Question

If A, B and C are the angles of a triangle such that sec (A – B), sec (A) and sec (A + B) are in arithmetic progression, then
$\left( a \right){{\operatorname{cosec}}^{2}}A=2{{\operatorname{cosec}}^{2}}\dfrac{B}{2}$
$\left( b \right)2se{{c}^{2}}A=se{{c}^{2}}\dfrac{B}{2}$
$\left( c \right)2{{\operatorname{cosec}}^{2}}A={{\operatorname{cosec}}^{2}}\dfrac{B}{2}$
$\left( d \right)2se{{c}^{2}}B=se{{c}^{2}}\dfrac{A}{2}$

Answer 1

In AP, the sum of the first and the last term is twice the middle term. So, we get that $2\sec A=\sec \left( A+B \right)+\sec \left( A-B \right),$ then we will use reciprocal identity, $\sec \theta =\dfrac{1}{\cos \theta }$ to simplify them and after this, we will expand cos (A + B) as $\cos A\cos B-\sin A\sin B$ and cos (A – B) as $\cos A\cos B+\sin A\sin B.$ We will solve to get a simplified value and at last we will convert $1+\cos B$ to $2{{\cos }^{2}}\dfrac{B}{2}$ using $\cos \dfrac{\theta }{2}=\dfrac{\sqrt{1+\cos \theta }}{2}$ to get the required answer.

Complete step-by-step answer:
We are given that A, B and C are the angles of triangles such that sin (A – B), sec A, and sec (A + B) are in AP. As A, B and C are the angles of the triangle, so their sum must be equal to ${{180}^{\circ }}$ by the angle sum property of the triangle.
$\Rightarrow A+B+C={{180}^{\circ }}......\left( i \right)$
We know that when the three terms are in the Arithmetic Progression, then the sum of the first and last sum is always equal to twice the middle term. As we have seen that sec (A – B), sec A, sec (A + B) are in AP, so the sum of sec (A – B) and sec (A + B) will be equal to two times sec A. So, we get,
$2\sec A=\sec \left( A-B \right)+\sec \left( A+B \right)$
Now, we know that,
$\sec \theta =\dfrac{1}{\cos \theta }$
So, using this in the above equality, we get,
$2\times \left( \dfrac{1}{\cos A} \right)=\dfrac{1}{\cos \left( A-B \right)}+\dfrac{1}{\cos \left( A+B \right)}$
Simplifying further, we get,
$\Rightarrow \dfrac{2}{\cos A}=\dfrac{\cos \left( A+B \right)+\cos \left( A-B \right)}{\cos \left( A+B \right)\cos \left( A-B \right)}$
As we have,
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
$\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$
So, we get,
$\Rightarrow \dfrac{2}{\cos A}=\dfrac{\left( \cos A\cos B-\sin A\sin B \right)+\left( \cos A\cos B+\sin A\sin B \right)}{\left( \cos A\cos B-\sin A\sin B \right)\left( \cos A\cos B+\sin A\sin B \right)}$
As we know that,
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
So, we get,
$\Rightarrow \dfrac{2}{\cos A}=\dfrac{\cos A\cos B-\sin A\sin B+\cos A\cos B+\sin A\sin B}{{{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B}$
Cancelling the like terms, we get,
$\Rightarrow \dfrac{2}{\cos A}=\dfrac{2\cos A\cos B}{{{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B}$
Cross multiplying, we get,
$\Rightarrow {{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B={{\cos }^{2}}A\cos B$
Now, we know that,
${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$
So, using this, we get,
${{\sin }^{2}}A=1-{{\cos }^{2}}A$
${{\sin }^{2}}B=1-{{\cos }^{2}}B$
Applying these formulas above, we get,
$\Rightarrow {{\cos }^{2}}A{{\cos }^{2}}B-\left( 1-{{\cos }^{2}}A \right)\left( 1-{{\cos }^{2}}B \right)={{\cos }^{2}}A\cos B$
Opening the bracket, we get,
$\Rightarrow {{\cos }^{2}}A{{\cos }^{2}}B-1+{{\cos }^{2}}B+{{\cos }^{2}}A-{{\cos }^{2}}A{{\cos }^{2}}B={{\cos }^{2}}A\cos B$
Now, cancelling the like terms, we get,
$\Rightarrow {{\cos }^{2}}A+{{\cos }^{2}}B-1={{\cos }^{2}}A\cos B$
Simplifying further, we get,
$\Rightarrow {{\cos }^{2}}A-{{\cos }^{2}}A\cos B=1-{{\cos }^{2}}B$
$\Rightarrow {{\cos }^{2}}A\left( 1-\cos B \right)=1-{{\cos }^{2}}B$
As,
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
We get,
$\Rightarrow {{\cos }^{2}}A\left( 1-\cos B \right)=\left( 1-\cos B \right)\left( 1+\cos B \right)$
Dividing both the sides by 1 – cos B, we get,
$\Rightarrow {{\cos }^{2}}A=1+\cos B$
Now as,
$\cos \dfrac{\theta }{2}=\sqrt{\dfrac{1+\cos \theta }{2}}$
So, applying this we get,
$\Rightarrow 1+\cos B=2{{\cos }^{2}}\dfrac{B}{2}$
Hence,
$\Rightarrow {{\cos }^{2}}A=2{{\cos }^{2}}\dfrac{B}{2}$
Again using reciprocal relation, $\sec \theta =\dfrac{1}{\cos \theta },$ we get,
$\Rightarrow {{\sec }^{2}}A=\dfrac{1}{2}{{\sec }^{2}}\dfrac{B}{2}$

So, the correct answer is “Option B”.

Note: While simplifying, remember $\cos \left( A+B \right)\ne \cos A+\cos B.$ But it is equal to $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B.$ Also, remember that, $1-{{\cos }^{2}}B=\left( 1-\cos B \right)\left( 1+\cos B \right)$ becomes ${{1}^{2}}=1,$ so $1-{{\cos }^{2}}B={{1}^{2}}-{{\cos }^{2}}B.$ Now using, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right),$ we have, $1-{{\cos }^{2}}B=\left( 1+\cos B \right)\left( 1-\cos B \right).$