Partial Solutions


  1. There is no solution provided for this question.
    1. Non-permissible values: \( \tan{(\theta)} \) is undefined when \( \cos{(\theta)}=0 \) so \( \theta \neq \frac{\pi}{2}+n\pi, n\in\mathbb{Z} \).
    2. \[\begin{align*} & 2\sin{\left(\dfrac{\pi}{6}\right)}\tan{\left(\dfrac{\pi}{6}\right)} -1 \\ & = 2\left(\dfrac{1}{2}\right)\left(\dfrac{1}{\sqrt{3}}\right)-1 \\ & = \dfrac{\sqrt{3}}{3}-1 \\ & = \dfrac{\sqrt{3}-3}{3}\\ \end{align*}\]
      \[\begin{align*} & \tan{\left(\dfrac{\pi}{6}\right)} -2 \sin{\left(\dfrac{\pi}{6}\right)} \\ & = \dfrac{1}{\sqrt{3}} - 2\left(\dfrac{1}{2}\right) \\ & = \dfrac{\sqrt{3}}{3}-1 \\ & = \dfrac{\sqrt{3}-3}{3}\\ \end{align*}\]
      So, \(2\sin{\left(\dfrac{\pi}{6}\right)}\tan{\left(\dfrac{\pi}{6}\right)} -1 = \tan{\left(\dfrac{\pi}{6}\right)} -2 \sin{\left(\dfrac{\pi}{6}\right)}\)
      \[\begin{align*} & 2\sin{\left(\dfrac{3\pi}{4}\right)}\tan{\left(\dfrac{3\pi}{4}\right)} -1 \\ & = 2\left(\dfrac{1}{\sqrt{2}}\right)\left(-1\right)-1 \\ & = -\sqrt{2}-1 \\ \end{align*}\]
      \[\begin{align*} & \tan{\left(\dfrac{3\pi}{4}\right)} -2 \sin{\left(\dfrac{3\pi}{4}\right)} \\ & = -1 - 2\left(\dfrac{1}{\sqrt{2}}\right) \\ & = -1 - \sqrt{2}\\ \end{align*}\]

      So, \(2\sin{\left(\dfrac{3\pi}{4}\right)}\tan{\left(\dfrac{3\pi}{4}\right)} -1 = \tan{\left(\dfrac{3\pi}{4}\right)} -2 \sin{\left(\dfrac{3\pi}{4}\right)}\). Therefore, the equation is true for \( x = \dfrac{\pi}{6} \) and \( x = \dfrac{3\pi}{4} \).

      This is not sufficient evidence to conclude that \( 2\sin{(x)}\tan{(x)} -1= \tan{(x)} -2 \sin{(x)} \) is an identity since this does not prove that the statement is true for all permissible values of the variable \( x \).

    3. The graphs of the two functions are different, although they share common points such as \( \left(\dfrac{\pi}{6},\dfrac{\sqrt{3}-3}{3}\right) \) and \( \left(\dfrac{3\pi}{4},-1-\sqrt{2}\right) \). Therefore, \( 2\sin{(x)}\tan{(x)} -1= \tan{(x)} -2 \sin{(x)} \) is not an identity (see below diagram). Graphs of both functions, not identical
  3. There is no solution provided for this question.
    1. \( \tan{(x)} \) is undefined when \( \cos{(x)}=0 \) so \( x \neq \dfrac{\pi}{2}+n\pi, n\in\mathbb{Z} \). The denominator \( 1+\cos{(x)} \) is zero when \( \cos{(x)}=-1 \), so \( x \neq \pi+2\pi n, n\in\mathbb{Z} \). Therefore, \( x \neq \dfrac{\pi}{2}+n\pi, n\in\mathbb{Z} \) and \( x \neq \pi+2\pi n\), \(n\in\mathbb{Z} \).

    2. Graph of f(x) = (tan(x)+sin(x))/(1+cos(x)); holes at non-permissible values given in part a

      Note: Holes in the graph of a function may not be obvious when graphing with technology.

