Question

Consider the given trigonometric expression: if sec(x - y), sec(x), sec(x + y) are in A.P. then $\cos x\sec \dfrac{y}{2}=$
(a) $\pm 2$
(b) $\pm \dfrac{1}{\sqrt{2}}$
(c) $\pm \dfrac{1}{2}$
(d) $\pm \sqrt{2}$

Answer 1

Hint: sec(x – y), sec(x), sec(x + y) are in A.P means, $2\sec x=\sec \left( x-y \right)+\sec \left( x+y \right)$.
Solve the above equation and find out the value of x and y. Then put those values in $\cos x\sec \dfrac{y}{2}$.

Complete step-by-step solution -
It is given in the question that sec(x – y), sec(x), sec(x + y) are in A.P. We know if a series a, b, c is in AP, then b – a = c – b. Applying this in the given series, we get
$\sec x-\sec (x-y)=\sec (x+y)-\sec x$
$\Rightarrow 2\sec x=\sec \left( x+y \right)+\sec \left( x-y \right)$
We know that, $\sec x=\dfrac{1}{\cos x}$. If we substitute cosine in the place of secant, we will get:
$\Rightarrow \dfrac{2}{\cos x}=\dfrac{1}{\cos \left( x+y \right)}+\dfrac{1}{\cos \left( x-y \right)}$
Now taking the LCM, we get
$\Rightarrow \dfrac{2}{\cos x}=\dfrac{\cos \left( x-y \right)+\cos \left( x+y \right)}{\cos \left( x+y \right)\cos \left( x-y \right)}$
By cross multiplying the above equation, we will get:
$\Rightarrow 2\cos \left( x-y \right)\cos \left( x+y \right)=\cos x\left[ \cos \left( x+y \right)+\cos \left( x-y \right) \right]$
Now we will apply the following formulas,
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B,\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ and
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
Therefore we have,
$\Rightarrow \cos \left( x-y+x+y \right)+\cos \left( x-y-x-y \right)=\cos x\left[ \cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y \right]$ By cancelling out the opposite terms we will get,
$\Rightarrow \cos \left( 2x \right)+\cos \left( -2y \right)=\cos x\left[ 2\cos x\cos y \right]$
$\Rightarrow \cos \left( 2x \right)+\cos \left( 2y \right)=2{{\cos }^{2}}x\cos y$, by applying $\cos \left( -\theta \right)=\cos \theta$.
Now we will apply the formula $\cos \left( 2A \right)=2{{\cos }^{2}}\left( A \right)-1$ in the above expression. Therefore,
$\Rightarrow 2{{\cos }^{2}}x-1+2{{\cos }^{2}}y-1=2{{\cos }^{2}}x\cos y$
We will take all the terms in left hand side,
$\Rightarrow 2{{\cos }^{2}}x+2{{\cos }^{2}}y-2{{\cos }^{2}}x\cos y-2=0$
We can take 2 common from the left hand side. Therefore,
$\Rightarrow 2\left( {{\cos }^{2}}x+{{\cos }^{2}}y-{{\cos }^{2}}x\cos y-1 \right)=0$
$\Rightarrow {{\cos }^{2}}x+{{\cos }^{2}}y-{{\cos }^{2}}x\cos y-1=0$
$\Rightarrow {{\cos }^{2}}x(1-\cos y)-\left( 1-{{\cos }^{2}}y \right)=0$
Now applying the formula, ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$, so the above equation can be written as,
$\Rightarrow {{\cos }^{2}}x\left( 1-\cos y \right)-\left( 1-\cos y \right)\left( 1+\cos y \right)=0$
We can take $1-\cos y$ common. Therefore we have,
$\Rightarrow \left( 1-\cos y \right)\left( {{\cos }^{2}}x-\left( 1+\cos y \right) \right)=0$
$\Rightarrow \left( 1-\cos y \right)\left( {{\cos }^{2}}x-\cos y-1 \right)=0$
Either,
$1-\cos y=0\Rightarrow \cos y=1$
But we know $\cos ({{0}^{{}^\circ }})=1$, so
$\Rightarrow y=0.....(1)$
Or,
${{\cos }^{2}}x-\cos y-1=0$
If we put $\cos y=1$, we will get,
$\Rightarrow {{\cos }^{2}}x-1-1=0$
$\Rightarrow {{\cos }^{2}}x=2$
Taking square root on both sides, we get
$\Rightarrow \cos x=\pm \sqrt{2}.......(2)$
Now we need to find out the value of $\cos x\sec \left( \dfrac{y}{2} \right)$.
We will use the values from (1) and (2). Therefore,
$\cos x\sec \left( \dfrac{y}{2} \right)=\pm \sqrt{2}\sec \left( \dfrac{0}{2} \right)=\pm \sqrt{2}\sec \left( 0 \right)=\pm \sqrt{2}\times 1=\pm \sqrt{2}$
Hence,
$\cos x\sec \left( \dfrac{y}{2} \right)=\pm \sqrt{2}$
Therefore, option (d) is correct.

Note: We can make mistake while simplifying the expression:
$2\sec x=\sec \left( x+y \right)+\sec \left( x-y \right)$
Apply the formulas very carefully. Even a single mistake in sign will change the value completely.
Another approach is,
$\Rightarrow \cos \left( 2x \right)+\cos \left( 2y \right)=2{{\cos }^{2}}x\cos y$
Here we can apply the formula, $\cos 2x=2{{\cos }^{2}}x-1$ , we get
$\begin{aligned} & \Rightarrow 2{{\cos }^{2}}x-1+2{{\cos }^{2}}y-1=2{{\cos }^{2}}x\cos y \\\ & \Rightarrow 2{{\cos }^{2}}x+2{{\cos }^{2}}y-2=2{{\cos }^{2}}x\cos y \\\ & \Rightarrow {{\cos }^{2}}x+{{\cos }^{2}}y-1={{\cos }^{2}}x\cos y \\\ & \Rightarrow {{\cos }^{2}}x+1-{{\sin }^{2}}y-1={{\cos }^{2}}x\cos y \\\ & \Rightarrow {{\cos }^{2}}x-{{\sin }^{2}}y={{\cos }^{2}}x\cos y \\\ \end{aligned}$
Solving this equation will lead to the desired result as above.