$$ \begin{array}{ll} ext { (e) } \sin ^{4} x-\cos ^{4} x=\sin ^{2} x-\cos ^{2} x & ext { (f) } \sin ^{2} heta+\sin ^{2} heta \cdot an ^{2} heta= an ^{2} heta \ ext { (g) } \cos heta(1+ an heta)=\cos heta+\sin heta & ext { (h) } \frac{1-\cos ^{2} heta}{\cos ^{2} heta+2 \cos heta+1}=\frac{1-\cos heta}{1+\cos heta} \ ext { (i) } \frac{1}{1-\cos heta}+\frac{1}{1+\cos heta}=\frac{2}{\sin ^{2} heta} & ext { (j) } \frac{1}{\sin heta}+\frac{1}{ an heta}=\frac{1+\cos heta}{\sin \left(180^{\circ}- heta ight)}\end{array} $$

Question

$$ \begin{array}{ll}	ext { (e) } \sin ^{4} x-\cos ^{4} x=\sin ^{2} x-\cos ^{2} x & 	ext { (f) } \sin ^{2} 	heta+\sin ^{2} 	heta \cdot 	an ^{2} 	heta=	an ^{2} 	heta \ 	ext { (g) } \cos 	heta(1+	an 	heta)=\cos 	heta+\sin 	heta & 	ext { (h) } \frac{1-\cos ^{2} 	heta}{\cos ^{2} 	heta+2 \cos 	heta+1}=\frac{1-\cos 	heta}{1+\cos 	heta} \ 	ext { (i) } \frac{1}{1-\cos 	heta}+\frac{1}{1+\cos 	heta}=\frac{2}{\sin ^{2} 	heta} & 	ext { (j) } \frac{1}{\sin 	heta}+\frac{1}{	an 	heta}=\frac{1+\cos 	heta}{\sin \left(180^{\circ}-	hetaight)}\end{array} $$

UpStudy AI · Accepted Answer

(e) We start with the left‐hand side:
  
$$
\sin^4 x - \cos^4 x = \left(\sin^2 x - \cos^2 x\right)\left(\sin^2 x + \cos^2 x\right).
$$
  
Using the Pythagorean identity $$\sin^2 x + \cos^2 x = 1$$, we have:
  
$$
\left(\sin^2 x - \cos^2 x\right)\cdot 1 = \sin^2 x - \cos^2 x.
$$
  
Thus, the identity holds.

---

(f) Write the left‐hand side as:
  
$$
\sin^2 \theta + \sin^2 \theta \cdot \tan^2 \theta = \sin^2 \theta (1+\tan^2 \theta).
$$
  
Recall the identity $$1+\tan^2 \theta = \sec^2 \theta$$, so:
  
$$
\sin^2 \theta \cdot \sec^2 \theta = \sin^2 \theta \cdot \frac{1}{\cos^2 \theta} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta.
$$
  
Thus, the equality is verified.

---

(g) Expand the left‐hand side:
  
$$
\cos \theta (1+\tan \theta) = \cos \theta + \cos \theta \cdot \tan \theta.
$$
  
Since $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$, it follows that:
  
$$
\cos \theta \cdot \tan \theta = \cos \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta.
$$
  
Therefore, we obtain:
  
$$
\cos \theta + \sin \theta,
$$
  
which is exactly the right‐hand side.

---

(h) Notice that the numerator can be factored using the difference of squares:
  
$$
1-\cos^2 \theta = (1-\cos \theta)(1+\cos \theta).
$$
  
Also, the denominator factors as:
  
$$
\cos^2 \theta + 2\cos \theta + 1 = (\cos \theta + 1)^2.
$$
  
Thus, the expression becomes:
  
$$
\frac{(1-\cos \theta)(1+\cos \theta)}{(\cos \theta+1)^2} = \frac{1-\cos \theta}{1+\cos \theta}.
$$

---

(i) To add the two fractions, first find a common denominator:
  
$$
\frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta} = \frac{(1+\cos \theta) + (1-\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}.
$$
  
Simplify the numerator:
  
$$
(1+\cos \theta) + (1-\cos \theta) = 2.
$$
  
Since
  
$$
(1-\cos \theta)(1+\cos \theta) = 1-\cos^2 \theta = \sin^2 \theta,
$$
  
the sum becomes:
  
$$
\frac{2}{\sin^2 \theta}.
$$

---

(j) Write the left‐hand side as:
  
$$
\frac{1}{\sin \theta} + \frac{1}{\tan \theta}.
$$
  
Since $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$, we have:
  
$$
\frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}.
$$
  
Thus, the expression becomes:
  
$$
\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \frac{1+\cos \theta}{\sin \theta}.
$$
  
Also, note that
  
$$
\sin(180^\circ-\theta) = \sin \theta.
$$
  
Therefore, the right‐hand side is
  
$$
\frac{1+\cos \theta}{\sin (180^\circ-\theta)} = \frac{1+\cos \theta}{\sin \theta},
$$
  
and the equality is established.
All the given trigonometric identities are verified as true through algebraic manipulation and the use of fundamental trigonometric identities.

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