Question

Solve the given function for k: $\begin{align}
  & f\left( \theta \right)=\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta },\text{ }for\text{ }\theta \ne \dfrac{\pi }{4} \\
 & =\dfrac{k}{5}\text{ }for\text{ }\theta =\dfrac{\pi }{4}
\end{align}$

Answer 1

Hint: Any function $f\left( x \right)$ will be continuous at $x\to c$ , if
$\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$
$\Rightarrow LHL=RHL=\text{ Exact value of f}\left( x \right)\text{ at }x=c$
Use the above result to get the value of k. It means get LHL or RHL of $f\left( \theta \right)$ at $x\to \dfrac{\pi }{4}$ by applying $x\to {{\dfrac{\pi }{4}}^{-}}$ for LHL or applying $x\to {{\dfrac{\pi }{4}}^{+}}$ for RHL. Don’t calculate both (both will be same). Equate the calculated LHL or RHL to $\dfrac{k}{5}$ to get the answer. Use the following results:
$\begin{align}
 & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\
 & \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \\
 & \cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2},\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} \\
\end{align}$

Complete step-by-step solution -
Given expression in the problem is
$\begin{align}
  & f\left( \theta \right)=\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta },\text{ }for\text{ }\theta \ne \dfrac{\pi }{4} \\
 & f\left( \theta \right)=\dfrac{k}{5},\text{ }for\text{ }\theta =\dfrac{\pi }{4} \\
\end{align}$
As we know any function will be continuous if the left-hand limit to any function and right-hand limit to the same function are equal to the exact value of that function at that point. It can be written mathematically as
$\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)..............\left( i \right)$
Now, as we are given a function $f\left( \theta \right)$ defined with respect to $\theta =\dfrac{\pi }{4}$ and $\theta \ne \dfrac{\pi }{4}$
So, we need to calculate value of $\underset{\theta \to \dfrac{\pi }{4}}{\mathop{\lim }}\,,\underset{\theta \to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,$and $\underset{\theta \to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,$ and hence, we need to equate them to get value of k.
So, LHL can be calculated as
$LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,f\left( \theta \right)$
As $f\left( \theta \right)=\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta }$ for $\theta \ne \dfrac{\pi }{4}$ .So, we need to replace $f\left( \theta \right)$ by $\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta }$
$LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta } \right)$
Now, we can put $\theta =\dfrac{\pi }{4}-h$ , where $h\to 0$ .Hence, we can get LHL w.r.t ‘h’ as
$LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\tan \left( \dfrac{\pi }{4}-h \right)}{1-\sqrt{2}\sin \left( \dfrac{\pi }{4}-h \right)}..........\left( ii \right)$
Now, as we know the identities of $\tan \left( A-B \right)$ and $\sin \left( A-B \right)$ are given as
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\operatorname{tanB}}............\left( iii \right)$
$\sin \left( A-B \right)=\sin A\operatorname{cosB}-\cos A\sin B.............\left( iv \right)$
So, we can simplify the equation (ii) with the help of above equations as
$\begin{align}
  & LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\left( \dfrac{\tan \dfrac{\pi }{4}-\tanh }{1+\tan \dfrac{\pi }{4}\tanh } \right)}{1-\sqrt{2}\left( \sin \dfrac{\pi }{4}\cosh -\cos \dfrac{\pi }{4}\sinh \right)} \\
 & LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+\tan \dfrac{\pi }{4}\tanh \right)-\left( \tan \dfrac{\pi }{4}-\tanh \right)}{\left( 1+\tan \dfrac{\pi }{4}\tanh \right)\left[ 1-\sqrt{2}\left( \sin \dfrac{\pi }{4}\cosh -\cos \dfrac{\pi }{4}\sinh \right) \right]} \\
\end{align}$
Now, as we know
$\tan \dfrac{\pi }{4}=1$ and $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
So, we get
\[\begin{align}
  & LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1+\tanh -1+\tanh }{\left( 1+\tanh \right)\left[ 1-\sqrt{2}\left( \dfrac{\cosh }{\sqrt{2}}-\dfrac{\sinh }{\sqrt{2}} \right) \right]} \\
 & LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\tanh }{\left( 1+\tanh \right)\left[ 1-\cosh +\sinh \right]} \\
\end{align}\]
We know
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
So, we can get LHL as
$LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sinh }{\cosh \left( 1+\tanh \right)\left[ 1-\cosh +\sinh \right]}$
We know
$\begin{align}
  & \cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2} \\
 & \sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} \\
\end{align}$
So, we can re-write LHL using the above identities as
$\begin{align}
  & LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\times 2\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{\cosh \left( 1+\tanh \right)\left[ 1-\left( 1-2{{\sin }^{2}}\dfrac{h}{2} \right)+2\sin \dfrac{h}{2}\cos \dfrac{h}{2} \right]} \\
 & \Rightarrow LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{4\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{\cosh \left( 1+\tanh \right)\left[ 2{{\sin }^{2}}\dfrac{h}{2}+2\sin \dfrac{h}{2}\cos \dfrac{h}{2} \right]} \\
 & \Rightarrow LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{4\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{2\cosh \left( 1+\tanh \right)\left( \sin \dfrac{h}{2} \right)\left[ \sin \dfrac{h}{2}+\cos \dfrac{h}{2} \right]} \\
\end{align}$
Now, cancelling out the term $2\sin \dfrac{h}{2}$ from numerator and denominator, we get LHL as
$LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \dfrac{h}{2}}{\cosh \left( 1+\tanh \right)\left[ \sin \dfrac{h}{2}+\cos \dfrac{h}{2} \right]}$

