If $$ \sin heta=p $$ and $$ heta $$ is obtuse, find expressions for $$ \cos heta $$ and $$ an heta $$ in of $$ p $$. If $$ an heta=\frac{a}{b} $$, where $$ heta $$ is an acute angle, find the values of (a) $$ \frac{a \cos heta-b^{\prime} \sin heta}{a \cos heta+b \sin heta} $$ (b) $$ \frac{b \cos heta-a \sin heta}{b \cos heta+a \sin heta} $$ Given that $$ an heta=\frac{2 t}{1-t^{2}} $$, and that $$ heta $$ is an acute angle, express sine $$ \cos heta $$ in terms of $$ t $$. Simplify (a) $$ \cos heta an heta $$ (b) $$ \frac{1-\cos ^{2} heta}{\sin ^{2} heta} $$ (c) $$ \left(1-\cos ^{2} heta ight) \sec ^{2} heta $$ (d) $$ \left(1+ an ^{2} heta ight) \cos ^{2} heta $$ (e) $$ \frac{1}{1-\sin heta}+\frac{1}{1+\sin heta} $$ (f) $$ \sec heta-\sin heta an heta $$ (g) $$ \frac{\sec heta}{\cos heta}-\frac{ an heta}{\cot heta} $$ 6 Show that

Question

If $$ \sin 	heta=p $$ and $$ 	heta $$ is obtuse, find expressions for $$ \cos 	heta $$ and $$ 	an 	heta $$ in of $$ p $$. If $$ 	an 	heta=\frac{a}{b} $$, where $$ 	heta $$ is an acute angle, find the values of (a) $$ \frac{a \cos 	heta-b^{\prime} \sin 	heta}{a \cos 	heta+b \sin 	heta} $$ (b) $$ \frac{b \cos 	heta-a \sin 	heta}{b \cos 	heta+a \sin 	heta} $$ Given that $$ 	an 	heta=\frac{2 t}{1-t^{2}} $$, and that $$ 	heta $$ is an acute angle, express sine $$ \cos 	heta $$ in terms of $$ t $$. Simplify (a) $$ \cos 	heta 	an 	heta $$ (b) $$ \frac{1-\cos ^{2} 	heta}{\sin ^{2} 	heta} $$ (c) $$ \left(1-\cos ^{2} 	hetaight) \sec ^{2} 	heta $$ (d) $$ \left(1+	an ^{2} 	hetaight) \cos ^{2} 	heta $$ (e) $$ \frac{1}{1-\sin 	heta}+\frac{1}{1+\sin 	heta} $$ (f) $$ \sec 	heta-\sin 	heta 	an 	heta $$ (g) $$ \frac{\sec 	heta}{\cos 	heta}-\frac{	an 	heta}{\cot 	heta} $$ 6 Show that

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part 1: Finding expressions for $$ \cos \theta $$ and $$ \tan \theta $$

Given that $$ \sin \theta = p $$ and $$ \theta $$ is obtuse, we know that:

1. **Using the Pythagorean identity**:
   $$
   \cos^2 \theta + \sin^2 \theta = 1
   $$
   Substituting $$ \sin \theta = p $$:
   $$
   \cos^2 \theta + p^2 = 1
   $$
   Therefore,
   $$
   \cos^2 \theta = 1 - p^2
   $$
   Since $$ \theta $$ is obtuse, $$ \cos \theta $$ is negative:
   $$
   \cos \theta = -\sqrt{1 - p^2}
   $$

2. **Finding $$ \tan \theta $$**:
   $$
   \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{p}{-\sqrt{1 - p^2}} = -\frac{p}{\sqrt{1 - p^2}}
   $$

### Part 2: Finding values of expressions involving $$ \tan \theta = \frac{a}{b} $$

Given $$ \tan \theta = \frac{a}{b} $$:

(a) We need to find:
$$
\frac{a \cos \theta - b' \sin \theta}{a \cos \theta + b \sin \theta}
$$
Using $$ \sin \theta = \frac{a}{\sqrt{a^2 + b^2}} $$ and $$ \cos \theta = \frac{b}{\sqrt{a^2 + b^2}} $$:
$$
\frac{a \cdot \frac{b}{\sqrt{a^2 + b^2}} - b' \cdot \frac{a}{\sqrt{a^2 + b^2}}}{a \cdot \frac{b}{\sqrt{a^2 + b^2}} + b \cdot \frac{a}{\sqrt{a^2 + b^2}}}
$$
This simplifies to:
$$
\frac{ab - b'a}{ab + ba} = \frac{ab - b'a}{2ab}
$$

