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if alpha and beta are the roots of the equation 3cosx+4sinx=5 then cot(alpha+beta) equals

if alpha and beta are the roots of the equation 3cosx+4sinx=5 then cot(alpha+beta) equals

Grade:11

3 Answers

Arun
25750 Points
4 years ago
3cosx + 4sinx = Asin(x + Ф)
 
Multiply and divide with = 5 in LHS
e.g., (3cosx + 4sinx)/5 = Asin(x + Ф)
⇒5(3/5.cosx + 4/5.sinx) = Asin(x + Ф)
 
Let cosФ= 3/5 then, sinФ = 4/5
Now, tanФ = sinФ/cosФ = 4/3
⇒ Ф = tan⁻¹(4/3) ----(1)
 
And 5(3/5.cosx + 4/5.sinx)= 5(sinФ.cosx + cosФ.sinx) = Asin(x + Ф)
⇒ 5sin(x + Ф) = Asin(x + Ф)
From equation (1),
⇒ 5sin(x + tan⁻¹(4/3)) = Asin(x + Ф)
Compare both sides,
A = 5 and Ф = tan⁻¹(4/3)
Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
Dear student

The simplification is

Sin ( x + 37 ) = 1= sin ( 90+ 2n*pi)

You can solve accordingly by evaluating alpha and beta

Regards
Aditya Gupta
2081 Points
4 years ago
note that aruns soln doesnt answer the ques.
we use the standard formulas:
cosx= (1 – t^2)/(1+t^2) and sinx= 2t/(1+t^2) where t= tan(x/2).
so eqn becomes
3(1 – t^2)+4(2t)= 5(1+t^2)
or 8t^2 – 8t + 2= 0 
or 4t^2 – 4t + 1= 0
or (2t – 1)^2= 0
or t= tan(x/2)= ½ so that tan(p)= tan(q)= ½ where p= alpha/2 and q= beta/2.
cot(alpha+beta)= 1/tan(alpha+beta)= 1/tan(2p + 2q)= (1 – tan2p*tan2q)/(tan2p + tan2q)
now, tan2p= 2tanp/(1 – tan^2p)= 4/3. similarly tan2q= 4/3
hence, cot(alpha+beta)= (1 – tan2p*tan2q)/(tan2p + tan2q)= – 7/24
kindly approve :))

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