$$ heta_{1}= heta_{2}= heta_{2}=2 $$ If $$ 1+\sin heta=9 \cos heta $$ and $$ 0^{\circ}

Question

$$ 	heta_{1}=	heta_{2}=	heta_{2}=2 $$ If $$ 1+\sin 	heta=9 \cos 	heta $$ and $$ 0^{\circ}<	heta<90^{\circ} $$, then analyse and assess the value of $$ 5(1-\cos 	heta) $$. $$ 1+\sin 	heta=9 \cos 	heta $$

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We are given the equation 
$$
1 + \sin \theta = 9 \cos \theta
$$
with $$0^\circ < \theta < 90^\circ$$ and need to find the value of 
$$
5(1 - \cos \theta).
$$

**Step 1. Solve for $$\sin \theta$$ and $$\cos \theta$$.**

Starting with
$$
1 + \sin \theta = 9 \cos \theta,
$$
we can rearrange this as
$$
\sin \theta = 9 \cos \theta - 1.
$$

**Step 2. Square both sides and use the Pythagorean identity.**

Square the equation:
$$
(1 + \sin \theta)^2 = (9 \cos \theta)^2.
$$
Expanding and simplifying, we have:
$$
1 + 2 \sin \theta + \sin^2 \theta = 81 \cos^2 \theta.
$$
Using the identity $$\sin^2 \theta + \cos^2 \theta = 1$$, substitute $$\sin^2 \theta = 1 - \cos^2 \theta$$:
$$
1 + 2 \sin \theta + (1 - \cos^2 \theta) = 81 \cos^2 \theta.
$$
This simplifies to:
$$
2 + 2 \sin \theta - \cos^2 \theta = 81 \cos^2 \theta.
$$
Bring the $$\cos^2 \theta$$ term to the right:
$$
2 + 2 \sin \theta = 82 \cos^2 \theta.
$$

Although we could proceed by expressing $$\cos^2 \theta$$ in terms of $$\sin \theta$$, a more straightforward method is to find $$\sin \theta$$ directly by another approach.

**Step 3. Solve by forming a quadratic in $$\sin \theta$$.**

Express the original equation $$1 + \sin \theta = 9 \cos \theta$$ in a different way by isolating $$\cos \theta$$:
$$
\cos \theta = \frac{1+\sin \theta}{9}.
$$
Now substitute this in the identity:
$$
\sin^2 \theta + \cos^2 \theta = 1.
$$
Thus,
$$
\sin^2 \theta + \left(\frac{1+\sin \theta}{9}\right)^2 = 1.
$$
Multiply through by $$81$$ (which is $$9^2$$) to eliminate the denominator:
$$
81 \sin^2 \theta + (1+\sin \theta)^2 = 81.
$$
Expand $$(1+\sin \theta)^2$$:
$$
81 \sin^2 \theta + 1 + 2 \sin \theta + \sin^2 \theta = 81.
$$
Combine like terms:
$$
82 \sin^2 \theta + 2 \sin \theta + 1 - 81 = 0,
$$
$$
82 \sin^2 \theta + 2 \sin \theta - 80 = 0.
$$
Divide the entire equation by $$2$$:
$$
41 \sin^2 \theta + \sin \theta - 40 = 0.
$$
Let $$t = \sin \theta$$. The quadratic is
$$
41t^2 + t - 40 = 0.
$$
The discriminant is:
$$
\Delta = 1^2 - 4(41)(-40) = 1 + 6560 = 6561.
$$
Since $$\sqrt{6561} = 81$$, the solutions for $$t$$ are:
$$
t = \frac{-1 \pm 81}{2 \cdot 41}.
$$
This gives:
$$
t = \frac{80}{82} = \frac{40}{41} \quad \text{or} \quad t = \frac{-82}{82} = -1.
$$
Since $$0^\circ < \theta < 90^\circ$$, we take the positive solution:
$$
\sin \theta = \frac{40}{41}.
$$

**Step 4. Find $$\cos \theta$$.**

Substitute the value of $$\sin \theta$$ back into the equation:
$$
1 + \sin \theta = 9 \cos \theta.
$$
Thus,
$$
1+\frac{40}{41} = 9 \cos \theta,
$$
$$
\frac{41+40}{41} = 9\cos \theta,
$$
$$
\frac{81}{41} = 9\cos \theta.
$$
So,
$$
\cos \theta = \frac{81}{41 \cdot 9} = \frac{9}{41}.
$$

**Step 5. Compute $$5(1-\cos \theta)$$.**

Now, substitute $$\cos \theta = \frac{9}{41}$$ into the expression:
$$
5(1-\cos \theta) = 5\left(1-\frac{9}{41}\right).
$$
Simplify inside the parentheses:
$$
1-\frac{9}{41} = \frac{41-9}{41} = \frac{32}{41}.
$$
Thus,
$$
5(1-\cos \theta) = 5 \cdot \frac{32}{41} = \frac{160}{41}.
$$

$$
\boxed{\frac{160}{41}}
$$
$$5(1 - \cos \theta) = \frac{160}{41}$$.

\( \theta_{1}=\theta_{2}=\theta_{2}=2 \) If \( 1+\sin \theta=9 \cos \theta \) and \( 0^{\circ}<\theta<90^{\circ} \), then analyse and assess the value of \( 5(1-\cos \theta) \). \( 1+\sin \theta=9 \cos \theta \)

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