Question

If cos(θ-α), cosθ, cos(θ+α) are in H. P, then prove that cos^2θ = 1 + cosα ​

Answer 1

Step-by-step explanation:

Given,

cos(θ-α) , cosθ , cos(θ+α) are in H.P.

⇒1/cos(θ-α)  , 1/cosθ  , 1/cos(θ+α) are in A.P.

If the above series is in A.P. then

$\frac{1}{\cos\theta}=\frac{\frac{1}{\cos(\theta-\alpha)}+\frac{1}{\cos(\theta+\alpha)}}{2}\\\;\\\frac{2}{\cos\theta}=\frac{1}{\cos(\theta-\alpha)}+\frac{1}{\cos(\theta+\alpha)}\\\;\\\frac{2}{\cos\theta}=\frac{\cos(\theta+\alpha)+\cos(\theta-\alpha)}{\cos(\theta-\alpha).\cos(\theta+\alpha)}\\\;\\\frac{2}{\cos\theta}=\frac{\cos\theta\cos\alpha-\sin\theta\sin\alpha+\cos\theta\cos\alpha+\sin\theta\sin\alpha}{cos^2\theta-sin^2\alpha}$

$\frac{2}{\cos\theta}=\frac{2\cos\theta\cos\alpha}{\cos^2\theta-\sin^2\alpha}\\\;\\\frac{1}{\cos\theta}=\frac{\cos\theta\cos\alpha}{cos^2\theta-\sin^2\alpha}\\\;\\\cos^2\theta-\sin^2\alpha=\cos^2\theta\cos\alpha\\\;\\\sin^2\alpha=\cos^2\theta-\cos^2\theta\cos\alpha\\\;\\\sin^2\alpha=\cos^2\theta(1-\cos\alpha)\\\;\\\frac{\sin^2\alpha}{1-\cos\alpha}=\cos^2\theta\\\;\\\cos^2\theta=\frac{\sin^2\alpha(1+\cos\alpha)}{(1-\cos\alpha)(1+\cos\alpha)}\\\;\\\cos^2\theta=\frac{\sin^2\alpha(1+\cos\alpha)}{1-\cos^2\alpha}$

$\cos^2\theta=\frac{\sin^2\alpha(1+\cos\alpha)}{\sin^2\alpha}\\\;\\\cos^2\theta=1+\cos\alpha$

Hence Proved.

Note:-

1) If a  , b ,  c are in H.P ; then 1/a , 1/b and 1/c  will be in A.P.

2) If a,b,c are in AP, then b = (a+c)/2

3) cos(a+b)cos(a-b) = cos²a - sin²b

4) 1 - cos²θ = sin²θ

5) cos(A+B) = cosAcosB - sinAsinB

6) cos(A-B) = cosAcosB + sinAsinB

Answer 2

Answer :

Proved!

Step-by-step explanation :

Given that :

$\tt \implies cos(\theta-\alpha), cos\theta, cos(\theta+\alpha)\;are\; in\ H. P\\ \\\implies \dfrac{1}{cos(\theta-\alpha)}, \dfrac{1}{cos\theta},\dfrac{1}{cos(\theta+\alpha )} are\ in\ A.P.$

So,

$\star\;\textbf{\underline{If the above series is in A.P, Then,}}\\ \\ \ \\ \tt \implies \dfrac{\dfrac{1}{\cos(\theta-\alpha)}+\dfrac{1}{\cos(\theta+\alpha)}}{2}=\dfrac{1}{\cos\theta}\\ \\ \\\implies \dfrac{1}{\cos(\theta-\alpha)}+\dfrac{1}{\cos(\theta+\alpha)}=\dfrac{2}{\cos\theta}\\ \\ \\ \\ \implies \dfrac{\cos(\theta+\alpha)+\cos(\theta-\alpha)}{\cos(\theta-\alpha)\cos(\theta+\alpha)}=\dfrac{2}{\cos\theta}$

$\tt \implies \dfrac{\cos\theta\cos\alpha-\sin\theta\sin\alpha+\cos\theta\cos\alpha+\sin\theta\sin\alpha}{cos^2\theta-sin^2\alpha} = \dfrac{2}{\cos\theta}$

$\tt \implies \dfrac{2\cos\theta\cos\alpha}{\cos^2\theta-\sin^2\alpha}=\dfrac{2}{\cos\theta}\\ \\ \\\implies \dfrac{\cos\theta\cos\alpha}{cos^2\theta-\sin^2\alpha}=\dfrac{1}{\cos\theta}$

$\tt \implies \cos^2\theta\cos\alpha=\cos^2\theta-\sin^2\alpha\\\\\ \implies\sin^2\alpha=\cos^2\theta-\cos^2\theta\cos\alpha\\\ \\ \implies \cos^2\theta(1-\cos\alpha)=\sin^2\alpha\\ \\\implies \dfrac{\sin^2\alpha}{1-\cos\alpha}=\cos^2\theta\\ \\ \implies\cos^2\theta=\dfrac{\sin^2\alpha(1+\cos\alpha)}{(1-\cos\alpha)(1+\cos\alpha)}\\ \\ \implies \dfrac{\sin^2\alpha(1+\cos\alpha)}{1-\cos^2\alpha}=\cos^2\theta$

$\tt \implies \cos^2\theta=\dfrac{\sin^2\alpha(1+\cos\alpha)}{\sin^2\alpha}\\ \\ \implies \cos^2\theta=1+\cos\alpha.\\ \\\bigstar\;\textbf{\underline{\underline{Hence, Proved.}}}$

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$\bigstar\;\textbf{\underline{\underline{Extra\;Information :-}}}\\ \\ \\\boxed{\begin{minipage}{7cm}{\underline{\underline{{\textbf {Fundamental Trignometric\:\: Indentities:}}}}}\\ \\ \sf \sin^{2}\theta+\cos^{2}\theta =1 \\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \end{minipage}}$