Now we can apply limit to the given expression i.e. $h\to 0$ using the results
$\begin{align}
  & \cos {{0}^{\circ }}=1 \\
 & \tan {{0}^{\circ }}=0 \\
 & \sin {{0}^{\circ }}=0 \\
\end{align}$
Hence, we get value of LHL as
$\begin{align}
  & LHL=\dfrac{2\times 1}{1\left( 1 \right)\left( 0+1 \right)}=2 \\
 & LHL=2.................\left( v \right) \\
\end{align}$
Now, we can calculate RHL by the same approach and using $\theta =h+\dfrac{\pi }{4}$ , where $h\to 0$ . We will get RHL as 2 as well. But we don not need to calculate RHL to get value of k. We can directly equate $\dfrac{k}{5}$ to 2 to get the value of k as LHL = RHL = value of function at that point. Hence, as we know
$\underset{\theta \to \dfrac{\pi }{4}}{\mathop{\lim }}\,f\left( \theta \right)=\dfrac{k}{5}$
Hence, we can write equation using equation (i) as
$\dfrac{k}{5}=2$
On cross-multiplying, we get k as
$k=10$
Hence, k = 10 is the answer of the problem.

Note: We do not need to calculate LHL or RHL i.e. any of them. It is done in the problem for better understanding of the problem. We can put $x\to \dfrac{\pi }{4}$ to the function $f\left( \theta \right)=\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta }$ directly. As it has not involvement of any special function like modulus, greatest integer etc. So, as the function will give same value for $x\to {{\dfrac{\pi }{4}}^{-}},x\to {{\dfrac{\pi }{4}}^{+}}$ and hence, for $x\to \dfrac{\pi }{4}$ as well. So, one may put $x\to \dfrac{\pi }{4}$ to $f\left( \theta \right)=\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta }$ directly and answer will be same. And use this concept for future reference as well.
In the given solution, we do not need to calculate RHL, as LHL = RHL, which is already mentioned in starting of the solution. So, do not waste your time for that. Hence, directly put $LHL=\dfrac{k}{5}$ to get the answer.
One may get confuse with the identities of $\cos \theta $ and $\sin \theta $ mentioned in the problem. As one may not be familiar with that. So, observe formula of $\cos 2\theta $ and $\sin 2\theta $ as
$\begin{align}
  & \cos 2\theta =1-2{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta -1={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
 & \sin 2\theta =2\sin \theta \cos \theta \\
\end{align}$
Now, replace $2\theta $ in the above identities by $\theta $ i.e. $\theta $ by $\dfrac{\theta }{2}$ in same ratio. So, you will get the same results written in the solution.