Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

$$\dfrac{secx-1}{secx+1}$$

Asked by Abhisek | 1 year ago |  55

##### Solution :-

Let f(x) = ​​$$\dfrac{secx-1}{secx+1}$$

$$f(x)=\dfrac{\dfrac{1}{cosx}-1}{\dfrac{1}{cosx}+1}=\dfrac{1-cosx}{1+cosx}$$

By quotient rule,

$$\dfrac{sinx+cosxsinx+sinx-sinxcosx}{(1+cos^2x)}$$

$$\dfrac{2sinx}{(1+cos^2x)}$$

$$\dfrac{2sinx}{(\dfrac{1}{1+secx})^2}=\dfrac{2sinx}{\dfrac{(secx+1)^2}{sec^2x}}$$

$$\dfrac{2sinxsec^2x}{(secx+1)^2}$$

$$\dfrac{\dfrac{2sinx}{cosx}secx}{(secx+1)^2}$$

$$\dfrac{2secxtanx}{(secx+1)}$$

Answered by Pragya Singh | 1 year ago

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