Let's solve the given trigonometric equation step-by-step:
The given equation to solve is:
[tex]\[ \frac{\operatorname{sin} \theta}{\csc \theta} + \frac{\cos \theta}{\sec \theta} = 1 \][/tex]
First, let's remember that the cosecant (csc) and secant (sec) functions are the reciprocals of the sine (sin) and cosine (cos) functions respectively. That is:
[tex]\[ \csc \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
Now, let's substitute these reciprocal identities back into the equation:
[tex]\[ \frac{\sin \theta}{\csc \theta} + \frac{\cos \theta}{\sec \theta} = \][/tex]
[tex]\[ \frac{\sin \theta}{\frac{1}{\sin \theta}} + \frac{\cos \theta}{\frac{1}{\cos \theta}} \][/tex]
Next, simplify each term:
[tex]\[ \frac{\sin \theta}{\frac{1}{\sin \theta}} = \sin \theta \cdot \sin \theta = \sin^2 \theta \][/tex]
[tex]\[ \frac{\cos \theta}{\frac{1}{\cos \theta}} = \cos \theta \cdot \cos \theta = \cos^2 \theta \][/tex]
So the equation now looks like:
[tex]\[ \sin^2 \theta + \cos^2 \theta \][/tex]
From trigonometric identities, we know that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Thus, substituting back, we have:
[tex]\[ 1 = 1 \][/tex]
Therefore, the given equation:
[tex]\[ \frac{\operatorname{sin} \theta}{\csc \theta} + \frac{\cos \theta}{\sec \theta} = 1 \][/tex]
is always satisfied for any value of [tex]\(\theta\)[/tex]. It means this is an identity, and the equation is true for all values of [tex]\(\theta\)[/tex].
