Question

Q1.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

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

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

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

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:

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