(b) \( \frac{\sin ^{3} x+\sin x \cdot \cos ^{2} x}{\cos x}=\tan x \) (d) \( 1-\frac{\cos ^{2} x}{1+\sin x}=\sin x \) (f) \( \frac{1}{(\cos x+1)(\cos x-1)}=\frac{-1}{\tan ^{2} x \cdot \cos ^{2} x} \) (h) \( \left(\tan y+\frac{1}{\tan y}\right)\left(1-\cos ^{2} y\right)=\tan y \)

Let's simplify and verify each of the given equations step by step. ### (b) \( \frac{\sin ^{3} x+\sin x \cdot \cos ^{2} x}{\cos x}=\tan x \) **Left Side:** \[ \frac{\sin^3 x + \sin x \cos^2 x}{\cos x} = \frac{\sin x (\sin^2 x + \cos^2 x)}{\cos x} \] Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \): \[ = \frac{\sin x \cdot 1}{\cos x} = \frac{\sin x}{\cos x} = \tan x \] **Conclusion:** The equation is true. --- ### (d) \( 1-\frac{\cos ^{2} x}{1+\sin x}=\sin x \) **Left Side:** \[ 1 - \frac{\cos^2 x}{1 + \sin x} \] Using the identity \( \cos^2 x = 1 - \sin^2 x \): \[ = 1 - \frac{1 - \sin^2 x}{1 + \sin x} = 1 - \frac{1}{1 + \sin x} + \frac{\sin^2 x}{1 + \sin x} \] Combining the terms: \[ = \frac{(1 + \sin x) - 1 + \sin^2 x}{1 + \sin x} = \frac{\sin x + \sin^2 x}{1 + \sin x} \] Factoring out \(\sin x\): \[ = \frac{\sin x(1 + \sin x)}{1 + \sin x} = \sin x \quad (\text{for } 1 + \sin x \neq 0) \] **Conclusion:** The equation is true. --- ### (f) \( \frac{1}{(\cos x+1)(\cos x-1)}=\frac{-1}{\tan ^{2} x \cdot \cos ^{2} x} \) **Left Side:** \[ \frac{1}{(\cos x + 1)(\cos x - 1)} = \frac{1}{\cos^2 x - 1} = \frac{1}{-\sin^2 x} = -\frac{1}{\sin^2 x} \] **Right Side:** \[ \frac{-1}{\tan^2 x \cdot \cos^2 x} = \frac{-1}{\frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x} = \frac{-1}{\sin^2 x} \] **Conclusion:** The equation is true. --- ### (h) \( \left(\tan y+\frac{1}{\tan y}\right)\left(1-\cos ^{2} y\right)=\tan y \) **Left Side:** \[ \left(\tan y + \frac{1}{\tan y}\right)(1 - \cos^2 y) = \left(\tan y + \frac{1}{\tan y}\right) \sin^2 y \] Using \( \tan y = \frac{\sin y}{\cos y} \): \[ = \left(\frac{\sin y}{\cos y} + \frac{\cos y}{\sin y}\right) \sin^2 y = \left(\frac{\sin^2 y + \cos^2 y}{\sin y \cos y}\right) \sin^2 y \] Using the identity \( \sin^2 y + \cos^2 y = 1 \): \[ = \frac{1}{\sin y \cos y} \sin^2 y = \frac{\sin^2 y}{\sin y \cos y} = \frac{\sin y}{\cos y} = \tan y \] **Conclusion:** The equation is true. --- ### Summary All equations (b), (d), (f), and (h) are verified to be true.