here we have added $ 2n\pi $ to the value of $ \theta $ , because sine is a periodic function and its value repeats after every $ 2n\pi $ radians. And an AP is a sequence in which every term starting from the second term is obtained by adding a fixed value to its previous term; this fixed value is called common difference whereas a GP is a sequence in which every term starting from the second term is obtained by multiplying a fixed value to its previous term; this fixed value is called common ratio. So do not confuse an AP with a GP.

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