(b) For:
$$
\frac{b \cos \theta - a \sin \theta}{b \cos \theta + a \sin \theta}
$$
This simplifies similarly:
$$
\frac{b \cdot \frac{b}{\sqrt{a^2 + b^2}} - a \cdot \frac{a}{\sqrt{a^2 + b^2}}}{b \cdot \frac{b}{\sqrt{a^2 + b^2}} + a \cdot \frac{a}{\sqrt{a^2 + b^2}}}
$$
This simplifies to:
$$
\frac{b^2 - a^2}{b^2 + a^2}
$$

### Part 3: Expressing $$ \sin \theta $$ and $$ \cos \theta $$ in terms of $$ t $$

Given $$ \tan \theta = \frac{2t}{1 - t^2} $$:
1. **Finding $$ \sin \theta $$**:
   $$
   \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}} = \frac{\frac{2t}{1 - t^2}}{\sqrt{1 + \left(\frac{2t}{1 - t^2}\right)^2}}
   $$

2. **Finding $$ \cos \theta $$**:
   $$
   \cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}} = \frac{1 - t^2}{\sqrt{(1 - t^2)^2 + 4t^2}}
   $$

### Part 4: Simplifying the given expressions

(a) $$ \cos \theta \tan \theta $$:
$$
\cos \theta \tan \theta = \cos \theta \cdot \frac{2t}{1 - t^2}
$$

(b) $$ \frac{1 - \cos^2 \theta}{\sin^2 \theta} $$:
Using $$ 1 - \cos^2 \theta = \sin^2 \theta $$:
$$
\frac{\sin^2 \theta}{\sin^2 \theta} = 1
$$

(c) $$ (1 - \cos^2 \theta) \sec^2 \theta $$:
$$
\sin^2 \theta \sec^2 \theta = \sin^2 \theta \cdot \frac{1}{\cos^2 \theta}
$$

(d) $$ (1 + \tan^2 \theta) \cos^2 \theta $$:
$$
\sec^2 \theta \cos^2 \theta = 1
$$

(e) $$ \frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} $$:
This can be simplified using a common denominator.

(f) $$ \sec \theta - \sin \theta \tan \theta $$:
This can be simplified using the definitions of secant and tangent.

(g) $$ \frac{\sec \theta}{\cos \theta} - \frac{\tan \theta}{\cot \theta} $$:
This simplifies to $$ \sec^2 \theta - 1 $$.

### Conclusion
The expressions have been derived and simplified step by step. If you need further calculations or specific simplifications, please let me know!
$$ \cos \theta = -\sqrt{1 - p^2} $$ and $$ \tan \theta = -\frac{p}{\sqrt{1 - p^2}} $$. For $$ \tan \theta = \frac{a}{b} $$: (a) $$ \frac{a \cos \theta - b' \sin \theta}{a \cos \theta + b \sin \theta} = \frac{ab - b'a}{2ab} $$ (b) $$ \frac{b \cos \theta - a \sin \theta}{b \cos \theta + a \sin \theta} = \frac{b^2 - a^2}{b^2 + a^2} $$ Given $$ \tan \theta = \frac{2t}{1 - t^2} $$: $$ \sin \theta = \frac{2t}{\sqrt{1 + \left(\frac{2t}{1 - t^2}\right)^2}} $$ $$ \cos \theta = \frac{1 - t^2}{\sqrt{(1 - t^2)^2 + 4t^2}} $$ Simplified expressions: (a) $$ \cos \theta \tan \theta = \cos \theta \cdot \frac{2t}{1 - t^2} $$ (b) $$ \frac{1 - \cos^2 \theta}{\sin^2 \theta} = 1 $$ (c) $$ (1 - \cos^2 \theta) \sec^2 \theta = \sin^2 \theta \cdot \frac{1}{\cos^2 \theta} $$ (d) $$ (1 + \tan^2 \theta) \cos^2 \theta = 1 $$ (e) $$ \frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} $$ (simplify using a common denominator) (f) $$ \sec \theta - \sin \theta \tan \theta $$ (simplify using definitions) (g) $$ \frac{\sec \theta}{\cos \theta} - \frac{\tan \theta}{\cot \theta} = \sec^2 \theta - 1 $$

If and is obtuse, find expressions for and in of .
If , where is an acute angle, find the values of
(a)
(b)

Given that , and that is an acute angle, express sine in terms of .
Simplify (a)
(b)
©
(d)
(e)
(f)
(g)

6 Show that

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Solución

Part 1: Finding expressions for and

Part 2: Finding values of expressions involving

Part 3: Expressing and in terms of

Part 4: Simplifying the given expressions

Conclusion

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Latest Trigonometry Questions

If and is obtuse, find expressions for and in of . If , where is an acute angle, find the values of (a) (b) Given that , and that is an acute angle, express sine in terms of . Simplify (a) (b) © (d) (e) (f) (g) 6 Show that