Question

If \[\cos \left( {\theta - \alpha } \right)\] , \[\cos \theta \] and \[\cos \left( {\theta + \alpha } \right)\] are in HP, then \[\cos \theta .\dfrac{{\sec \alpha }}{2}\] is equal to
\[\left( 1 \right){\text{ }} \pm \sqrt 2 \]
\[\left( 2 \right){\text{ }} \pm \sqrt 3 \]
\[\left( 3 \right){\text{ }} \pm \dfrac{1}{{\sqrt 2 }}\]
\[\left( 4 \right)\] None of these

Answer 1

Hint: First use harmonic mean formula for the given terms. Then use the cosine addition formulas for further simplifications. This question is all based on trigonometric formulas and identities and at every step you have to check which formula is suitable for that equation. On further solving you will finally get your answer.

Complete step-by-step answer:
As we know that if a, b and c form an H.P. then 1/a, 1/b and 1/c form an A.P. And then \[\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}\] that is \[b = \dfrac{{2ac}}{{a + c}}\] where b is known as the harmonic mean of a and c.
Similarly, in the question it is given to us that \[\cos \left( {\theta - \alpha } \right)\] , \[\cos \theta \] and \[\cos \left( {\theta + \alpha } \right)\] are in HP . Therefore their harmonic mean will be
\[\cos \theta = \dfrac{{2\cos \left( {\theta - \alpha } \right)\cos \left( {\theta + \alpha } \right)}}{{\cos \left( {\theta - \alpha } \right) + \cos \left( {\theta + \alpha } \right)}}\] ----------- (i)
Now by using formulas \[\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\] and \[\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\] in the numerator and denominator, the equation (i) becomes
\[ \Rightarrow \cos \theta = \dfrac{{2\left( {\cos \theta \cos \alpha + \sin \theta \sin \alpha } \right)\left( {\cos \theta \cos \alpha - \sin \theta \sin \alpha } \right)}}{{\left( {\cos \theta \cos \alpha + \sin \theta \sin \alpha } \right) + \left( {\cos \theta \cos \alpha - \sin \theta \sin \alpha } \right)}}\]
Now by using identity \[\left( {{a^2} + {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\] in the numerator and by doing addition in the denominator we get
\[ \Rightarrow \cos \theta = \dfrac{{2\left( {{{\cos }^2}\theta {{\cos }^2}\alpha - {{\sin }^2}\theta {{\sin }^2}\alpha } \right)}}{{2\cos \theta \cos \alpha }}\]
Now both \[2\] present in the numerator and the denominator will cancel out and the above expression becomes
\[ \Rightarrow \cos \theta = \dfrac{{{{\cos }^2}\theta {{\cos }^2}\alpha - {{\sin }^2}\theta {{\sin }^2}\alpha }}{{\cos \theta \cos \alpha }}\]
As we know that \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] therefore from this \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \] .Put this value in the above expression
\[ \Rightarrow \cos \theta = \dfrac{{{{\cos }^2}\theta {{\cos }^2}\alpha - \left( {1 - {{\cos }^2}\theta } \right){{\sin }^2}\alpha }}{{\cos \theta \cos \alpha }}\]
Now multiply \[{\sin ^2}\alpha \] with \[\left( {1 - {{\cos }^2}\theta } \right)\] to open the bracket,
\[ \Rightarrow \cos \theta = \dfrac{{{{\cos }^2}\theta {{\cos }^2}\alpha - \left( {{{\sin }^2}\alpha - {{\cos }^2}\theta {{\sin }^2}\alpha } \right)}}{{\cos \theta \cos \alpha }}\]
Multiply the bracket term by the minus sign,
