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Question

If $$y = \tan^{-1} (\ secx+ \ tanx)$$ then $$\dfrac{dy}{dx} = $$


A
1
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B
12
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C
1
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D
0
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Solution

The correct option is C $$\dfrac{1}{2}$$
Given $$y=tan^{-1}(secx+tanx)$$

Differentiate on both sides w.r.t x

  $$\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1}{1+(secx+tanx)^{2}}\dfrac{\mathrm{d} }{\mathrm{d} x}(secx+tanx)$$         $$\because \dfrac{\mathrm{d} }{\mathrm{d} x}(tan^{-1}(x)=\dfrac{1}{1+x^{2}})$$
      
         $$=\dfrac{secxtanx+sec^{2}x}{1+sec^{2}x+tan^{2}x+2secxtanx}$$

         $$=\dfrac{secxtanx+sec^{2}x}{2sec^{2}x+2secxtanx}$$        $$\because \left ( 1+tan^{2}x=sec^{2}x \right )$$

         $$=\dfrac{secxtanx+sec^{2}x}{2[sec^{2}x+secxtanx]}$$

  $$\therefore \dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1}{2}$$

Mathematics

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