Question

If $\cos \left( \theta -\alpha \right),\cos \left( \theta \right),\cos \left( \theta +\alpha \right)$ are in H.P, then $\cos \left( \theta \right)\sec \dfrac{\alpha }{2}$ is equal to
[a] $-\sqrt{2}$
[b] $\sqrt{2}$
[c] $\dfrac{1}{2}$
[d] $\dfrac{-1}{2}$

Answer 1

Hint: Use the fact that if numbers are in Harmonic progression, then their inverses are in Arithmetic progression. Use the fact that if a,b and c are in H.P, then $b=\dfrac{2ac}{a+c}$. Hence prove that $\cos \theta =\dfrac{2\left( \cos \left( \theta -\alpha \right)\cos \left( \theta +\alpha \right) \right)}{\cos \left( \theta -\alpha \right)+\cos \left( \theta +\alpha \right)}$. Use $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos \left( A-B \right)\cos \left( A+B \right)={{\cos }^{2}}A-{{\sin }^{2}}B$.

Complete step-by-step solution -
We have $\cos \left( \theta -\alpha \right),\cos \left( \theta \right),\cos \left( \theta +\alpha \right)$ are in H.P
We know that if a,b,c are in H.P, then
$b=\dfrac{2ac}{a+c}$
Hence we have $\cos \theta =\dfrac{2\cos \left( \theta -\alpha \right)\cos \left( \theta +\alpha \right)}{\cos \left( \theta -\alpha \right)+\cos \left( \theta +\alpha \right)}$
Simplifying the numerator:
We know that $\cos \left( A+B \right)\cos \left( A-B \right)={{\cos }^{2}}A-{{\sin }^{2}}B$
Hence, we have
Numerator $=2\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\alpha \right)$
Simplifying the denominator:
We know that $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Put $A=\theta +\alpha $ and $B=\theta -\alpha $, we have
$\cos \left( \theta +\alpha \right)+\cos \left( \theta -\alpha \right)=2\cos \dfrac{\theta +\alpha +\theta -\alpha }{2}\cos \dfrac{\theta +\alpha -\theta +\alpha }{2}=2\cos \theta \cos \alpha $
Hence, we have
$\cos \theta =\dfrac{2\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\alpha \right)}{2\cos \theta \cos \alpha }$
Multiplying both sides by $\cos \theta \cos \alpha $, we get
${{\cos }^{2}}\theta \cos \alpha ={{\cos }^{2}}\theta -{{\sin }^{2}}\alpha $
Transposing ${{\sin }^{2}}\alpha $ on LHS and ${{\cos }^{2}}\theta \cos \alpha $ on RHS, we get
${{\sin }^{2}}\alpha ={{\cos }^{2}}\theta -{{\cos }^{2}}\theta \cos \alpha $
Taking ${{\cos }^{2}}\theta $ common on RHS, we get
${{\sin }^{2}}\alpha ={{\cos }^{2}}\theta \left( 1-\cos \alpha \right)$
We know that $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$
Hence, we have
${{\left( 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} \right)}^{2}}={{\cos }^{2}}\theta \left( 2{{\sin }^{2}}\dfrac{\alpha }{2} \right)$
Hence, we have
$4{{\sin }^{2}}\dfrac{\alpha }{2}{{\cos }^{2}}\dfrac{\alpha }{2}=2{{\cos }^{2}}\theta {{\sin }^{2}}\dfrac{\alpha }{2}$
When $\sin \dfrac{\alpha }{2}\ne 0,$ we have
$2{{\cos }^{2}}\dfrac{\alpha }{2}={{\cos }^{2}}\theta $
Multiplying both sides by ${{\sec }^{2}}\dfrac{\alpha }{2}$, we get
${{\cos }^{2}}\theta {{\sec }^{2}}\dfrac{\alpha }{2}=2$
Hence, we have
$\cos \theta \sec \dfrac{\alpha }{2}=\pm \sqrt{2}$
Hence option [a] and [b] are correct.

Note: [1] In this question, we are taking $\sin \dfrac{\alpha }{2}\ne 0$ as we want to get information regarding $\cos \theta \sec \dfrac{\alpha }{2}$. Nothing can be said about $\cos \theta \sec \dfrac{\alpha }{2}$, when $\sin \dfrac{\alpha }{2}=0$
[2] Aid to memory:
[i] S+S = 2SC
[ii] S-S = 2CS
[iii] C+C = 2CC
[iv] C-C = -2SS
Each one of the above parts helps to memorise two formula
Like from [ii], we have S-S = 2CS.
Hence, we have $\sin \left( A \right)-\sin \left( B \right)=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$ and $2\cos x\sin y=\sin \left( x+y \right)-\sin \left( x-y \right)$
Hence by memorising the above mnemonic, we can memorise 8 different formulae.