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\( \begin{array}{ll}\text { (g) } \cos \theta(1+\tan \theta)=\cos \theta+\sin \theta & \text { (h) } \frac{1-\cos ^{2} \theta}{\cos ^{2} \theta+2 \cos \theta+1}=\frac{1-\cos \theta}{1+\cos \theta} \\ \text { (i) } \frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=\frac{2}{\sin ^{2} \theta} & \text { (j) } \frac{1}{\sin \theta}+\frac{1}{\tan \theta}=\frac{1+\cos \theta}{\sin \left(180^{\circ}-\theta\right)}\end{array} \)

Ask by Rodriguez Schneider. in South Africa
Mar 15,2025

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All the given trigonometric identities are true. ---

Solution

**(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.

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The Deep Dive

Did you know that the trigonometric identities you’re working with have deep historical roots? Ancient civilizations like the Greeks and Indians were the first to explore the relationships between angles and lengths in relation to the circle. Figures like Hipparchus and Aryabhata were pioneers in developing trigonometric tables that laid the groundwork for modern mathematics as we know it! In real life, trigonometric identities are incredibly useful beyond just classroom problems. Engineers use them when designing structures, pilots rely on them to calculate flight paths, and game developers implement them for animations and graphics. So the next time you’re wrestling with a tricky identity, remember that mathematicians and engineers just like you have transformed these abstract concepts into practical applications that shape our world!

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