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Question: Solve the given function for k: \(\begin{aligned} & f\left( \theta \right)=\dfrac{1-\tan \theta ...

Solve the given function for k: f(θ)=1−tan⁡θ1−2sin⁡θ, for θ≠π4 =k5 for θ=π4\begin{aligned} & 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{aligned}

Explanation

Solution

Hint: Any function f(x)f\left( x \right) will be continuous at x→cx\to c , if
lim⁡x→c− f(x)=lim⁡x→c+ f(x)=lim⁡x→c f(x)\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)
⇒LHL=RHL= Exact value of f(x) at x=c\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(θ)f\left( \theta \right) at x→π4x\to \dfrac{\pi }{4} by applying x→π4−x\to {{\dfrac{\pi }{4}}^{-}} for LHL or applying x→π4+x\to {{\dfrac{\pi }{4}}^{+}} for RHL. Don’t calculate both (both will be same). Equate the calculated LHL or RHL to k5\dfrac{k}{5} to get the answer. Use the following results:
tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B cos⁡θ=1−2sin⁡2θ2,sin⁡θ=2sin⁡θ2cos⁡θ2 \begin{aligned} & \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{aligned}

Complete step-by-step solution -
Given expression in the problem is
f(θ)=1−tan⁡θ1−2sin⁡θ, for θ≠π4 f(θ)=k5, for θ=π4 \begin{aligned} & 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{aligned}
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
lim⁡x→c− f(x)=lim⁡x→c+ f(x)=lim⁡x→c f(x)..............(i)\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(θ)f\left( \theta \right) defined with respect to θ=π4\theta =\dfrac{\pi }{4} and θ≠π4\theta \ne \dfrac{\pi }{4}
So, we need to calculate value of lim⁡θ→π4 ,lim⁡θ→π4− \underset{\theta \to \dfrac{\pi }{4}}{\mathop{\lim }}\,,\underset{\theta \to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,and lim⁡θ→π4+ \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=lim⁡x→π4− f(θ)LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,f\left( \theta \right)
As f(θ)=1−tan⁡θ1−2sin⁡θf\left( \theta \right)=\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta } for θ≠π4\theta \ne \dfrac{\pi }{4} .So, we need to replace f(θ)f\left( \theta \right) by 1−tan⁡θ1−2sin⁡θ\dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta }
LHL=lim⁡x→π4− (1−tan⁡θ1−2sin⁡θ)LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \dfrac{1-\tan \theta }{1-\sqrt{2}\sin \theta } \right)
Now, we can put θ=π4−h\theta =\dfrac{\pi }{4}-h , where h→0h\to 0 .Hence, we can get LHL w.r.t ‘h’ as
LHL=lim⁡h→0 1−tan⁡(π4−h)1−2sin⁡(π4−h)..........(ii)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⁡(A−B)\tan \left( A-B \right) and sin⁡(A−B)\sin \left( A-B \right) are given as
tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡AtanB⁡............(iii)\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\operatorname{tanB}}............\left( iii \right)
sin⁡(A−B)=sin⁡AcosB⁡−cos⁡Asin⁡B.............(iv)\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
LHL=lim⁡h→0 1−(tan⁡π4−tanh⁡1+tan⁡π4tanh⁡)1−2(sin⁡π4cosh⁡−cos⁡π4sinh⁡) LHL=lim⁡h→0 (1+tan⁡π4tanh⁡)−(tan⁡π4−tanh⁡)(1+tan⁡π4tanh⁡)[1−2(sin⁡π4cosh⁡−cos⁡π4sinh⁡)] \begin{aligned} & 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{aligned}
Now, as we know
tan⁡π4=1\tan \dfrac{\pi }{4}=1 and sin⁡π4=cos⁡π4=12\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}
So, we get

& 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{aligned}$$ 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 $\begin{aligned} & \cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2} \\\ & \sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} \\\ \end{aligned}$ So, we can re-write LHL using the above identities as $\begin{aligned} & LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\times 2\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{\cosh \left( 1+\tanh \right)\left[ 1-\left( 1-2{{\sin }^{2}}\dfrac{h}{2} \right)+2\sin \dfrac{h}{2}\cos \dfrac{h}{2} \right]} \\\ & \Rightarrow LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{4\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{\cosh \left( 1+\tanh \right)\left[ 2{{\sin }^{2}}\dfrac{h}{2}+2\sin \dfrac{h}{2}\cos \dfrac{h}{2} \right]} \\\ & \Rightarrow LHL=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{4\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{2\cosh \left( 1+\tanh \right)\left( \sin \dfrac{h}{2} \right)\left[ \sin \dfrac{h}{2}+\cos \dfrac{h}{2} \right]} \\\ \end{aligned}$ 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 $\begin{aligned} & \cos {{0}^{\circ }}=1 \\\ & \tan {{0}^{\circ }}=0 \\\ & \sin {{0}^{\circ }}=0 \\\ \end{aligned}$ Hence, we get value of LHL as $\begin{aligned} & LHL=\dfrac{2\times 1}{1\left( 1 \right)\left( 0+1 \right)}=2 \\\ & LHL=2.................\left( v \right) \\\ \end{aligned}$ 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 $\begin{aligned} & \cos 2\theta =1-2{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta -1={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\\ & \sin 2\theta =2\sin \theta \cos \theta \\\ \end{aligned}$ 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.