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Question: If a is not a multiple of \(\pi \), then show that the function\(g\left( x \right)=\csc x\) is diffe...

If a is not a multiple of π\pi , then show that the functiong(x)=csc⁡xg\left( x \right)=\csc x is differentiable at a and g′(a)=−csc⁡acot⁡ag'\left( a \right)=-\csc a\cot a. In general, g′(x)=−csc⁡xcot⁡xg'\left( x \right)=-\csc x\cot x for all xx ≠\ne nπn\pi , where n∈Zn\in Z.

Explanation

Solution

Hint: Use the fundamental definition for proving any function to be differentiable or not which is given as. If any function f(x)f\left( x \right)is differentiable at point ‘cc’ then LHD and RHD should be equal which are given by relation

Complete step-by-step answer:
LHD =lim⁡λ→c− f(x)−f(c)x−c=\underset{\lambda \to {{c}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c} and RHD =lim⁡x→c+ f(x)−f(c)x−c=\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c}

As we know that any function f(x)f\left( x \right)is differentiable at any point c, if it’s Left hand derivative (LHD) and Right hand derivative (RHD) are equal to each other and equal to f′(c)f'\left( c \right) as well.
LHD and RHD of function f(x)f\left( x \right) at any point ‘c’ can be given as
LHD =lim⁡λ→c− f(x)−f(c)x−c=\underset{\lambda \to {{c}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c} ………………………………………………(i)
RHD =lim⁡x→c+ f(x)−f(c)x−c=\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( c \right)}{x-c} ………………………………………………(ii)
Hence, any function f(x)f\left( x \right)is differentiable at point ‘c’ if
LHD ==RHD =f′(c)=f'\left( c \right) ……………………………………………………(iii)
Now coming to the question, we have a functiong(x)=csc⁡xg\left( x \right)=\csc x where we need to prove it is differentiable at x=ax=awhere aais not multiple of π\pi i.e. aa ≠\ne nπn\pi .
And, we have given in question that g′(x)=−csc⁡xcot⁡xg'\left( x \right)=-\csc x\cot x, so g′(a)=−csc⁡acot⁡ag'\left( a \right)=-\csc a\cot a
For all xx ≠\ne nπn\pi , where n∈Zn\in Z.
So, let us calculate LHD and RHD of g(x)=csc⁡xg\left( x \right)=\csc x at point a from equation (i) and (ii)
Hence, LHD can be given as
LHD =lim⁡x→a− g(x)−g(a)x−a=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}
Since, g(x)=csc⁡xg\left( x \right)=\csc x, so , g(a)=csc⁡ag\left( a \right)=\csc a
Hence, we get
LHD =lim⁡x→a− csc⁡x−csc⁡ax−a=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\dfrac{\csc x-\csc a}{x-a}
Now, we can replace a−{{a}^{-}} by (a−h)\left( a-h \right) where h→0h\to 0, so, we get
LHD =lim⁡h→0 csc⁡(a−h)−csc⁡aa−h−a=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\csc \left( a-h \right)-\csc a}{a-h-a}
or LHD =lim⁡h→0 csc⁡(a−h)−csc⁡a−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\csc \left( a-h \right)-\csc a}{-h}
We know that csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}. Hence, we get
LHD =lim⁡h→0  −1h[1sin⁡(a−h)−1sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{1}{\sin \left( a-h \right)}-\dfrac{1}{\sin a} \right]
LHD =lim⁡h→0  −1h[sin⁡a−sin⁡(a−h)sin⁡(a−h)sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{\sin a-\sin \left( a-h \right)}{\sin \left( a-h \right)\sin a} \right]
Now, we can use trigonometric identity as
sin⁡C−sin⁡D=2sin⁡(C−D2)cos⁡(C+D2)\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)
Hence, above equation becomes
LHD =lim⁡h→0  −1h[2sin⁡(a−a+h2)cos⁡(a+h2)sin⁡(a−h)sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{a-a+h}{2} \right)\cos \left( \dfrac{a+h}{2} \right)}{\sin \left( a-h \right)\sin a} \right]
LHD =lim⁡h→0  −1h[2sin⁡(h2)cos⁡(2a−h2)sin⁡(a−h)sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{h}{2} \right)\cos \left( \dfrac{2a-h}{2} \right)}{\sin \left( a-h \right)\sin a} \right]
or
LHD =lim⁡h→0  −sin⁡(h2)(h2)(cos⁡(2a−h2)sin⁡(a−h)sin⁡a)=\underset{h\to 0}{\mathop{\lim }}\,\ -\dfrac{\sin \left( \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)}\left( \dfrac{\cos \left( \dfrac{2a-h}{2} \right)}{\sin \left( a-h \right)\sin a} \right)
