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# Prove that $\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0$.

Hint: In this problem we are to prove $\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0$. In our left hand side we will use $\sin (90^\circ - \theta ) = \cos \theta$ and $\cos (90^\circ - \theta ) = \sin \theta$ formulas to get ahead with the problem. Then we get both terms are equal and subtracting them we will give the required result.

We have to prove, $\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0$
We are given as L.H.S,
$\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta )$
Now, as, $\sin (90^\circ - \theta ) = \cos \theta$ and $\cos (90^\circ - \theta ) = \sin \theta$,
So we get,
$= \sin \theta \cos \theta - \cos \theta \sin \theta$
Taking, $\sin \theta$ common from the term, we get,
$= \sin \theta (\cos \theta - \cos \theta )$
On simplifying we get,
$= \sin \theta .0$
$= 0$
So, we have our answer as, $\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0$.
Hence, proved.

Note: You may have noticed that the words sine and cosine sound similar. That's because they're cofunctions! The way cofunctions work is exactly what you saw above. In general, if f and g are cofunctions, then,
$f(90^\circ - \theta ) = g(\theta )$and $g(90^\circ - \theta ) = f(\theta )$.