Question

If $\cos h\alpha =secx$, then ${\tan }^2\left(\frac{x}{2}\right)=?$

• A. ${\cos }^2\left(\frac{\alpha }{2}\right)$
• B. ${\sin }^2\left(\frac{\alpha }{2}\right)$
• C. $\mathrm{co}{t}^2\left(\frac{\alpha }{2}\right)$
• D. $\tan h^2\left(\frac{\alpha }{2}\right)$

Answer 1

The correct option is D

$\tan h^2\left(\frac{\alpha }{2}\right)$

Find the value of $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\left(\frac{x}{2}\right)$:

Given, $\cos h\alpha =secx$

As we know that,

$\begin{array}{rcl}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{h}\mathit{\alpha }\mathbf{}& \mathbf{=}& \mathbf{}\frac{{\mathbf{e}}^{\alpha }\mathbf{}\mathbf{+}\mathbf{}{\mathbf{e}}^{\mathbf{-}\mathit{\alpha }}}{\mathbf{2}}\\ & =& secx\end{array}$

Also we know that, $\mathit{t}\mathit{a}\mathit{n}\mathbf{}{\mathit{h}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\alpha }}{\mathbf{2}}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{\left[}\frac{{\mathbf{e}}^{{\displaystyle \raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\mathbf{-}\mathbf{}{\mathbf{e}}^{\raisebox{1ex}{$\mathbf{-}\mathbf{\alpha }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}}{{\mathbf{e}}^{\raisebox{1ex}{$\mathbf{\alpha }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{e}}^{\raisebox{1ex}{$\mathbf{-}\mathbf{\alpha }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}}\mathbf{\right]}}^{\mathbf{2}}$

$$\begin{gathered} =\frac{e^{\alpha }+e^{-\alpha }-2}{e^{\alpha }+e^{-\alpha }+2} \\ =\frac{2secx-2}{2secx+2} \\ =\frac{secx-1}{secx+1} \\ =\frac{{\displaystyle \frac{1}{\cos x}}-1}{{\displaystyle \frac{1}{\cos x}}+1} \\ =\frac{1-\cos x}{1+\cos x} \\ =\frac{1-\left(1-2{\sin }^2{\displaystyle \frac{x}{2}}\right)}{1+\left(2{\cos }^2\frac{x}{2}-1\right)} \\ =\frac{{\sin }^2\frac{x}{2}}{{\cos }^2\frac{x}{2}} \\ ={\tan }^2\left(\frac{x}{2}\right) \end{gathered}$$

$\therefore \mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{tan}\mathbf{}{\mathit{h}}^{\mathbf{2}}\left(\frac{\alpha }{2}\right)$

Hence, Option ‘D’ is Correct.