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

Step-by-step explanation:

Given :-

cos(θ-α), cosθ, cos(θ+α) are in H. P

To find :-

Prove that cos²θ = 1 + cosα .

Solution :-

Given that

cos(θ-α), cosθ, cos(θ+α) are in H. P

We know that

If a, b,c are in the A.P. then 1/a, 1/b ,1/c are in the H.P.

=> 1/[cos(θ-α)] ,1/cosθ,1/[ cos(θ+α) ] are in A. P.

We know that

If a,b,c are in the A.P then 2b = a+c

We have ,

a = 1/[cos(θ-α)]

b = 1/cosθ

c = 1/[ cos(θ+α) ]

Now,

2b = a+c

=> 2(1/cosθ) = 1/[cos(θ-α)] + 1/[ cos(θ+α) ]

=>2/cosθ=[cos(θ+α)+cos(θ-α)]/[cos(θ-α)cos(θ+α)]

We know that

Cos(A+B) = Cos A Cos B - Sin A Sin B

cos(θ+α) = Cos θ Cos α - Sin θ Sin α

and

Cos (A-B) = Cos A Cos B + Sin A Sin B

cos(θ-α) = Cos θ Cos α + Sin θ Sin α

Now,

cos(θ+α)+cos(θ-α)

=> CosθCosα-SinθSinα+CosθCosα+SinθSinα

=> CosθCosα + CosθCosα

=> 2CosθCosα

and

cos(θ+α)cos(θ-α)

=> (CosθCosα-SinθSinα)(CosθCosα+SinθSinα)

=> (CosθCosα)²-(SinθSinα)²

=> Cos²θCos²α-Sin²θSin²α

=> Cos² θ(1-sin²α)-(1-cos²θ)Sin²α

=> Cos² θ- cos² θsin²α+cos²θ-Sin²α

=> Cos² θ-Sin²α

Now,

2/cosθ=[cos(θ+α)+cos(θ-α)]/[cos(θ-α)cos(θ+α)]

=>2/Cos θ =2CosθCosα/(Cos² θ-Sin²α)

=> 1/Cos θ =CosθCosα/(Cos² θ-Sin²α)

On applying cross multiplication then

=> Cos² θ-Sin²α = CosθCosα Cos θ

=> Cos² θ-Sin²α = Cos²θCosα

=> Cos² θ= Sin²α + Cos²θCosα

=> Cos² θ-Cos²θCosα= Sin²α

=> Cos² θ(1-Cosα) = Sin²α

=> Cos² θ= Sin²α /(1-Cosα)

On multiplying both numerator and denominator with (1+Cosα) then

=> Cos² θ= Sin²α( 1+Cosα)/(1-Cosα) (1+Cosα)

=> Cos² θ= Sin²α( 1+Cosα)/(1²-Cos²α)

Since (a+b)(a-b) = a²-b²

=> Cos² θ= Sin²α( 1+Cosα)/(1-Cos²α)

=> Cos² θ= Sin²α( 1+Cosα)/(Sin²α)

On cancelling Sin²α in RHS

=> Cos² θ= 1+Cosα

Hence, Proved.

Used formulae:-

→ (a+b)(a-b) = a²-b²

→ Sin² A + Cos² A = 1

→ Cos(A+B) = Cos A Cos B - Sin A Sin B

→ Cos (A-B) = Cos A Cos B + Sin A Sin B

→ If a, b,c are in the A.P. then 1/a, 1/b ,1/c are in the H.P.

→ If a,b,c are in the A.P then 2b = a+c