Question

If $A,B$ and $C$ are the angles of a triangle, such that $sec(A-B)$ , $secA$ and $sec(A+B)$ are in arithmetic progression, then

• A. $\cos e{c}^{2}A=2\cos e{c}^2\left(\frac{B}{2}\right)$
• B. $2se{c}^{2}A=se{c}^2\left(\frac{B}{2}\right)$
• C. $2\cos e{c}^{2}A=\cos e{c}^2\left(\frac{B}{2}\right)$
• D. $2se{c}^{2}A=2se{c}^2\left(\frac{A}{2}\right)$

Answer 1

The correct option is B

$2se{c}^{2}A=se{c}^2\left(\frac{B}{2}\right)$

Explanation for the correct option:

Step 1. Given that, $A,B$ and $C$ are the angles of a triangle

$sec(A-B)$ , $secA$ and $sec(A+B)$ are in A.P

$\Rightarrow$$2secA=sec(A–B)+sec(A+B)$

$\Rightarrow$$2secA=\frac{1}{\cos (A–B)}+\frac{1}{\cos (A+B)}$

$\Rightarrow$ $secA=\frac{\cos (A+B)+\cos (A–B)}{2\cos (A+B)\cos (A–B)}$

Step 2. Using identities $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\left(\theta +\varphi \right)\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\left(\theta -\varphi \right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\varphi }$ ,

$\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{(}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\varphi }\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\varphi }$, $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\left(\theta -\varphi \right)\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\varphi }$

$\Rightarrow$ $\frac{1}{\cos A}=\frac{(2\cos A\cos B)}{2({\cos }^{2}A{\cos }^{2}B–{\sin }^{2}A{\sin }^{2}B)}$

$\Rightarrow$ $\frac{({\cos }^{2}A\cos B)}{{\cos }^{2}A{\cos }^{2}B–(1–{\cos }^{2}A)(1–{\cos }^{2}B)}=1$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$${\cos }^{2}A{\cos }^{2}B–1+{\cos }^{2}A+{\cos }^{2}B-{\cos }^{2}A{\cos }^{2}B={\cos }^{2}A\cos B$

$\Rightarrow$ ${\cos }^{2}A–{\cos }^{2}A\cos B=1–{\cos }^{2}B$

$\Rightarrow$ ${\cos }^{2}A(1–\cos B)=(1+\cos B)(1–\cos B)$

$\Rightarrow$ ${\cos }^{2}A=1+\cos B$

$\Rightarrow$ ${\cos }^{2}A=2{\cos }^2\left(\frac{B}{2}\right)$

$\Rightarrow$ $se{c}^{2}A=\left(\frac{1}{2}\right)se{c}^2\left(\frac{B}{2}\right)$

$\mathbf{\therefore }\mathbf{2}\mathbf{}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\left(\frac{B}{2}\right)$

Hence, option ‘B’ is correct.