    3. Since the graph of the function appears to be equivalent to the graph of \( y=\tan{(x)} \) for permissible values of \( x \), a possible identity could be \( \dfrac{\tan{(x)} + \sin{(x)}}{1 + \cos{(x)}}= \tan{(x)} \).
    4.  \[\begin{align*} L.S. &= \dfrac{\tan{(x)} +\sin{(x)}}{1+\cos{(x)}} \\ &= \dfrac{\frac{\sin{(x)}}{\cos{(x)}}+\sin{(x)}}{1+\cos{(x)}}\\ &= \left(\dfrac{\sin{(x)}+\sin{(x)}\cos{(x)}}{\cos{(x)}}\right) \times \dfrac{1}{1+\cos{(x)}}\\ &= \dfrac{\sin{(x)}\left(1+\cos{(x)}\right)}{\cos{(x)}\left(1+\cos{(x)}\right)} \\ &= \tan{(x)}\\ &= R.S. \end{align*}\]
  5. There is no solution provided for this question.
  6. Proofs may vary.
    1.  \[\begin{align*} L.S. &= \csc^2{(x)} - \csc{(x)}\cot{(x)} \\ &= \dfrac{1}{\sin^2{(x)}} - \dfrac{1}{\sin{(x)}}\left(\dfrac{\cos{(x)}}{\sin{(x)}}\right) \\ &= \dfrac{1-\cos{(x)}}{\sin^2{(x)}} \\ &= \dfrac{1-\cos{(x)}}{1-\cos^2{(x)}}\\ &= \dfrac{1-\cos{(x)}}{(1-\cos{(x)})(1+\cos{(x)})}\\ &= \dfrac{1}{1+\cos{(x)}}\\ &= R.S. \end{align*}\]
    2.  \[\begin{align*} R.S. &= \Big(\tan{(\theta)} + \sec{(\theta)}\Big)^{2} \\ &= \left(\dfrac{\sin{(\theta)}}{\cos{(\theta)}} + \dfrac{1}{\cos{(\theta)}}\right)^{2} \\ &= \left(\dfrac{\sin{(\theta)}+1}{\cos{(\theta)}} \right)^{2} \\ &= \dfrac{\left(\sin{(\theta)} + 1\right)^2}{\cos^2{(\theta)}}\\ &= \dfrac{\left(\sin{(\theta)} + 1\right)^2}{1-\sin^2{(\theta)}}\\ &= \dfrac{(\sin{(x)} + 1)^{2}}{(1+\sin{(x)})(1-\sin{(x)})}\\ &= \dfrac{\sin{(x)} + 1}{1-\sin{(x)}}\\ &= L.S. \end{align*}\]
    3.  \[\begin{align*} L.S. &= \dfrac{\tan{(\beta)} - \sin{(\beta)}}{\sin^3{(\beta)}} \\ &= \left(\dfrac{\sin{(\beta)}}{\cos{(\beta)}} - \sin{(\beta)}\right) \times \dfrac{1}{\sin^3{(\beta)}} \\ &= \dfrac{\sin{(\beta)} - \sin{(\beta)}\cos{(\beta)}}{\cos{(\beta)}\sin^3{(\beta)}} \\ &= \dfrac{\sin{(\beta)}\left(1-\cos{(\beta)}\right)}{\cos{(\beta)} \sin^3{(\beta)}} \\ &= \dfrac{1-\cos{(\beta)}}{\cos{(\beta)} \sin^2{(\beta)}} \\ &= \dfrac{(1-\cos{(\beta)}) }{\cos{(\beta)}\left(1-\cos^2{(\beta)}\right)} \\ &= \dfrac{(1-\cos{(\beta)})}{\cos{(\beta)}(1-\cos{(\beta)})(1+\cos{(\beta)})} \\ &= \dfrac{1}{\cos{(\beta)}} \times \dfrac{1}{(1+\cos{(\beta)})}\\ &= \dfrac{\sec{(\beta)}}{(1+\cos{(\beta)})}\\ &= R.S. \end{align*}\]
    4. Factoring the sum of cubes, \(\cos^3{(\theta)}+\sin^3{(\theta)}\), using \(a^3+b^3 = a^2 -ab +b^2\), where \( a = \cos{(\theta)} \) and \( b = \sin {(\theta)}\),\[\begin{align*} L.S. &= \dfrac{\cos^3{(\theta)}+\sin^3{(\theta)}}{\sin{(\theta)}+\cos{(\theta)}} \\ &= \dfrac{\left(\cos{(\theta)}+\sin{(\theta)}\right)\left(\cos^2{(\theta)}-\sin{(\theta)}\cos{(\theta)} + \sin^2{(\theta)}\right)}{\sin{(\theta)}+\cos{(\theta)}} \\ &= \sin^2{(\theta)}+ \cos^2{(\theta)} -\sin{(\theta)}\cos{(\theta)} \\ &= 1 - \sin{(\theta)}\cos{(\theta)}\\ &= R.S. \end{align*}\]
