Question

If \[\sec (A-2B),\sec A,\sec (A+2B)\]are in A.P. then the value of \[\left( \dfrac{{{\cos }^{2}}A}{{{\cos }^{2}}B} \right)\] is less than,
A. \[1\]
B. \[\dfrac{1}{2}\]
C. \[2\]
D. \[3\]

Answer 1

Hint: In A.P. if three terms are in A.P., e.g.:-a, b, c, then they can be written as \[2b=a+c\]. Apply this concept in the A.P. given in the question and form an expression. Using trigonometric identities, simplify the expression.

Complete step-by-step answer:
Arithmetic progression (AP) or arithmetic sequence is a sequence of numbers in which each term after the first is obtained by adding a constant, d, to the preceding term. The constant d is called common difference.

An arithmetic progression (AP) can be given by \[a,a+d,a+2d,a+3d....\]where a = first term, d = common difference.

Now if a, b, c are in A.P. then they can be written as \[2b=a+c\].

It is given that \[\sec (A-2B),\sec A,\sec (A+2B)\] are in A.P.

\[\therefore a=\sec (A-2B),b=secA,c=sec(A+2B)\]

\[2b=a+c\]can be written as \[2\sec A=\sec (A-2B)+\sec (A+2B)\].

We know from basic trigonometry that,

\[\sec A={}^{1}/{}_{\cos A.}\]

\[\therefore \]The expression becomes,

\[\dfrac{2}{\cos A}=\dfrac{1}{\cos (A-2B)}+\dfrac{1}{\cos (A+2B)}.\]

Now let us cross multiply and rearrange the expression.

\[\dfrac{2}{\cos A}=\dfrac{\cos (A+2B)+\cos (A-2B)}{\cos (A-2B)\cos (A+2B)}........(2)\]

We know that,

\[\begin{align}

  & \cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right) \\

 & \therefore \cos (A+2B)+\cos (A-2B)=2\cos \left( \dfrac{A+2B+A-2B}{2} \right)\cos \left( \dfrac{A+2B-A+2B}{2} \right) \\

 & \Rightarrow 2\cos \left( \dfrac{2A}{2} \right)\cos \left( \dfrac{4B}{2} \right) \\

 & \Rightarrow 2\cos A\cos 2B.........(3) \\

\end{align}\]

Similarly, we know the trigonometric identities of,

\[\begin{align}

  & \cos (a-b)=\cos a\cos b+\sin A\sin B \\

 & \Rightarrow \cos (A-2B)=\cos A\cos 2B+\sin A\sin 2B. \\

\end{align}\]

Similarly,

\[\begin{align}

  & \cos (a+b)=\cos a\cos b-\operatorname{sinAsinB}. \\

 & \therefore cos(A+2B)=cosAcos2B-sinAsin2B. \\

\end{align}\]

Now let us find out the value of the denominator \[\cos (A-2B)\cos (A+2B)\]

\[\cos (A-2B)\cos (A+2B)=(cosAcos2B+sinAsin2B)(cosAcos2B-sinAsin2B)\]

Open the brackets and multiply them.

\[={{\cos }^{2}}A{{\cos }^{2}}B-\cos A\cos 2B\operatorname{sinA}\sin 2B+\sin A\sin 2B\cos A\cos 2B-{{\sin }^{2}}A{{\sin }^{2}}2B\]

Cancel out the like terms,

\[={{\cos }^{2}}A{{\cos }^{2}}2B-{{\sin }^{2}}A{{\sin }^{2}}2B\]

We know \[\begin{align}

  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1, \\

 & \Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta . \\

\end{align}\]

Substitute \[{{\sin }^{2}}A=1-{{\cos }^{2}}A\] and \[{{\sin }^{2}}2B=1-{{\cos }^{2}}2B\] in the above equation.

\[\Rightarrow {{\cos }^{2}}A{{\cos }^{2}}2B-\left[ \left( 1-{{\cos }^{2}}A \right)(1-{{\cos }^{2}}B \right].\]

Open brackets and simplify them.

\[\begin{align}

  & ={{\cos }^{2}}A{{\cos }^{2}}2B-\left[ 1-{{\cos }^{2}}2B-{{\cos }^{2}}A+{{\cos }^{2}}A{{\cos }^{2}}2B \right] \\

 & ={{\cos }^{2}}A{{\cos }^{2}}B-1+{{\cos }^{2}}2B+{{\cos }^{2}}A-{{\cos }^{2}}A{{\cos }^{2}}2B \\

\end{align}\]

Cancel out like terms.

\[\begin{align}

  & =-1+{{\cos }^{2}}2B+{{\cos }^{2}}A \\

 & ={{\cos }^{2}}A-1+{{\cos }^{2}}2B \\

\end{align}\]

We know,

\[\begin{align}

  & {{\sin }^{2}}2B+{{\cos }^{2}}2B=1 \\

 & \Rightarrow -1+{{\cos }^{2}}2B=-{{\sin }^{2}}2B. \\

 & \therefore \cos (A-2B)\cos (A+2B)={{\cos }^{2}}A-{{\sin }^{2}}2B.....(4) \\

\end{align}\]

Substitute the values of equation (3) and (4) in equation (2).

\[\dfrac{2}{\cos A}=\dfrac{2\cos A\cos 2B}{{{\cos }^{2}}A-{{\sin }^{2}}2B}\Rightarrow

 \dfrac{1}{\cos A}=\dfrac{\cos A\cos 2B}{{{\cos }^{2}}A-{{\sin }^{2}}2B}.\]

Let us cross multiply the above expression.

\[\begin{align}

  & {{\cos }^{2}}A-{{\sin }^{2}}2B={{\cos }^{2}}A\cos 2B \\

 & {{\cos }^{2}}A-{{\cos }^{2}}A\cos 2B={{\sin }^{2}}2B \\

 & {{\cos }^{2}}A(1-\cos 2B)={{\sin }^{2}}2B \\

\end{align}\]

We know the formula \[1-\cos 2x=2{{\sin }^{2}}x.\]

Similarly,

\[\begin{align}

  & 1-\cos 2B=2{{\sin }^{2}}B \\

 & \therefore {{\cos }^{2}}A(2{{\sin }^{2}}B)={{(\sin 2B)}^{2}} \\

 & \Rightarrow 2{{\cos }^{2}}A{{\sin }^{2}}B={{(2\sin B\cos B)}^{2}} \\

 & \Rightarrow 2{{\cos }^{2}}A{{\sin }^{2}}B=4{{\sin }^{2}}B{{\cos }^{2}}B. \\

\end{align}\] and \[\begin{align}

  & \sin 2\theta =2\sin \theta \cos \theta . \\

 & \therefore \sin 2B=2\sin B\cos B \\

\end{align}\]

Cancel out \[{{\sin }^{2}}B\] on LHS and RHS.

\[\begin{align}

  & 2{{\cos }^{2}}A=4{{\cos }^{2}}B \\

 & \Rightarrow \dfrac{{{\cos }^{2}}A}{{{\cos }^{2}}B}=\dfrac{4}{2}=2 \\

\end{align}\] [Cross multiply]

Thus we got \[\left( \dfrac{{{\cos }^{2}}A}{{{\cos }^{2}}B} \right)=2.\]

\[\therefore \]Option C is the correct answer.

Note: From this question you can say that by remembering the trigonometric identities you can solve it easily. We have used at least seven trigonometric identities. Similarly, it’s important to form an expression using the 3 terms or you won’t be able to find the answer. So if any 3 terms are in AP, then form expressions using \[2b=a+c\].