Answer :
Certainly! Let's go through the given mathematical expression step-by-step to simplify it and verify the given equality:
[tex]\[ \operatorname{cosec}(180^\circ + \theta) \cdot \sin(360^\circ - \theta) - \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \cos^2 \theta \][/tex]
1. Simplify [tex]\(\operatorname{cosec}(180^\circ + \theta)\)[/tex]:
Recall the identity, [tex]\(\operatorname{cosec}(180^\circ + \theta) = \frac{1}{\sin(180^\circ + \theta)}\)[/tex].
Since [tex]\(\sin(180^\circ + \theta) = -\sin(\theta)\)[/tex], we get:
[tex]\[ \operatorname{cosec}(180^\circ + \theta) = \frac{1}{-\sin(\theta)} = -\frac{1}{\sin(\theta)} \][/tex]
2. Simplify [tex]\(\sin(360^\circ - \theta)\)[/tex]:
Using the identity [tex]\(\sin(360^\circ - \theta) = -\sin(\theta)\)[/tex], we have:
[tex]\[ \sin(360^\circ - \theta) = -\sin(\theta) \][/tex]
3. Simplify [tex]\(\left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ}\)[/tex]:
First, use the identity for [tex]\(\sin(180^\circ + \theta) = -\sin(\theta)\)[/tex], so:
[tex]\[ \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \left[ -\sin(\theta) \right]^{\sec 60^\circ} \][/tex]
Next, note that [tex]\(\sec(60^\circ) = \frac{1}{\cos(60^\circ)} = \frac{1}{\frac{1}{2}} = 2\)[/tex]. Thus, we get:
[tex]\[ \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \left[ -\sin(\theta) \right]^2 = \sin^2(\theta) \][/tex]
4. Combine all parts to form the left-hand side (LHS) of the original expression:
[tex]\[ \operatorname{cosec}(180^\circ + \theta) \cdot \sin(360^\circ - \theta) - \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} \][/tex]
Substitute the simplified terms:
[tex]\[ -\frac{1}{\sin(\theta)} \cdot (-\sin(\theta)) - \sin^2(\theta) \][/tex]
Simplify this product:
[tex]\[ 1 - \sin^2(\theta) \][/tex]
5. Compare the LHS with the right-hand side (RHS) of the equation:
The given RHS is [tex]\(\cos^2(\theta)\)[/tex], and we know from the Pythagorean identity that:
[tex]\[ 1 - \sin^2(\theta) = \cos^2(\theta) \][/tex]
Thus, we have:
[tex]\[ 1 - \sin^2(\theta) = \cos^2(\theta) \][/tex]
And therefore, the given equation holds true:
[tex]\[ \operatorname{cosec}(180^\circ + \theta) \cdot \sin(360^\circ - \theta) - \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \cos^2 \theta \][/tex]
So, the result is:
[tex]\((-1/\sin(\theta), -\sin(\theta), \sin(\theta)^2, 1 - \sin(\theta)^2, \cos(\theta)^2, -\sin(\theta)^2 - \cos(\theta)^2 + 1, \text{True})\)[/tex]
Yes, the equation is true!
[tex]\[ \operatorname{cosec}(180^\circ + \theta) \cdot \sin(360^\circ - \theta) - \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \cos^2 \theta \][/tex]
1. Simplify [tex]\(\operatorname{cosec}(180^\circ + \theta)\)[/tex]:
Recall the identity, [tex]\(\operatorname{cosec}(180^\circ + \theta) = \frac{1}{\sin(180^\circ + \theta)}\)[/tex].
Since [tex]\(\sin(180^\circ + \theta) = -\sin(\theta)\)[/tex], we get:
[tex]\[ \operatorname{cosec}(180^\circ + \theta) = \frac{1}{-\sin(\theta)} = -\frac{1}{\sin(\theta)} \][/tex]
2. Simplify [tex]\(\sin(360^\circ - \theta)\)[/tex]:
Using the identity [tex]\(\sin(360^\circ - \theta) = -\sin(\theta)\)[/tex], we have:
[tex]\[ \sin(360^\circ - \theta) = -\sin(\theta) \][/tex]
3. Simplify [tex]\(\left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ}\)[/tex]:
First, use the identity for [tex]\(\sin(180^\circ + \theta) = -\sin(\theta)\)[/tex], so:
[tex]\[ \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \left[ -\sin(\theta) \right]^{\sec 60^\circ} \][/tex]
Next, note that [tex]\(\sec(60^\circ) = \frac{1}{\cos(60^\circ)} = \frac{1}{\frac{1}{2}} = 2\)[/tex]. Thus, we get:
[tex]\[ \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \left[ -\sin(\theta) \right]^2 = \sin^2(\theta) \][/tex]
4. Combine all parts to form the left-hand side (LHS) of the original expression:
[tex]\[ \operatorname{cosec}(180^\circ + \theta) \cdot \sin(360^\circ - \theta) - \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} \][/tex]
Substitute the simplified terms:
[tex]\[ -\frac{1}{\sin(\theta)} \cdot (-\sin(\theta)) - \sin^2(\theta) \][/tex]
Simplify this product:
[tex]\[ 1 - \sin^2(\theta) \][/tex]
5. Compare the LHS with the right-hand side (RHS) of the equation:
The given RHS is [tex]\(\cos^2(\theta)\)[/tex], and we know from the Pythagorean identity that:
[tex]\[ 1 - \sin^2(\theta) = \cos^2(\theta) \][/tex]
Thus, we have:
[tex]\[ 1 - \sin^2(\theta) = \cos^2(\theta) \][/tex]
And therefore, the given equation holds true:
[tex]\[ \operatorname{cosec}(180^\circ + \theta) \cdot \sin(360^\circ - \theta) - \left[ \sin(180^\circ + \theta) \right]^{\sec 60^\circ} = \cos^2 \theta \][/tex]
So, the result is:
[tex]\((-1/\sin(\theta), -\sin(\theta), \sin(\theta)^2, 1 - \sin(\theta)^2, \cos(\theta)^2, -\sin(\theta)^2 - \cos(\theta)^2 + 1, \text{True})\)[/tex]
Yes, the equation is true!
