$$ \begin{array}{ll} ext { (g) } \cos heta(1+ an heta)=\cos heta+\sin heta & ext { (h) } \frac{1-\cos ^{2} heta}{\cos ^{2} heta+2 \cos heta+1}=\frac{1-\cos heta}{1+\cos heta} \ ext { (i) } \frac{1}{1-\cos heta}+\frac{1}{1+\cos heta}=\frac{2}{\sin ^{2} heta} & ext { (j) } \frac{1}{\sin heta}+\frac{1}{ an heta}=\frac{1+\cos heta}{\sin \left(180^{\circ}- heta ight)}\end{array} $$

Question

$$ \begin{array}{ll}	ext { (g) } \cos 	heta(1+	an 	heta)=\cos 	heta+\sin 	heta & 	ext { (h) } \frac{1-\cos ^{2} 	heta}{\cos ^{2} 	heta+2 \cos 	heta+1}=\frac{1-\cos 	heta}{1+\cos 	heta} \ 	ext { (i) } \frac{1}{1-\cos 	heta}+\frac{1}{1+\cos 	heta}=\frac{2}{\sin ^{2} 	heta} & 	ext { (j) } \frac{1}{\sin 	heta}+\frac{1}{	an 	heta}=\frac{1+\cos 	heta}{\sin \left(180^{\circ}-	hetaight)}\end{array} $$

UpStudy AI · Accepted Answer

**(g)**  
Starting with the left-hand side,  
$$
\cos \theta(1+\tan \theta)=\cos \theta+\cos \theta \tan \theta.
$$  
Recall that  
$$
\tan \theta=\frac{\sin \theta}{\cos \theta},
$$  
so  
$$
\cos \theta\tan \theta=\cos \theta\left(\frac{\sin \theta}{\cos \theta}\right)=\sin \theta.
$$  
Thus,  
$$
\cos \theta(1+\tan \theta)=\cos \theta+\sin \theta,
$$  
which equals the right-hand side.

---

**(h)**  
Consider the expression  
$$
\frac{1-\cos ^{2} \theta}{\cos ^{2} \theta+2 \cos \theta+1}.
$$  
Notice that the numerator is a difference of squares:  
$$
1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta).
$$  
The denominator can be written as a perfect square:  
$$
\cos^2\theta+2\cos\theta+1=(\cos \theta+1)^2.
$$  
Thus, the expression becomes  
$$
\frac{(1-\cos\theta)(1+\cos\theta)}{(\cos\theta+1)^2}=\frac{1-\cos\theta}{1+\cos\theta},
$$  
assuming $$\cos\theta\neq -1$$. This matches the right-hand side.

---

**(i)**  
Examine the left-hand side:  
$$
\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}.
$$  
To combine the fractions, find a common denominator:  
$$
(1-\cos\theta)(1+\cos\theta)=1-\cos^2\theta.
$$  
Thus,  
$$
\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=\frac{(1+\cos\theta)+(1-\cos\theta)}{1-\cos^2\theta}.
$$  
Simplify the numerator:  
$$
(1+\cos\theta)+(1-\cos\theta)=2.
$$  
Also, use the Pythagorean identity  
$$
1-\cos^2\theta=\sin^2\theta.
$$  
Therefore,  
$$
\frac{2}{\sin^2\theta},
$$  
which is the simplified form on the right-hand side.

---

**(j)**  
For the left-hand side, we have  
$$
\frac{1}{\sin \theta}+\frac{1}{\tan \theta}.
$$  
Write $$\frac{1}{\tan \theta}$$ in terms of sine and cosine:  
$$
\frac{1}{\tan \theta}=\frac{\cos \theta}{\sin \theta}.
$$  
Now,  
$$
\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\frac{1+\cos \theta}{\sin \theta}.
$$  
On the right-hand side, we have  
$$
\frac{1+\cos \theta}{\sin (180^\circ-\theta)}.
$$  
Recall that  
$$
\sin (180^\circ-\theta)=\sin \theta.
$$  
Thus,  
$$
\frac{1+\cos \theta}{\sin (180^\circ-\theta)}=\frac{1+\cos \theta}{\sin \theta},
$$  
which is identical to the expression obtained for the left-hand side.

---

Each identity is verified as true.
All the given trigonometric identities are true. ---

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