  7. There is no solution provided for this question.
  8. Proof may vary.\[\begin{align*} L.S. &= (\csc{(\theta)}\sec{(\theta)})^2 - \dfrac{\left(1-\tan^2{(\theta)}\right)^2}{\tan^2{(\theta)}}\\ &= \dfrac{1}{\sin^2{(\theta)}\cos^2{(\theta)}}- \dfrac{1-2\tan^2{(\theta)}+\tan^4{(\theta)}}{\tan^2{(\theta)}}\\ &= \dfrac{1}{\sin^2{(\theta)}\cos^2{(\theta)}} - \dfrac{\left(1-\frac{2\sin^2{(\theta)}}{\cos^2{(\theta)}}+\frac{\sin^4{(\theta)}}{\cos^4{(\theta)}}\right)\times \cos^4{(\theta)}}{\left(\frac{\sin^2{(\theta)}}{\cos^2{(\theta)}}\right)\times \cos^4{(\theta)}}\\ &= \dfrac{1}{\sin^2{(\theta)}\cos^2{(\theta)}} - \dfrac{\cos^4{(\theta)}-2\sin^2{(\theta)}\cos^2{(\theta)}+\sin^4{(\theta)}}{\sin^2{(\theta)}\cos^2{(\theta)}} \\ &= \dfrac{1-\cos^4{(\theta)} + 2\left(1-\cos^2{(\theta)}\right)\cos^2{(\theta)}-\left(1-\cos^2{(\theta)}\right)^2}{\sin^2{(\theta)}\cos^2{(\theta)}} \\ &= \dfrac{1-\cos^4{(\theta)} + 2\cos^2{(\theta)}-2\cos^4{(\theta)}-1+2\cos^2{(\theta)}-\cos^4{(\theta)}}{\sin^2{(\theta)}\cos^2{(\theta)}} \\ &= \dfrac{4\cos^2{(\theta)}-4\cos^4{(\theta)}}{\sin^2{(\theta)}\cos^2{(\theta)}} \\ &= \dfrac{4\cos^2{(\theta)}\left(1-\cos^2{(\theta)}\right)}{\sin^2{(\theta)}\cos^2{(\theta)}} \\ &= \dfrac{4\cos^2{(\theta)}\sin^2{(\theta)}}{\sin^2{(\theta)}\cos^2{(\theta)}} \\ &= 4\\ &= R.S. \end{align*}\]
  9. There is no solution provided for this question.
  10. Proof may vary.\[\begin{align*} L.S. &= \dfrac{\tan{(\beta)}+\sec{(\beta)}-1}{\tan{(\beta)}-\sec{(\beta)}+1}\\ &= \dfrac{\dfrac{\sin{(\beta)}}{\cos{(\beta)}}+\dfrac{1}{\cos{(\beta)}}-1}{\dfrac{\sin{(\beta)}}{\cos{(\beta)}}-\dfrac{1}{\cos{(\beta)}}+1}\\ &= \dfrac{\sin{(\beta)}+ 1 - \cos{(\beta)}}{\cos{(\beta)}} \times \dfrac{\cos{(\beta)}}{\sin{(\beta)} - 1 + \cos{(\beta)}} \\ &= \dfrac{\sin{(\beta)}- \cos{(\beta)}+1}{\sin{(\beta)} + \cos{(\beta)}- 1} \\ &= \dfrac{\sin{(\beta)}- (\cos{(\beta)}-1)}{\sin{(\beta)} + (\cos{(\beta)}- 1)} \times \dfrac{\sin{(\beta)}- (\cos{(\beta)}-1)}{\sin{(\beta)}- (\cos{(\beta)}-1)} \\ &= \dfrac{\sin^2{(\beta)} - 2\sin{(\beta)} (\cos{(\beta)}-1) + (\cos{(\beta)} - 1)^2}{\sin^2{(\beta)} - (\cos{(\beta)} - 1)^2} \\ &= \dfrac{\sin^2{(\beta)} - 2\sin{(\beta)}\cos{(\beta)}+2\sin{(\beta)} + \cos^2{(\beta)} - 2\cos{(\beta)} +1}{\sin^2{(\beta)} - \cos^2{(\beta)} +2\cos{(\beta)} -1} \\ &= \dfrac{2 - 2\sin{(\beta)}\cos{(\beta)}+2\sin{(\beta)} - 2\cos{(\beta)} }{1-\cos^2{(\beta)} - \cos^2{(\beta)} +2\cos{(\beta)} -1} \\ &= \dfrac{2\sin{(\beta)} - 2\sin{(\beta)}\cos{(\beta)}+ 2 - 2\cos{(\beta)} }{ 2\cos{(\beta)}-2\cos^2{(\beta)} } \\ &= \dfrac{2\sin{(\beta)}(1 - \cos{(\beta)})+ 2(1 - \cos{(\beta)})}{ 2\cos{(\beta)}(1-\cos{(\beta)}) } \\ &= \dfrac{2(\sin{(\beta)}+1)(1 - \cos{(\beta)})}{ 2\cos{(\beta)}(1-\cos{(\beta)}) } \\ &= \dfrac{\sin{(\beta)} + 1}{\cos{(\beta)}}\\ &= R.S. \end{align*}\]