Question

If we have an expression as \[{{U}_{n}}=\sin n\theta {{\sec }^{n}}\theta ,\text{ }{{V}_{n}}=\cos n\theta {{\sec }^{n}}\theta \ne 1\] , then $\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}}$ is equal to
A. 0
B. $\tan \theta $
C. $-\tan \theta +\dfrac{\tan n\theta }{n}$
D. $\tan \theta +\dfrac{\tan n\theta }{n}$

Answer 1

Hint: To find the value of $\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}}$ , we will first find \[{{U}_{n-1}}\text{ and }{{V}_{n-1}}\] by substituting $n-1$ for $n$ in \[{{U}_{n}}=\sin n\theta {{\sec }^{n}}\theta \text{ and }{{V}_{n}}=\cos n\theta {{\sec }^{n}}\theta \] . Then substitute these in $\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}}$ ., that is, $\dfrac{\cos n\theta {{\sec }^{n}}\theta -\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }$ . Solve this using trigonometric identities. We will get $\dfrac{{{\sec }^{n-1}}\theta \left[ \cos n\theta \sec \theta -\cos \left( \left( n-1 \right)\theta \right) \right]}{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }$ . Solving this further, and writing sec in terms of cos, we will get \[\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta }{\cos n\theta }\] . Change the equation to use the formula $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ . We will get \[\dfrac{\cos n\theta -\dfrac{1}{2}\left[ \cos n\theta +\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \] . Let us solve this to get \[\dfrac{\dfrac{1}{2}\left[ \cos n\theta -\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \] . Use the identity $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and solve to get \[-\tan \theta +\dfrac{1}{n}\tan n\theta \] .

Complete step-by-step solution:
We have to find the value of $\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}}$ . It is given that
 \[{{U}_{n}}=\sin n\theta {{\sec }^{n}}\theta ...(i)\]
Let us substitute $n-1$ for $n$ . we will get
\[{{U}_{n-1}}=\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta ...(ii)\]
It is also given that
\[{{V}_{n}}=\cos n\theta {{\sec }^{n}}\theta ...(iii)\]
Let us substitute $n-1$ for $n$ . we will get
\[{{V}_{n-1}}=\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta ...(iv)\]
We have $\dfrac{{{V}_{n}}-{{V}_{n-1}}}{{{U}_{n-1}}}+\dfrac{1}{n}\dfrac{{{U}_{n}}}{{{V}_{n}}}$
Now, we can substitute the (i),(ii),(iii) and (iv) in the above equation. That is
$\dfrac{\cos n\theta {{\sec }^{n}}\theta -\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }$
Let us multiply and divide the $\cos n\theta {{\sec }^{n}}\theta $ by $\sec \theta $ . We will get
$\dfrac{\cos n\theta {{\sec }^{n}}\theta \times \dfrac{\sec \theta }{\sec \theta }-\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }$
We know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ . Hence, the above equation can be written as
$\dfrac{\cos n\theta {{\sec }^{n-1}}\theta \times \sec \theta -\cos \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }$
Let us take ${{\sec }^{n-1}}\theta $ common from the numerator of first term. We will get
$\dfrac{{{\sec }^{n-1}}\theta \left[ \cos n\theta \sec \theta -\cos \left( \left( n-1 \right)\theta \right) \right]}{\sin \left( \left( n-1 \right)\theta \right){{\sec }^{n-1}}\theta }+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }$
Now, we can cancel ${{\sec }^{n-1}}\theta $ from the numerator and denominator of the first term. That is
 $\dfrac{\cos n\theta \sec \theta -\cos \left( \left( n-1 \right)\theta \right)}{\sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }$
We know that $\sec \theta =\dfrac{1}{\cos \theta }$ . Hence, the above equation becomes
\[\dfrac{\dfrac{\cos n\theta }{\cos \theta }-\cos \left( \left( n-1 \right)\theta \right)}{\sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }\]
This can be written as
\[\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta {{\sec }^{n}}\theta }{\cos n\theta {{\sec }^{n}}\theta }\]
Meanwhile, let us solve the second term. \[{{\sec }^{n}}\theta \] can be cancelled from numerator and denominator. We will get
\[\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\dfrac{\sin n\theta }{\cos n\theta }\]
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . hence, the above equation becomes
\[\dfrac{\cos n\theta -\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
Let us multiply and divide \[\cos \theta \cos \left( \left( n-1 \right)\theta \right)\] by 2. We will get
\[\dfrac{\cos n\theta -\dfrac{2}{2}\cos \theta \cos \left( \left( n-1 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
We know that $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ . Hence, we can write the above form as
\[\dfrac{\cos n\theta -\dfrac{1}{2}\left[ \cos n\theta +\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
Let us now expand the term \[\dfrac{1}{2}\left[ \cos n\theta +\cos \left( \left( n-2 \right)\theta \right) \right]\] . We will get
\[\dfrac{\cos n\theta -\dfrac{1}{2}\cos n\theta -\dfrac{1}{2}\cos \left( \left( n-2 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
Solving the numerator of first term, we will get
\[\dfrac{\dfrac{1}{2}\cos n\theta -\dfrac{1}{2}\cos \left( \left( n-2 \right)\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
Now let us take $\dfrac{1}{2}$ common from the numerator of first term. We will get
\[\dfrac{\dfrac{1}{2}\left[ \cos n\theta -\cos \left( \left( n-2 \right)\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
We know that $\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)$
\[\dfrac{\dfrac{1}{2}\left[ 2\sin \left[ \dfrac{n\theta +\left( n-2 \right)\theta }{2} \right]\sin \left( \dfrac{\left( n-2 \right)\theta -n\theta }{2} \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
This can be written as
\[\begin{align}
& \dfrac{\dfrac{1}{2}\left[ 2\sin \left( \dfrac{2n\theta -2\theta }{2} \right)\sin \left( \dfrac{-2\theta }{2} \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \\
& =\dfrac{\dfrac{1}{2}\left[ 2\sin \left( n\theta -\theta \right)\sin \left( -\theta \right) \right]}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \\
\end{align}\]
Let us solve this further by cancelling 2 from the numerator of first term and taking $\theta $ outside from \[\sin \left( n\theta -\theta \right)\] .
\[\dfrac{\sin \left( \left( n-1 \right)\theta \right)\sin \left( -\theta \right)}{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
We know that $\sin \left( -\theta \right)=-\sin \theta $ . Hence, the above equation becomes
\[\dfrac{\sin \left( \left( n-1 \right)\theta \right)\times -\sin \theta }{\cos \theta \sin \left( \left( n-1 \right)\theta \right)}+\dfrac{1}{n}\tan n\theta \]
Let us cancel the common terms from the first term. We will get
\[\dfrac{-\sin \theta }{\cos \theta }+\dfrac{1}{n}\tan n\theta \]
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . Hence, we will get
\[-\tan \theta +\dfrac{1}{n}\tan n\theta \]
\[=-\tan \theta +\dfrac{\tan n\theta }{n}\]
Hence, the correct option is C.

Note: You must know the trigonometric identities and basic conversions to solve this question. You may write the formula for $\cos A+\cos B$ as $\cos A+\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ . This is wrong. You may also make mistake when writing the formula for $\cos A-\cos B$ as $\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ . Also you may write $\sin \left( -\theta \right)=-\cos \theta $ .