Question

If $-\frac{\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2}$, then the value of $\log \left(secx\right)$ is

• A. $2cot{h}^{-1}\left(\cos e{c}^2\left(\left(\frac{x}{2}\right)-1\right)\right)$
• B. $2\mathrm{co}t{h}^{-1}\left(\cos e{c}^2\left(\left(\frac{x}{2}\right)-1\right)\right)$
• C. $2\cos ec{h}^{-1}\left(\mathrm{co}{t}^2\left(\left(\frac{x}{2}\right)-1\right)\right)$
• D. $2\cos ec{h}^{-1}\left(\mathrm{co}{t}^2\left(\left(\frac{x}{2}\right)+1\right)\right)$

Answer 1

The correct option is A

$2cot{h}^{-1}\left(\cos e{c}^2\left(\left(\frac{x}{2}\right)-1\right)\right)$

Explantion for the correct option:

Step 1: Apply the given condition $-\frac{\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2}$

Let us consider, $\log \left(secx\right)=p$

$\Rightarrow$ $secx=e^p$

$\Rightarrow$ $\frac{1}{\cos x}=e^p$

Write $e^p=e^{\frac{p}{2}+\frac{p}{2}}$ in the above equation

$\Rightarrow$$\frac{1}{\cos x}=e^{\frac{p}{2}+\frac{p}{2}}$

$\Rightarrow$$\frac{1}{\cos x}=\frac{e^{\frac{p}{2}}}{e^{-\frac{p}{2}}}$

Step 2: Apply component - dividend rule,

$\Rightarrow$$\frac{1+\cos x}{1-\cos x}=\frac{e^{\frac{p}{2}}+e^{-\frac{p}{2}}}{e^{\frac{p}{2}}-e^{-\frac{p}{2}}}$ $\left[\because 1+\cos x=2{\cos }^2\left(\frac{x}{2}\right)\mathbf{and}\mathbf{}1-\mathrm{cosx}=2{\sin }^2\left(\frac{\mathrm{x}}{2}\right)\right]$

$\Rightarrow$$\frac{2{\cos }^2{\displaystyle \left(\frac{x}{2}\right)}}{2{\sin }^2{\displaystyle \left(\frac{x}{2}\right)}}=\frac{e^{\frac{p}{2}}+e^{-\frac{p}{2}}}{e^{\frac{p}{2}}-e^{-\frac{p}{2}}}$

$\Rightarrow$$\mathrm{co}{t}^2\left(\frac{x}{2}\right)=coth\left(\frac{p}{2}\right)$ $\left[\because \cot h\left(\frac{p}{2}\right)=\frac{e^{\frac{p}{2}}+e^{-\frac{p}{2}}}{e^{\frac{p}{2}}-e^{-\frac{p}{2}}}\right]$

From the above equation, we can write

$\mathbf{\Rightarrow }$ $coth\left(\frac{p}{2}\right)=\mathrm{co}{t}^2\left(\frac{x}{2}\right)$

$\mathbf{\Rightarrow }$ $\frac{p}{2}=cot{h}^{-1}\left(\mathrm{co}{t}^2\left(\frac{x}{2}\right)\right)$

$\Rightarrow$ $p=2cot{h}^{-1}\left(\cos e{c}^2\left(\left(\frac{x}{2}\right)-1\right)\right)$ ; $\left[\therefore \mathrm{co}{t}^2\left(\frac{x}{2}\right)=\cos e{c}^2\left(\frac{x}{2}\right)-1\right]$

$\Rightarrow$ $\log \left(secx\right)=2cot{h}^{-1}\left(\cos e{c}^2\left(\frac{x}{2}\right)-1\right)$ ; $\left[\because \log \left(secx\right)=p\right]$

Therefore, the value of $\log \left(secx\right)=2cot{h}^{-1}\left(\cos e{c}^2\left(\frac{x}{2}\right)-1\right)$.

Hence, the correct answer is an option (A).