(An important result to remember) (d) \( \frac{\tan x \cdot \cos x}{\sin x}=1 \) (f) \( \sin ^{2} \theta+\sin ^{2} \theta \cdot \tan ^{2} \theta=\tan ^{2} \theta \) (h) \( \frac{1-\cos ^{2} \theta}{\cos ^{2} \theta+2 \cos \theta+1}=\frac{1-\cos \theta}{1+\cos \theta} \) (i) \( \frac{1}{\sin \theta}+\frac{1}{\tan \theta}=\frac{1+\cos \theta}{\sin \left(180^{\circ}-\theta\right)} \)

Let's analyze and simplify each of the given equations step by step. ### (d) \( \frac{\tan x \cdot \cos x}{\sin x}=1 \) 1. **Start with the left side**: \[ \frac{\tan x \cdot \cos x}{\sin x} \] Recall that \( \tan x = \frac{\sin x}{\cos x} \). Substitute this into the equation: \[ \frac{\frac{\sin x}{\cos x} \cdot \cos x}{\sin x} \] 2. **Simplify**: The \( \cos x \) in the numerator cancels with the \( \cos x \) in the denominator: \[ \frac{\sin x}{\sin x} = 1 \] Thus, the equation holds true. ### (f) \( \sin ^{2} \theta+\sin ^{2} \theta \cdot \tan ^{2} \theta=\tan ^{2} \theta \) 1. **Start with the left side**: \[ \sin^2 \theta + \sin^2 \theta \cdot \tan^2 \theta \] Recall that \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \). Substitute this into the equation: \[ \sin^2 \theta + \sin^2 \theta \cdot \frac{\sin^2 \theta}{\cos^2 \theta} \] 2. **Factor out \( \sin^2 \theta \)**: \[ \sin^2 \theta \left(1 + \frac{\sin^2 \theta}{\cos^2 \theta}\right) \] 3. **Combine the terms**: \[ \sin^2 \theta \left(\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\right) \] Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \sin^2 \theta \cdot \frac{1}{\cos^2 \theta} = \tan^2 \theta \] Thus, the equation holds true. ### (h) \( \frac{1-\cos ^{2} \theta}{\cos ^{2} \theta+2 \cos \theta+1}=\frac{1-\cos \theta}{1+\cos \theta} \) 1. **Start with the left side**: \[ \frac{1 - \cos^2 \theta}{\cos^2 \theta + 2 \cos \theta + 1} \] Recall that \( 1 - \cos^2 \theta = \sin^2 \theta \) and \( \cos^2 \theta + 2 \cos \theta + 1 = (\cos \theta + 1)^2 \): \[ \frac{\sin^2 \theta}{(\cos \theta + 1)^2} \] 2. **Now simplify the right side**: \[ \frac{1 - \cos \theta}{1 + \cos \theta} \] Multiply numerator and denominator by \( 1 - \cos \theta \): \[ \frac{(1 - \cos \theta)^2}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{(1 - 2\cos \theta + \cos^2 \theta)}{\sin^2 \theta} \] This shows that both sides are equivalent, thus the equation holds true. ### (i) \( \frac{1}{\sin \theta}+\frac{1}{\tan \theta}=\frac{1+\cos \theta}{\sin \left(180^{\circ}-\theta\right)} \) 1. **Start with the left side**: \[ \frac{1}{\sin \theta} + \frac{1}{\tan \theta} \] Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \): \[ \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta} \] 2. **Now simplify the right side**: \[ \frac{1 + \cos \theta}{\sin(180^{\circ} - \theta)} = \frac{1 + \cos \theta}{\sin \theta} \] Thus, both sides are equal, confirming the equation holds true. In conclusion, all the equations provided are valid and hold true under the given trigonometric identities.