Now, we can use relation
lim⁡x→0  sin⁡xx=1\underset{x\to 0}{\mathop{\lim }}\,\ \dfrac{\sin x}{x}=1 with lim⁡h→0 sin⁡(h2)h2\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}},
We get after applying limit h→0h\to 0
LHD =−1cos⁡asin⁡acos⁡a=-1\dfrac{\cos a}{\sin a\cos a}
We know that cos⁡asin⁡a\dfrac{\cos a}{\sin a} == cot⁡a\cot aand 1sin⁡a=csc⁡a\dfrac{1}{\sin a}=\csc a, Hence, we get
LHD =−cot⁡acsc⁡a=-\cot a\csc a ………………………………………………(iv)
Now we can calculate RHD from equation (ii), we get
RHD =lim⁡x→a+ g(x)−g(a)x−a=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( a \right)}{x-a}
As we have g(x)=csc⁡xg\left( x \right)=\csc x, so g(a)=csc⁡ag\left( a \right)=\csc a
Hence, we get
RHD =lim⁡x→a+ csc⁡x−csc⁡ax−a=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\dfrac{\csc x-\csc a}{x-a}
Now replace a+{{a}^{+}} by a+ha+h where h→0h\to 0
Hence, we get
RHD =lim⁡h→0 −csc⁡(a+h)−csc⁡aa+h−a=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-\csc \left( a+h \right)-\csc a}{a+h-a}
or RHD =lim⁡h→0 −csc⁡(a+h)−csc⁡ah=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-\csc \left( a+h \right)-\csc a}{h}
We know that csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}, Hence, we get
RHD =lim⁡h→0  [1sin⁡(a+h)−1sin⁡a]h=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{\left[ \dfrac{1}{\sin \left( a+h \right)}-\dfrac{1}{\sin a} \right]}{h}
RHD =lim⁡h→0  1h[sin⁡a−sin⁡(a+h)sin⁡(a+h)sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{1}{h}\left[ \dfrac{\sin a-\sin \left( a+h \right)}{\sin \left( a+h \right)\sin a} \right]
Now, we can use trigonometric identity as
sin⁡C−sin⁡D=2sin⁡(C−D2)cos⁡(C+D2)\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)
Hence, RHD can be re-written as
RHD =lim⁡h→0  1h[2sin⁡(a−a−h2)cos⁡(a+a+h2)sin⁡(a+h)sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{a-a-h}{2} \right)\cos \left( \dfrac{a+a+h}{2} \right)}{\sin \left( a+h \right)\sin a} \right]
RHD =lim⁡h→0  1h[2sin⁡(−h2)cos⁡(2a+h2)sin⁡(a+h)sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ \dfrac{1}{h}\left[ \dfrac{2\sin \left( \dfrac{-h}{2} \right)\cos \left( \dfrac{2a+h}{2} \right)}{\sin \left( a+h \right)\sin a} \right]
or
RHD =lim⁡h→0  −[(sin⁡(h2)(h2))cos⁡(2a+h2)sin⁡(a+h)sin⁡a]=\underset{h\to 0}{\mathop{\lim }}\,\ -\left[ \left( \dfrac{\sin \left( \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)} \right)\dfrac{\cos \left( \dfrac{2a+h}{2} \right)}{\sin \left( a+h \right)\sin a} \right]
where, we know sin⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x
Now, using the relation lim⁡x→0  sin⁡xx=1\underset{x\to 0}{\mathop{\lim }}\,\ \dfrac{\sin x}{x}=1, we get after putting limit h→0h\to 0 to RHD;
RHD =−[(1)cos⁡asin⁡asin⁡a]=-\left[ \left( 1 \right)\dfrac{\cos a}{\sin a\sin a} \right]
Hence, RHD =−cot⁡acsc⁡a=-\cot a\csc a …………………………………………(v)
As, it is given that g(x)=−cot⁡xcsc⁡xg\left( x \right)=-\cot x\csc x , and henceg′(a)=−cot⁡acsc⁡ag'\left( a \right)=-\cot a\csc a, Therefore, we get
LHD = RHD =g′(a)g'\left( a \right)
Hence, the given functiong(x)=csc⁡xg\left( x \right)=\csc x is differentiable at x=ax=a from equation (iii) where aa ≠\ne nπn\pi .
Note: Don’t get confused with the term statement ‘xx ≠\ne nπn\pi ’ or ‘a is not multiple of π\pi ’. It is used because we cannot put x=nπx=n\pi in csc⁡x\csc x. It will give positive, infinite or negative for x→nπ+x\to n{{\pi }^{+}} or x→nπ−x\to n{{\pi }^{-}}. Hence csc⁡x\csc x is not continuous at x=nπx=n\pi . That’s why we cannot put x=nπx=n\pi .
One can get confused with the identity sin⁡C−sin⁡D\sin C-\sin D , so be clear with the trigonometric identities with these kinds of questions.
One can use the L' Hospital rule for calculating LHD and RHD as LHD and RHD are of the form 00\dfrac{0}{0}. So, we need to use identities; we can use L’ Hospital as well.