\[ \Rightarrow \cos \theta = \dfrac{{{{\cos }^2}\theta {{\cos }^2}\alpha + {{\cos }^2}\theta {{\sin }^2}\alpha - {{\sin }^2}\alpha }}{{\cos \theta \cos \alpha }}\]
Take out \[{\cos ^2}\theta \] common from the terms \[{\cos ^2}\theta {\cos ^2}\alpha \] and \[{\cos ^2}\theta {\sin ^2}\alpha \] ,
\[ \Rightarrow \cos \theta = \dfrac{{{{\cos }^2}\theta \left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) - {{\sin }^2}\alpha }}{{\cos \theta \cos \alpha }}\]
By using trigonometric identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] we get
\[ \Rightarrow \cos \theta = \dfrac{{{{\cos }^2}\theta \left( 1 \right) - {{\sin }^2}\alpha }}{{\cos \theta \cos \alpha }}\]
\[ \Rightarrow \cos \theta = \dfrac{{{{\cos }^2}\theta - {{\sin }^2}\alpha }}{{\cos \theta \cos \alpha }}\]
On multiplying \[\cos \theta \cos \alpha \] on both sides we get
\[ \Rightarrow \cos \theta .\cos \theta \cos \alpha = {\cos ^2}\theta - {\sin ^2}\alpha \]
We can write the above equation as
\[ \Rightarrow {\cos ^2}\theta \cos \alpha = {\cos ^2}\theta - {\sin ^2}\alpha \]
On shifting \[{\sin ^2}\alpha \] to the left hand side and \[{\cos ^2}\theta \cos \alpha \] to the right hand side we get
\[ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\theta - {\cos ^2}\theta \cos \alpha \]
Take out \[{\cos ^2}\theta \] at the right hand side,
\[ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\theta \left( {1 - \cos \alpha } \right)\]
By using a formula \[\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\] at the left hand side and by using formula \[1 - \cos x = 2{\sin ^2}\dfrac{x}{2}\] at the right hand side we get
\[ \Rightarrow {\left( {2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}} \right)^2} = {\cos ^2}\theta \left( {2{{\sin }^2}\dfrac{\alpha }{2}} \right)\]
Now open the brackets for further calculation,
\[ \Rightarrow 4{\sin ^2}\dfrac{\alpha }{2}{\cos ^2}\dfrac{\alpha }{2} = 2{\cos ^2}\theta {\sin ^2}\dfrac{\alpha }{2}\]
Now shift \[2{\sin ^2}\dfrac{\alpha }{2}\] to the left hand side,
\[ \Rightarrow \dfrac{{4{{\sin }^2}\dfrac{\alpha }{2}{{\cos }^2}\dfrac{\alpha }{2}}}{{2{{\sin }^2}\dfrac{\alpha }{2}}} = {\cos ^2}\theta \]
So on further solving we get
\[ \Rightarrow 2{\cos ^2}\dfrac{\alpha }{2} = {\cos ^2}\theta \]
As secant is the reciprocal of cosine. Therefore we can write the above expression as
\[ \Rightarrow 2.\dfrac{1}{{{{\sec }^2}\dfrac{\alpha }{2}}} = {\cos ^2}\theta \] or \[ \Rightarrow \dfrac{2}{{{{\sec }^2}\dfrac{\alpha }{2}}} = {\cos ^2}\theta \]
Now the shift the term in the denominator to the right hand side,
 \[ \Rightarrow 2 = {\cos ^2}\theta .{\sec ^2}\dfrac{\alpha }{2}\]
Or we can write it as
\[ \Rightarrow {\left( {\cos \theta .\sec \dfrac{\alpha }{2}} \right)^2} = 2\]
This will be
 \[ \Rightarrow \cos \theta .\sec \dfrac{\alpha }{2} = {\left( 2 \right)^{\dfrac{1}{2}}}\]
That is
 \[ \Rightarrow \cos \theta .\sec \dfrac{\alpha }{2} = \pm \sqrt 2 \]
Hence the correct option is \[\left( 1 \right){\text{ }} \pm \sqrt 2 \]
So, the correct answer is “Option 1”.

Note: Keep in mind the formula of harmonic as it is your first step to start the solution. Remember the trigonometric identities and formulas. Avoid any kind of mistake. After completing the solution again, recheck your steps so that if there is any kind of mistake it will be improved to get the right answer.